Solve the following equation using quadratic formula:
x2 - 4x + 1 = 0
Answer
Comparing equation x2 - 4x + 1 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = -4 and c = 1.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 4 ) ± ( − 4 ) 2 − 4 × 1 × 1 2 × 1 = 4 ± 16 − 4 2 = 4 ± 12 2 = 4 ± 4 × 3 2 = 4 ± 2 3 2 = 2 ( 2 ± 3 ) 2 = 2 ± 3 = 2 + 3 or 2 − 3 . \Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times 1}}{2 \times 1} \\[1em] = \dfrac{4 \pm \sqrt{16 - 4}}{2} \\[1em] = \dfrac{4 \pm \sqrt{12}}{2} \\[1em] = \dfrac{4 \pm \sqrt{4 \times 3}}{2} \\[1em] = \dfrac{4 \pm 2\sqrt{3}}{2} \\[1em] = \dfrac{2(2 \pm \sqrt{3})}{2} \\[1em] = 2 \pm \sqrt{3} \\[1em] = 2 + \sqrt{3} \text{ or } 2 - \sqrt{3}. ⇒ x = 2 × 1 − ( − 4 ) ± ( − 4 ) 2 − 4 × 1 × 1 = 2 4 ± 16 − 4 = 2 4 ± 12 = 2 4 ± 4 × 3 = 2 4 ± 2 3 = 2 2 ( 2 ± 3 ) = 2 ± 3 = 2 + 3 or 2 − 3 .
Hence, x = 2 + 3 , 2 − 3 x = {2 + \sqrt{3}, 2 - \sqrt{3}} x = 2 + 3 , 2 − 3
Solve the following equation using quadratic formula:
9x2 + 7x - 2 = 0
Answer
Comparing equation 9x2 + 7x - 2 = 0 with ax2 + bx + c = 0, we get :
a = 9, b = 7 and c = -2.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 7 ) ± ( 7 ) 2 − 4 × 9 × ( − 2 ) 2 ( 9 ) = − 7 ± 49 + 72 18 ⇒ x = − 7 ± 121 18 = − 7 ± 11 18 = − 7 + 11 18 or − 7 − 11 18 = 4 18 or − 18 18 = 2 9 or − 1. \Rightarrow x = \dfrac{-(7) \pm \sqrt{(7)^2 - 4 \times 9 \times (-2)}}{2(9)} \\[1em] = \dfrac{-7 \pm \sqrt{49 + 72}}{18} \\[1em] \Rightarrow x = \dfrac{-7 \pm \sqrt{121}}{18} \\[1em] = \dfrac{-7 \pm 11}{18} \\[1em] = \dfrac{-7 + 11}{18} \text{ or } \dfrac{-7 - 11}{18}\\[1em] = \dfrac{4}{18} \text{ or } \dfrac{-18}{18}\\[1em] = \dfrac{2}{9} \text{ or } -1. ⇒ x = 2 ( 9 ) − ( 7 ) ± ( 7 ) 2 − 4 × 9 × ( − 2 ) = 18 − 7 ± 49 + 72 ⇒ x = 18 − 7 ± 121 = 18 − 7 ± 11 = 18 − 7 + 11 or 18 − 7 − 11 = 18 4 or 18 − 18 = 9 2 or − 1.
Hence, x = { 2 9 , − 1 } x = \Big\lbrace\dfrac{2}{9}, -1\Big\rbrace x = { 9 2 , − 1 }
Solve the following equation using quadratic formula:
3 4 \dfrac{3}{4} 4 3 2 - x - 1 = 0
Answer
Comparing equation 3 4 \dfrac{3}{4} 4 3 2 - x - 1 = 0 with ax2 + bx + c = 0, we get :
a = 3 4 \dfrac{3}{4} 4 3
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 3 4 ) ( − 1 ) 2 ( 3 4 ) = 1 ± 1 + 3 ( 3 2 ) = 1 ± 4 ( 3 2 ) = 1 ± 2 ( 3 2 ) = 2 ( 1 ± 2 ) 3 = 2 ± 4 3 = 2 + 4 3 or 2 − 4 3 = 6 3 or − 2 3 = 2 or − 2 3 . \Rightarrow x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4\Big(\dfrac{3}{4}\Big)(-1)}}{2\Big(\dfrac{3}{4}\Big)} \\[1em] = \dfrac{1 \pm \sqrt{1 + 3}}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{1 \pm \sqrt{4}}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{1 \pm 2}{\Big(\dfrac{3}{2}\Big)} \\[1em] = \dfrac{2(1 \pm 2)}{3} \\[1em] = \dfrac{2 \pm 4}{3} \\[1em] = \dfrac{2 + 4}{3} \text{ or } \dfrac{2 - 4}{3} \\[1em] = \dfrac{6}{3} \text{ or } \dfrac{-2}{3} \\[1em] = 2 \text{ or } \dfrac{-2}{3}. ⇒ x = 2 ( 4 3 ) − ( − 1 ) ± ( − 1 ) 2 − 4 ( 4 3 ) ( − 1 ) = ( 2 3 ) 1 ± 1 + 3 = ( 2 3 ) 1 ± 4 = ( 2 3 ) 1 ± 2 = 3 2 ( 1 ± 2 ) = 3 2 ± 4 = 3 2 + 4 or 3 2 − 4 = 3 6 or 3 − 2 = 2 or 3 − 2 .
Hence, x = { 2 , − 2 3 } x = \Big\lbrace2, \dfrac{-2}{3}\Big\rbrace x = { 2 , 3 − 2 }
Solve the following equation using quadratic formula:
4 - 11x = 3x2
Answer
⇒ 3x2 + 11x - 4 = 0
Comparing equation 3x2 + 11x - 4 = 0 with ax2 + bx + c = 0, we get :
a = 3, b = 11 and c = -4.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 11 ) ± ( 11 ) 2 − 4 × ( 3 ) × ( − 4 ) 2 ( 3 ) = − 11 ± 121 + 48 6 = − 11 ± 169 6 = − 11 ± 13 6 = − 11 + 13 6 or − 11 − 13 6 = 2 6 or − 24 6 = 1 3 or − 4. \Rightarrow x = \dfrac{-(11) \pm \sqrt{(11)^2 - 4 \times (3) \times (-4)}}{2(3)} \\[1em] = \dfrac{-11 \pm \sqrt{121 + 48}}{6} \\[1em] = \dfrac{-11 \pm \sqrt{169}}{6} \\[1em] = \dfrac{-11 \pm 13}{6} \\[1em] = \dfrac{-11 + 13}{6} \text{ or } \dfrac{-11 - 13}{6} \\[1em] = \dfrac{2}{6} \text{ or } \dfrac{-24}{6} \\[1em] = \dfrac{1}{3} \text{ or } -4. ⇒ x = 2 ( 3 ) − ( 11 ) ± ( 11 ) 2 − 4 × ( 3 ) × ( − 4 ) = 6 − 11 ± 121 + 48 = 6 − 11 ± 169 = 6 − 11 ± 13 = 6 − 11 + 13 or 6 − 11 − 13 = 6 2 or 6 − 24 = 3 1 or − 4.
Hence, x = { 1 3 , − 4 } x = \Big\lbrace\dfrac{1}{3}, -4\Big\rbrace x = { 3 1 , − 4 }
Solve the following equation using quadratic formula:
25x2 + 30x + 7 = 0
Answer
Comparing equation 25x2 + 30x + 7 = 0 with ax2 + bx + c = 0, we get :
a = 25, b = 30 and c = 7.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 30 ) ± ( 30 ) 2 − 4 × 25 × 7 2 ( 25 ) = − 30 ± 900 − 700 50 = − 30 ± 200 50 = − 30 ± 2 × 100 50 = − 30 ± 10 2 50 = 10 ( − 3 ± 2 ) 50 = ( − 3 ± 2 ) 5 = ( − 3 + 2 ) 5 or ( − 3 − 2 ) 5 . \Rightarrow x = \dfrac{-(30) \pm \sqrt{(30)^2 - 4 \times 25 \times 7}}{2(25)} \\[1em] = \dfrac{-30 \pm \sqrt{900 - 700}}{50} \\[1em] = \dfrac{-30 \pm \sqrt{200}}{50} \\[1em] = \dfrac{-30 \pm \sqrt{2 \times 100}}{50} \\[1em] = \dfrac{-30 \pm 10\sqrt{2}}{50} \\[1em] = \dfrac{10(-3 \pm \sqrt{2})}{50} \\[1em] = \dfrac{(-3 \pm \sqrt{2})}{5} \\[1em] = \dfrac{(-3 + \sqrt{2})}{5} \text{ or } \dfrac{(-3 - \sqrt{2})}{5}. ⇒ x = 2 ( 25 ) − ( 30 ) ± ( 30 ) 2 − 4 × 25 × 7 = 50 − 30 ± 900 − 700 = 50 − 30 ± 200 = 50 − 30 ± 2 × 100 = 50 − 30 ± 10 2 = 50 10 ( − 3 ± 2 ) = 5 ( − 3 ± 2 ) = 5 ( − 3 + 2 ) or 5 ( − 3 − 2 ) .
Hence, x = { − 3 + 2 5 , − 3 − 2 5 } x = \Big\lbrace\dfrac{-3 + \sqrt{2}}{5}, \dfrac{-3 - \sqrt{2}}{5}\Big\rbrace x = { 5 − 3 + 2 , 5 − 3 − 2 }
Solve the following equation using quadratic formula:
5x2 - 19x + 17 = 0
Answer
Comparing equation 5x2 - 19x + 17 = 0 with ax2 + bx + c = 0, we get :
a = 5, b = -19 and c = 17.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 19 ) ± ( − 19 ) 2 − 4 × 5 × 17 2 × 5 = 19 ± 361 − 340 10 = 19 ± 21 10 = 19 + 21 10 or 19 − 21 10 . \Rightarrow x = \dfrac{-(-19) \pm \sqrt{(-19)^2 - 4 \times 5 \times 17}}{2 \times 5} \\[1em] = \dfrac{19 \pm \sqrt{361 - 340}}{10} \\[1em] = \dfrac{19 \pm \sqrt{21}}{10} \\[1em] = \dfrac{19 + \sqrt{21}}{10} \text{ or } \dfrac{19 - \sqrt{21}}{10}. ⇒ x = 2 × 5 − ( − 19 ) ± ( − 19 ) 2 − 4 × 5 × 17 = 10 19 ± 361 − 340 = 10 19 ± 21 = 10 19 + 21 or 10 19 − 21 .
Hence, x = { 19 + 21 10 , 19 − 21 10 } x = \Big\lbrace\dfrac{19 + \sqrt{21}}{10}, \dfrac{19 - \sqrt{21}}{10}\Big\rbrace x = { 10 19 + 21 , 10 19 − 21 }
Solve the following equation using quadratic formula:
3x2 - 8x + 2 = 0
Answer
Comparing equation 3x2 - 8x + 2 = 0 with ax2 + bx + c = 0, we get :
a = 3, b = -8 and c = 2.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 8 ) ± ( − 8 ) 2 − 4 × 3 × 2 2 × 3 = 8 ± 64 − 24 6 = 8 ± 40 6 = 8 ± 4 × 10 6 = 8 ± 2 10 6 = 2 ( 4 ± 10 ) 6 = ( 4 ± 10 ) 3 = 4 + 10 3 or 4 − 10 3 . \Rightarrow x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2\times 3} \\[1em] = \dfrac{8 \pm \sqrt{64 - 24}}{6} \\[1em] = \dfrac{8 \pm \sqrt{40}}{6} \\[1em] = \dfrac{8 \pm \sqrt{4 \times 10}}{6} \\[1em] = \dfrac{8 \pm 2\sqrt{10}}{6} \\[1em] = \dfrac{2(4 \pm \sqrt{10})}{6} \\[1em] = \dfrac{(4 \pm \sqrt{10})}{3} \\[1em] = \dfrac{4 + \sqrt{10}}{3} \text{ or} \dfrac{4 - \sqrt{10}}{3}. ⇒ x = 2 × 3 − ( − 8 ) ± ( − 8 ) 2 − 4 × 3 × 2 = 6 8 ± 64 − 24 = 6 8 ± 40 = 6 8 ± 4 × 10 = 6 8 ± 2 10 = 6 2 ( 4 ± 10 ) = 3 ( 4 ± 10 ) = 3 4 + 10 or 3 4 − 10 .
Hence, x = { 4 + 10 3 , 4 − 10 3 } x = \Big\lbrace\dfrac{4 + \sqrt{10}}{3}, \dfrac{4 - \sqrt{10}}{3}\Big\rbrace x = { 3 4 + 10 , 3 4 − 10 }
Solve the following equation using quadratic formula:
3 x 2 + 10 x − 8 3 \sqrt{3}x^2 + 10x - 8\sqrt{3} 3 x 2 + 10 x − 8 3
Answer
Comparing equation 3 x 2 + 10 x − 8 3 \sqrt{3}x^2 + 10x - 8\sqrt{3} 3 x 2 + 10 x − 8 3 2 + bx + c = 0, we get :
a = 3 \sqrt{3} 3 − 8 3 -8\sqrt{3} − 8 3
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 10 ) ± ( 10 ) 2 − 4 × 3 × ( − 8 3 ) 2 × ( 3 ) = − 10 ± 100 + 96 2 3 = − 10 ± 196 2 3 = − 10 ± 14 2 3 = 2 ( − 5 ± 7 ) 2 3 = − 5 ± 7 3 = − 5 + 7 3 or − 5 − 7 3 = 2 3 or − 12 3 = 2 3 or − 4 × 3 3 = 2 3 or − 4 3 . \Rightarrow x = \dfrac{-(10) \pm \sqrt{(10)^2 - 4 \times \sqrt{3} \times (-8\sqrt{3})}}{2\times(\sqrt{3})} \\[1em] = \dfrac{-10 \pm \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] = \dfrac{-10 \pm \sqrt{196}}{2\sqrt{3}} \\[1em] = \dfrac{-10 \pm 14}{2\sqrt{3}} \\[1em] = \dfrac{2(-5 \pm 7)}{2\sqrt{3}} \\[1em] = \dfrac{-5 \pm 7}{\sqrt{3}} \\[1em] = \dfrac{-5 + 7}{\sqrt{3}} \text{ or } \dfrac{-5 - 7}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-4 \times 3}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } -4\sqrt{3}. ⇒ x = 2 × ( 3 ) − ( 10 ) ± ( 10 ) 2 − 4 × 3 × ( − 8 3 ) = 2 3 − 10 ± 100 + 96 = 2 3 − 10 ± 196 = 2 3 − 10 ± 14 = 2 3 2 ( − 5 ± 7 ) = 3 − 5 ± 7 = 3 − 5 + 7 or 3 − 5 − 7 = 3 2 or 3 − 12 = 3 2 or 3 − 4 × 3 = 3 2 or − 4 3 .
Hence, x = { 2 3 , − 4 3 } x = \Big\lbrace\dfrac{2}{\sqrt{3}}, -4\sqrt{3}\Big\rbrace x = { 3 2 , − 4 3 }
Solve the following equation using quadratic formula:
2x2 + 7 x \sqrt{7}x 7 x
Answer
Comparing equation 2x2 + 7 x \sqrt{7}x 7 x 2 + bx + c = 0, we get :
a = 2, b = 7 \sqrt{7} 7
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 7 ) ± ( 7 ) 2 − 4 × 2 × ( − 7 ) 2 × 2 = − 7 ± 7 + 56 4 = − 7 ± 63 4 = − 7 ± 7 × 9 4 = − 7 ± 3 7 4 = − 7 + 3 7 4 or − 7 − 3 7 4 = 2 7 4 or − 4 7 4 = 7 2 or − 7 . \Rightarrow x = \dfrac{-(\sqrt{7}) \pm \sqrt{(\sqrt{7})^2 - 4 \times 2 \times (-7)}}{2 \times 2} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{7 + 56}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{63}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{7 \times 9}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm 3\sqrt{7}}{4} \\[1em] = \dfrac{-\sqrt{7} + 3\sqrt{7}}{4} \text{ or } \dfrac{-\sqrt{7} - 3\sqrt{7}}{4} \\[1em] = \dfrac{2\sqrt{7}}{4} \text{ or } \dfrac{-4\sqrt{7}}{4} \\[1em] = \dfrac{\sqrt{7}}{2} \text{ or } -\sqrt{7}. ⇒ x = 2 × 2 − ( 7 ) ± ( 7 ) 2 − 4 × 2 × ( − 7 ) = 4 − 7 ± 7 + 56 = 4 − 7 ± 63 = 4 − 7 ± 7 × 9 = 4 − 7 ± 3 7 = 4 − 7 + 3 7 or 4 − 7 − 3 7 = 4 2 7 or 4 − 4 7 = 2 7 or − 7 .
Hence, x = { − 7 , 7 2 } x = \Big\lbrace-\sqrt{7}, \dfrac{\sqrt{7}}{2}\Big\rbrace x = { − 7 , 2 7 }
Solve the following equation using quadratic formula:
6x2 - 31x = 105
Answer
⇒ 6x2 - 31x - 105 = 0
Comparing equation 6x2 - 31x - 105 = 0 with ax2 + bx + c = 0, we get :
a = 6, b = -31 and c = -105.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 31 ) ± ( − 31 ) 2 − 4 × 6 × ( − 105 ) 2 × ( 6 ) = 31 ± 961 + 2520 12 = 31 ± 3481 12 = 31 ± 59 12 = 31 + 59 12 or 31 − 59 12 = 90 12 or − 28 12 = 15 2 or − 7 3 . \Rightarrow x = \dfrac{-(-31) \pm \sqrt{(-31)^2 - 4 \times 6 \times(-105)}}{2\times(6)} \\[1em] = \dfrac{31 \pm \sqrt{961 + 2520}}{12} \\[1em] = \dfrac{31 \pm \sqrt{3481}}{12} \\[1em] = \dfrac{31 \pm 59}{12} \\[1em] = \dfrac{31 + 59}{12} \text{ or } \dfrac{31 - 59}{12} \\[1em] = \dfrac{90}{12} \text{ or } \dfrac{-28}{12} \\[1em] = \dfrac{15}{2} \text{ or } \dfrac{-7}{3}. ⇒ x = 2 × ( 6 ) − ( − 31 ) ± ( − 31 ) 2 − 4 × 6 × ( − 105 ) = 12 31 ± 961 + 2520 = 12 31 ± 3481 = 12 31 ± 59 = 12 31 + 59 or 12 31 − 59 = 12 90 or 12 − 28 = 2 15 or 3 − 7 .
Hence, x = { 15 2 , − 7 3 } x = \Big\lbrace\dfrac{15}{2}, \dfrac{-7}{3}\Big\rbrace x = { 2 15 , 3 − 7 }
Solve the following equation using quadratic formula:
x + 3 2 x + 3 = x + 1 3 x + 2 \dfrac{x + 3}{2x + 3} = \dfrac{x + 1}{3x + 2} 2 x + 3 x + 3 = 3 x + 2 x + 1
Answer
⇒ x + 3 2 x + 3 = x + 1 3 x + 2 ⇒ ( x + 3 ) ( 3 x + 2 ) = ( x + 1 ) ( 2 x + 3 ) ⇒ ( 3 x 2 + 2 x + 9 x + 6 ) = ( 2 x 2 + 3 x + 2 x + 3 ) ⇒ 3 x 2 + 11 x + 6 = 2 x 2 + 5 x + 3 ⇒ 3 x 2 − 2 x 2 + 11 x − 5 x + 6 − 3 = 0 ⇒ x 2 + 6 x + 3 = 0. \Rightarrow \dfrac{x + 3}{2x + 3} = \dfrac{x + 1}{3x + 2} \\[1em] \Rightarrow (x + 3)(3x + 2) = (x + 1)(2x + 3) \\[1em] \Rightarrow (3x^2 + 2x + 9x + 6) = (2x^2 + 3x + 2x + 3) \\[1em] \Rightarrow 3x^2 + 11x + 6 = 2x^2 + 5x + 3 \\[1em] \Rightarrow 3x^2 - 2x^2 + 11x - 5x + 6 - 3 = 0 \\[1em] \Rightarrow x^2 + 6x + 3 = 0. ⇒ 2 x + 3 x + 3 = 3 x + 2 x + 1 ⇒ ( x + 3 ) ( 3 x + 2 ) = ( x + 1 ) ( 2 x + 3 ) ⇒ ( 3 x 2 + 2 x + 9 x + 6 ) = ( 2 x 2 + 3 x + 2 x + 3 ) ⇒ 3 x 2 + 11 x + 6 = 2 x 2 + 5 x + 3 ⇒ 3 x 2 − 2 x 2 + 11 x − 5 x + 6 − 3 = 0 ⇒ x 2 + 6 x + 3 = 0.
Comparing equation x2 + 6x + 3 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = 6 and c = 3.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 6 ) ± ( 6 ) 2 − 4 × ( 1 ) × ( 3 ) 2 × 1 = − 6 ± 36 − 12 2 = − 6 ± 24 2 = − 6 ± 6 × 4 2 = − 6 ± 2 6 2 = 2 ( − 3 ± 6 ) 2 = − 3 ± 6 = − 3 + 6 or − 3 − 6 . \Rightarrow x = \dfrac{-(6) \pm \sqrt{(6)^2 - 4 \times (1) \times (3)}}{2 \times 1} \\[1em] = \dfrac{-6 \pm \sqrt{36 - 12}}{2} \\[1em] = \dfrac{-6 \pm \sqrt{24}}{2} \\[1em] = \dfrac{-6 \pm \sqrt{6 \times 4}}{2} \\[1em] = \dfrac{-6 \pm 2\sqrt{6}}{2} \\[1em] = \dfrac{2(-3 \pm \sqrt{6})}{2} \\[1em] = -3 \pm \sqrt{6} \\[1em] = -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}. ⇒ x = 2 × 1 − ( 6 ) ± ( 6 ) 2 − 4 × ( 1 ) × ( 3 ) = 2 − 6 ± 36 − 12 = 2 − 6 ± 24 = 2 − 6 ± 6 × 4 = 2 − 6 ± 2 6 = 2 2 ( − 3 ± 6 ) = − 3 ± 6 = − 3 + 6 or − 3 − 6 .
Hence, x = { ( − 3 + 6 ) , ( − 3 − 6 ) } x = \Big\lbrace(-3 + \sqrt{6}), (-3 - \sqrt{6})\Big\rbrace x = { ( − 3 + 6 ) , ( − 3 − 6 ) }
Solve the following equation using quadratic formula:
x − 1 x − 2 + x − 3 x − 4 = 3 1 3 \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3} x − 2 x − 1 + x − 4 x − 3 = 3 3 1
Answer
⇒ x − 1 x − 2 + x − 3 x − 4 = 3 1 3 ⇒ ( x − 1 ) ( x − 4 ) + ( x − 3 ) ( x − 2 ) ( x − 2 ) ( x − 4 ) = 10 3 ⇒ x 2 − 4 x − x + 4 + ( x 2 − 2 x − 3 x + 6 ) x 2 − 4 x − 2 x + 8 = 10 3 ⇒ x 2 − 5 x + 4 + ( x 2 − 5 x + 6 ) x 2 − 6 x + 8 = 10 3 ⇒ 2 x 2 − 10 x + 10 x 2 − 6 x + 8 = 10 3 ⇒ 3 ( 2 x 2 − 10 x + 10 ) = 10 ( x 2 − 6 x + 8 ) ⇒ 6 x 2 − 30 x + 30 = 10 x 2 − 60 x + 80 ⇒ 10 x 2 − 60 x + 80 − ( 6 x 2 − 30 x + 30 ) = 0 ⇒ 10 x 2 − 60 x + 80 − 6 x 2 + 30 x − 30 = 0 ⇒ 4 x 2 − 30 x + 50 = 0 ⇒ 2 ( 2 x 2 − 15 x + 25 ) = 0 ⇒ 2 x 2 − 15 x + 25 = 0. \Rightarrow \dfrac{x - 1}{x - 2} + \dfrac{x - 3}{x - 4} = 3\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 - 4x - x + 4 + (x^2 - 2x - 3x + 6)}{x^2 - 4x - 2x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 - 5x + 4 + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{2x^2 - 10x + 10 }{x^2 - 6x + 8} = \dfrac{10}{3} \\[1em] \Rightarrow 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \\[1em] \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \\[1em] \Rightarrow 10x^2 - 60x + 80 - (6x^2 - 30x + 30 ) = 0 \\[1em] \Rightarrow 10x^2 - 60x + 80 - 6x^2 + 30x - 30 = 0 \\[1em] \Rightarrow 4x^2 - 30x + 50 = 0 \\[1em] \Rightarrow 2(2x^2 - 15x + 25) = 0 \\[1em] \Rightarrow 2x^2 - 15x + 25 = 0. ⇒ x − 2 x − 1 + x − 4 x − 3 = 3 3 1 ⇒ ( x − 2 ) ( x − 4 ) ( x − 1 ) ( x − 4 ) + ( x − 3 ) ( x − 2 ) = 3 10 ⇒ x 2 − 4 x − 2 x + 8 x 2 − 4 x − x + 4 + ( x 2 − 2 x − 3 x + 6 ) = 3 10 ⇒ x 2 − 6 x + 8 x 2 − 5 x + 4 + ( x 2 − 5 x + 6 ) = 3 10 ⇒ x 2 − 6 x + 8 2 x 2 − 10 x + 10 = 3 10 ⇒ 3 ( 2 x 2 − 10 x + 10 ) = 10 ( x 2 − 6 x + 8 ) ⇒ 6 x 2 − 30 x + 30 = 10 x 2 − 60 x + 80 ⇒ 10 x 2 − 60 x + 80 − ( 6 x 2 − 30 x + 30 ) = 0 ⇒ 10 x 2 − 60 x + 80 − 6 x 2 + 30 x − 30 = 0 ⇒ 4 x 2 − 30 x + 50 = 0 ⇒ 2 ( 2 x 2 − 15 x + 25 ) = 0 ⇒ 2 x 2 − 15 x + 25 = 0.
Comparing equation 2x2 - 15x + 25 = 0 with ax2 + bx + c = 0, we get :
a = 2, b = -15 and c = 25.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 15 ) ± ( − 15 ) 2 − 4 ( 2 ) ( 25 ) 2 ( 2 ) = 15 ± 225 − 200 4 = 15 ± 25 4 = 15 ± 5 4 = 15 + 5 4 or 15 − 5 4 = 20 4 or 10 4 = 5 or 5 2 . \Rightarrow x = \dfrac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(25)}}{2(2)} \\[1em] = \dfrac{15 \pm \sqrt{225 - 200}}{4} \\[1em] = \dfrac{15 \pm \sqrt{25}}{4} \\[1em] = \dfrac{15 \pm 5}{4} \\[1em] = \dfrac{15 + 5}{4} \text{ or } \dfrac{15 - 5}{4} \\[1em] = \dfrac{20}{4} \text{ or } \dfrac{10}{4} \\[1em] = 5 \text{ or } \dfrac{5}{2}. ⇒ x = 2 ( 2 ) − ( − 15 ) ± ( − 15 ) 2 − 4 ( 2 ) ( 25 ) = 4 15 ± 225 − 200 = 4 15 ± 25 = 4 15 ± 5 = 4 15 + 5 or 4 15 − 5 = 4 20 or 4 10 = 5 or 2 5 .
Hence, x = { 5 , 5 2 } \Big\lbrace5, \dfrac{5}{2}\Big\rbrace { 5 , 2 5 }
Solve the following equation using quadratic formula:
x2 - 10x + 6 = 0
Answer
Given,
⇒ x2 - 10x + 6 = 0
Comparing equation x2 - 10x + 6 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = -10 and c = 6.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 10 ) ± ( − 10 ) 2 − 4 × 1 × ( 6 ) 2 × ( 1 ) = 10 ± 100 − 24 2 = 10 ± 76 2 = 10 ± 19 × 4 2 = 10 ± 2 19 2 = 10 + 2 19 2 or 10 − 2 19 2 = 2 ( 5 + 19 ) 2 or 2 ( 5 − 19 ) 2 = 5 + 19 or 5 − 19 = 5 + 4.36 or 5 − 4.36 = 9.36 or 0.64 \Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 1 \times (6)}}{2 \times (1)} \\[1em] = \dfrac{10 \pm \sqrt{100 - 24}}{2} \\[1em] = \dfrac{10 \pm \sqrt{76}}{2} \\[1em] = \dfrac{10 \pm \sqrt{19 \times 4}}{2} \\[1em] = \dfrac{10 \pm 2\sqrt{19}}{2} \\[1em] = \dfrac{10 + 2\sqrt{19}}{2} \text{ or } \dfrac{10 - 2\sqrt{19}}{2} \\[1em] = \dfrac{2(5 + \sqrt{19})}{2} \text{ or } \dfrac{2(5 - \sqrt{19})}{2} \\[1em] = 5 + \sqrt{19} \text{ or } 5 - \sqrt{19} \\[1em] = 5 + 4.36 \text{ or } 5 - 4.36 \\[1em] = 9.36 \text{ or } 0.64 ⇒ x = 2 × ( 1 ) − ( − 10 ) ± ( − 10 ) 2 − 4 × 1 × ( 6 ) = 2 10 ± 100 − 24 = 2 10 ± 76 = 2 10 ± 19 × 4 = 2 10 ± 2 19 = 2 10 + 2 19 or 2 10 − 2 19 = 2 2 ( 5 + 19 ) or 2 2 ( 5 − 19 ) = 5 + 19 or 5 − 19 = 5 + 4.36 or 5 − 4.36 = 9.36 or 0.64
Hence, x = {9.36, 0.64}.
Solve the following equation using quadratic formula:
2x2 - 6x + 3 = 0
Answer
Given,
⇒ 2x2 - 6x + 3 = 0
Comparing equation 2x2 - 6x + 3 = 0 with ax2 + bx + c = 0, we get :
a = 2, b = -6 and c = 3.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 6 ) ± ( − 6 ) 2 − 4 × 2 × ( 3 ) 2 × ( 2 ) = 6 ± 36 − 24 4 = 6 ± 12 4 = 6 ± 3 × 4 4 = 6 ± 2 3 4 = 6 + 2 3 4 or 6 − 2 3 4 = 2 ( 3 + 3 ) 4 or 2 ( 3 − 3 ) 4 = 3 + 3 2 or 3 − 3 2 = 3 + 1.73 2 or 3 − 1.73 2 = 4.73 2 or 1.27 2 = 2.365 or 0.635 ≈ 2.37 or 0.64 \Rightarrow x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 2 \times (3)}}{2 \times (2)} \\[1em] = \dfrac{6 \pm \sqrt{36 - 24}}{4} \\[1em] = \dfrac{6 \pm \sqrt{12}}{4} \\[1em] = \dfrac{6 \pm \sqrt{3 \times 4}}{4} \\[1em] = \dfrac{6 \pm 2\sqrt{3}}{4} \\[1em] = \dfrac{6 + 2\sqrt{3}}{4} \text{ or } \dfrac{6 - 2\sqrt{3}}{4} \\[1em] = \dfrac{2(3 + \sqrt{3})}{4} \text{ or } \dfrac{2(3 - \sqrt{3})}{4} \\[1em] = \dfrac{3 + \sqrt{3}}{2} \text{ or } \dfrac{3 - \sqrt{3}}{2} \\[1em] = \dfrac{3 + 1.73}{2} \text{ or } \dfrac{3 - 1.73}{2} \\[1em] = \dfrac{4.73}{2} \text{ or } \dfrac{1.27}{2} \\[1em] = 2.365 \text{ or } 0.635 \\[1em] \approx 2.37 \text{ or } 0.64 ⇒ x = 2 × ( 2 ) − ( − 6 ) ± ( − 6 ) 2 − 4 × 2 × ( 3 ) = 4 6 ± 36 − 24 = 4 6 ± 12 = 4 6 ± 3 × 4 = 4 6 ± 2 3 = 4 6 + 2 3 or 4 6 − 2 3 = 4 2 ( 3 + 3 ) or 4 2 ( 3 − 3 ) = 2 3 + 3 or 2 3 − 3 = 2 3 + 1.73 or 2 3 − 1.73 = 2 4.73 or 2 1.27 = 2.365 or 0.635 ≈ 2.37 or 0.64
Hence, x = {2.37, 0.64}.
Solve the following equation using quadratic formula:
3x2 - 32x + 12 = 0
Answer
Given,
⇒ 3x2 - 32x + 12 = 0
Comparing equation 3x2 - 32x + 12 = 0 with ax2 + bx + c = 0, we get :
a = 3, b = -32 and c = 12.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 32 ) ± ( − 32 ) 2 − 4 × 3 × ( 12 ) 2 × ( 3 ) = 32 ± 1024 − 144 6 = 32 ± 880 6 = 32 ± 4 × 220 6 = 32 ± 2 220 6 = 32 + 2 220 6 or 32 − 2 220 6 = 2 ( 16 + 220 ) 6 or 2 ( 16 − 220 ) 6 = 16 + 220 3 or 16 − 220 3 = 16 + 14.83 3 or 16 − 14.83 3 = 30.83 3 or 1.17 3 = 10.28 or 0.39 \Rightarrow x = \dfrac{-(-32) \pm \sqrt{(-32)^2 - 4 \times 3 \times (12)}}{2 \times (3)} \\[1em] = \dfrac{32 \pm \sqrt{1024 - 144}}{6} \\[1em] = \dfrac{32 \pm \sqrt{880}}{6} \\[1em] = \dfrac{32 \pm \sqrt{4 \times 220}}{6} \\[1em] = \dfrac{32 \pm 2\sqrt{220}}{6} \\[1em] = \dfrac{32 + 2\sqrt{220}}{6} \text{ or } \dfrac{32 - 2\sqrt{220}}{6} \\[1em] = \dfrac{2(16 + \sqrt{220})}{6} \text{ or } \dfrac{2(16 - \sqrt{220})}{6} \\[1em] = \dfrac{16 + \sqrt{220}}{3} \text{ or } \dfrac{16 - \sqrt{220}}{3} \\[1em] = \dfrac{16 + 14.83}{3} \text{ or } \dfrac{16 - 14.83}{3} \\[1em] = \dfrac{30.83}{3} \text{ or } \dfrac{1.17}{3} \\[1em] = 10.28 \text{ or } 0.39 ⇒ x = 2 × ( 3 ) − ( − 32 ) ± ( − 32 ) 2 − 4 × 3 × ( 12 ) = 6 32 ± 1024 − 144 = 6 32 ± 880 = 6 32 ± 4 × 220 = 6 32 ± 2 220 = 6 32 + 2 220 or 6 32 − 2 220 = 6 2 ( 16 + 220 ) or 6 2 ( 16 − 220 ) = 3 16 + 220 or 3 16 − 220 = 3 16 + 14.83 or 3 16 − 14.83 = 3 30.83 or 3 1.17 = 10.28 or 0.39
Hence, x = {10.28, 0.39}.
Solve the following equation using quadratic formula:
x2 + 7x = 7
Answer
Given,
⇒ x2 + 7x - 7 = 0
Comparing equation x2 + 7x - 7 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = 7 and c = -7.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 7 ) ± ( 7 ) 2 − 4 × ( 1 ) × ( − 7 ) 2 × ( 1 ) = − 7 ± 49 + 28 2 = − 7 ± 77 2 = − 7 + 77 2 or − 7 − 77 2 = − 7 + 8.77 2 or − 7 − 8.77 2 = 1.77 2 or − 15.77 2 = 0.89 or − 7.89. \Rightarrow x = \dfrac{-(7) \pm \sqrt{(7)^2 - 4 \times (1) \times (-7)}}{2 \times (1)} \\[1em] = \dfrac{-7 \pm \sqrt{49 + 28}}{2} \\[1em] = \dfrac{-7 \pm \sqrt{77}}{2} \\[1em] = \dfrac{-7 + \sqrt{77}}{2} \text{ or } \dfrac{-7 - \sqrt{77}}{2} \\[1em] = \dfrac{-7 + 8.77}{2} \text{ or } \dfrac{-7 - 8.77}{2} \\[1em] = \dfrac{1.77}{2} \text{ or } \dfrac{-15.77}{2} \\[1em] = 0.89 \text{ or } -7.89. ⇒ x = 2 × ( 1 ) − ( 7 ) ± ( 7 ) 2 − 4 × ( 1 ) × ( − 7 ) = 2 − 7 ± 49 + 28 = 2 − 7 ± 77 = 2 − 7 + 77 or 2 − 7 − 77 = 2 − 7 + 8.77 or 2 − 7 − 8.77 = 2 1.77 or 2 − 15.77 = 0.89 or − 7.89.
Hence, x = {0.89, -7.89}.
Solve the following equation using quadratic formula:
3x2 - x - 7 = 0
Answer
Given,
⇒ 3x2 - x - 7 = 0
Comparing equation 3x2 - x - 7 = 0 with ax2 + bx + c = 0, we get :
a = 3, b = -1 and c = -7.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 1 ) ± ( − 1 ) 2 − 4 × ( 3 ) × ( − 7 ) 2 × ( 3 ) = 1 ± 1 + 84 6 = 1 ± 85 6 = 1 + 85 6 or 1 − 85 6 = 1 + 9.2195 6 or 1 − 9.2195 6 = 10.2195 6 or − 8.2195 6 = 1.703 ≈ 1.70 or − 1.369 ≈ − 1.37. \Rightarrow x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4 \times (3) \times (-7)}}{2 \times (3)} \\[1em] = \dfrac{1 \pm \sqrt{1 + 84}}{6} \\[1em] = \dfrac{1 \pm \sqrt{85}}{6} \\[1em] = \dfrac{1 + \sqrt{85}}{6} \text{ or } \dfrac{1 - \sqrt{85}}{6} \\[1em] = \dfrac{1 + 9.2195}{6} \text{ or } \dfrac{1 - 9.2195}{6} \\[1em] = \dfrac{10.2195}{6} \text{ or } \dfrac{-8.2195}{6} \\[1em] = 1.703 \approx 1.70 \text{ or } -1.369 \approx -1.37. ⇒ x = 2 × ( 3 ) − ( − 1 ) ± ( − 1 ) 2 − 4 × ( 3 ) × ( − 7 ) = 6 1 ± 1 + 84 = 6 1 ± 85 = 6 1 + 85 or 6 1 − 85 = 6 1 + 9.2195 or 6 1 − 9.2195 = 6 10.2195 or 6 − 8.2195 = 1.703 ≈ 1.70 or − 1.369 ≈ − 1.37.
Hence, x = {1.70, -1.37}.
Solve the following equation using quadratic formula:
4x2 - 7x + 2 = 0
Answer
Given,
⇒ 4x2 - 7x + 2 = 0
Comparing equation 4x2 - 7x + 2 = 0 with ax2 + bx + c = 0, we get :
a = 4, b = -7 and c = 2.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 7 ) ± ( − 7 ) 2 − 4 × ( 4 ) × ( 2 ) 2 × ( 4 ) = 7 ± 49 − 32 8 = 7 ± 17 8 = 7 + 17 8 or 7 − 17 8 = 7 + 4.12 8 or 7 − 4.12 8 = 11.12 8 or 2.88 8 = 1.39 or 0.36 \Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times (4) \times (2)}}{2 \times (4)} \\[1em] = \dfrac{7 \pm \sqrt{49 - 32}}{8} \\[1em] = \dfrac{7 \pm \sqrt{17}}{8} \\[1em] = \dfrac{7 + \sqrt{17}}{8} \text{ or } \dfrac{7 - \sqrt{17}}{8} \\[1em] = \dfrac{7 + 4.12}{8} \text{ or } \dfrac{7 - 4.12}{8} \\[1em] = \dfrac{11.12}{8} \text{ or } \dfrac{2.88}{8} \\[1em] = 1.39 \text{ or } 0.36 ⇒ x = 2 × ( 4 ) − ( − 7 ) ± ( − 7 ) 2 − 4 × ( 4 ) × ( 2 ) = 8 7 ± 49 − 32 = 8 7 ± 17 = 8 7 + 17 or 8 7 − 17 = 8 7 + 4.12 or 8 7 − 4.12 = 8 11.12 or 8 2.88 = 1.39 or 0.36
Hence, x = {1.39, 0.36}.
Solve the following equation using quadratic formula:
x2 - 7x + 3 = 0
Answer
Given,
⇒ x2 - 7x + 3 = 0
Comparing equation x2 - 7x + 3 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = -7 and c = 3.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 7 ) ± ( − 7 ) 2 − 4 × ( 1 ) × ( 3 ) 2 × ( 1 ) = 7 ± 49 − 12 2 = 7 ± 37 2 = 7 + 37 2 or 7 − 37 2 = 7 + 6.08 2 or 7 − 6.08 2 = 13.08 2 or 0.92 2 = 6.54 or 0.46 \Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times (1) \times (3)}}{2 \times (1)} \\[1em] = \dfrac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \dfrac{7 \pm \sqrt{37}}{2} \\[1em] = \dfrac{7 + \sqrt{37}}{2} \text{ or } \dfrac{7 - \sqrt{37}}{2} \\[1em] = \dfrac{7 + 6.08}{2} \text{ or } \dfrac{7 - 6.08}{2} \\[1em] = \dfrac{13.08}{2} \text{ or } \dfrac{0.92}{2} \\[1em] = 6.54 \text{ or } 0.46 ⇒ x = 2 × ( 1 ) − ( − 7 ) ± ( − 7 ) 2 − 4 × ( 1 ) × ( 3 ) = 2 7 ± 49 − 12 = 2 7 ± 37 = 2 7 + 37 or 2 7 − 37 = 2 7 + 6.08 or 2 7 − 6.08 = 2 13.08 or 2 0.92 = 6.54 or 0.46
Hence, x = {6.54, 0.46}.
Solve the following equation using quadratic formula:
x2 - 5x - 10 = 0
Answer
Comparing x2 - 5x - 10 = 0 with ax2 + bx + c = 0 we get,
a = 1, b = -5 and c = -10.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 5 ) ± ( − 5 ) 2 + 4 × ( 1 ) × ( − 10 ) 2 × ( 1 ) ⇒ x = 5 ± 25 + 40 2 ⇒ x = 5 ± 65 2 ⇒ x = 5 ± 8.06 2 ⇒ x = 5 + 8.06 2 or 5 − 8.06 2 ⇒ x = 13.06 2 or − 3.06 2 ⇒ x = 6.53 or − 1.53. \Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 + 4 \times (1) \times (-10)}}{2 \times (1)} \\[1em] \Rightarrow x = \dfrac{5 \pm \sqrt{25 + 40}}{2} \\[1em] \Rightarrow x = \dfrac{5 \pm \sqrt{65}}{2} \\[1em] \Rightarrow x = \dfrac{5 \pm 8.06}{2} \\[1em] \Rightarrow x = \dfrac{5 + 8.06}{2} \text{ or } \dfrac{5 - 8.06}{2} \\[1em] \Rightarrow x = \dfrac{13.06}{2} \text{ or } \dfrac{-3.06}{2} \\[1em] \Rightarrow x = 6.53 \text{ or } -1.53. ⇒ x = 2 × ( 1 ) − ( − 5 ) ± ( − 5 ) 2 + 4 × ( 1 ) × ( − 10 ) ⇒ x = 2 5 ± 25 + 40 ⇒ x = 2 5 ± 65 ⇒ x = 2 5 ± 8.06 ⇒ x = 2 5 + 8.06 or 2 5 − 8.06 ⇒ x = 2 13.06 or 2 − 3.06 ⇒ x = 6.53 or − 1.53.
Hence, x = 6.53 and -1.53
Solve for x the quadratic equation x2 - 4x - 8 = 0. Give your answer correct to three significant figures.
Answer
Given,
⇒ x2 - 4x - 8 = 0
Comparing equation x2 - 4x - 8 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = -4 and c = -8.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( − 4 ) ± ( − 4 ) 2 − 4 × ( 1 ) × ( − 8 ) 2 × ( 1 ) = 4 ± 16 + 32 2 = 4 ± 48 2 = 4 + 48 2 or 4 − 48 2 = 4 + 6.928 2 or 4 − 6.928 2 = 10.928 2 or − 2.928 2 = 5.464 or − 1.464 \Rightarrow x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times (1) \times (-8)}}{2 \times (1)} \\[1em] = \dfrac{4 \pm \sqrt{16 + 32}}{2} \\[1em] = \dfrac{4 \pm \sqrt{48}}{2} \\[1em] = \dfrac{4 + \sqrt{48}}{2} \text{ or } \dfrac{4 - \sqrt{48}}{2} \\[1em] = \dfrac{4 + 6.928}{2} \text{ or } \dfrac{4 - 6.928}{2} \\[1em] = \dfrac{10.928}{2} \text{ or } \dfrac{-2.928}{2} \\[1em] = 5.464 \text{ or } -1.464 ⇒ x = 2 × ( 1 ) − ( − 4 ) ± ( − 4 ) 2 − 4 × ( 1 ) × ( − 8 ) = 2 4 ± 16 + 32 = 2 4 ± 48 = 2 4 + 48 or 2 4 − 48 = 2 4 + 6.928 or 2 4 − 6.928 = 2 10.928 or 2 − 2.928 = 5.464 or − 1.464
Correcting the value of x to three significant figures.
Hence, x = {5.46, -1.46}.
Solve the following quadratic equation : x2 + 4x - 8 = 0. Give your answer correct to one decimal place.
Answer
Given,
⇒ x2 + 4x - 8 = 0
Comparing equation x2 + 4x - 8 = 0 with ax2 + bx + c = 0, we get :
a = 1, b = 4 and c = -8.
By formula,
x = − b ± b 2 − 4 a c 2 a \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} 2 a − b ± b 2 − 4 a c
Substituting values we get :
⇒ x = − ( 4 ) ± ( 4 ) 2 − 4 × ( 1 ) × ( − 8 ) 2 × ( 1 ) = − 4 ± 16 + 32 2 = − 4 ± 48 2 = − 4 + 48 2 or − 4 − 48 2 = − 4 + 6.9 2 or − 4 − 6.9 2 = 2.9 2 or − 10.9 2 = 1.45 or − 5.45 \Rightarrow x = \dfrac{-(4) \pm \sqrt{(4)^2 - 4 \times (1) \times (-8)}}{2 \times (1)} \\[1em] = \dfrac{-4 \pm \sqrt{16 + 32}}{2} \\[1em] = \dfrac{-4 \pm \sqrt{48}}{2} \\[1em] = \dfrac{-4 + \sqrt{48}}{2} \text{ or } \dfrac{-4 - \sqrt{48}}{2} \\[1em] = \dfrac{-4 + 6.9}{2} \text{ or } \dfrac{-4 - 6.9}{2} \\[1em] = \dfrac{2.9}{2} \text{ or } \dfrac{-10.9}{2} \\[1em] = 1.45 \text{ or } -5.45 ⇒ x = 2 × ( 1 ) − ( 4 ) ± ( 4 ) 2 − 4 × ( 1 ) × ( − 8 ) = 2 − 4 ± 16 + 32 = 2 − 4 ± 48 = 2 − 4 + 48 or 2 − 4 − 48 = 2 − 4 + 6.9 or 2 − 4 − 6.9 = 2 2.9 or 2 − 10.9 = 1.45 or − 5.45
Correcting the value of x to one decimal place.
Hence, x = {1.5, -5.5}.
