Quadratic Equations — Exercise 5(C)

If x² - 3x + 2 = 0, values of x correct to one decimal place are :

1. 2.0 and 1.0

2. 2.0 or 1.0

3. 3.0 and 2.0

4. 3.0 or 2.0

Answer

Comparing equation x² - 3x + 2 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -3 and c = 2.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times 2}}{2 \times 1} \\[1em] = \dfrac{3 \pm \sqrt{9 - 8}}{2} \\[1em] = \dfrac{3 \pm \sqrt{1}}{2} \\[1em] = \dfrac{3 \pm 1}{2} \\[1em] = \dfrac{3 + 1}{2} \text{ or } \dfrac{3 - 1}{2} \\[1em] = \dfrac{4}{2} \text{ or } \dfrac{2}{2} \\[1em] = 2.0 \text{ or } 1.0$

Hence, Option 2 is the correct option.

If x² - 4x - 5 = 0, values of x correct to two decimal places are :

1. 5.00 or -1.00

2. 5 or 1

3. 5.0 and -1.0

4. -5.00 and -1.00

Answer

Comparing equation x² - 4x - 5 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -4 and c = -5.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times -5}}{2 \times 1} \\[1em] = \dfrac{4 \pm \sqrt{16 + 20}}{2} \\[1em] = \dfrac{4 \pm \sqrt{36}}{2} \\[1em] = \dfrac{4 \pm 6}{2} \\[1em] = \dfrac{4 + 6}{2} \text{ or } \dfrac{4 - 6}{2} \\[1em] = \dfrac{10}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 5.00 \text{ or } -1.00$

Hence, Option 1 is the correct option.

If x² - 8x - 9 = 0; values of x correct to one significant figure are :

1. 9 and -1

2. 9 or -1

3. 9.0 or -1.0

4. 9.00 or -1.00

Answer

Comparing equation x² - 8x - 9 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -8 and c = -9.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times -9}}{2 \times 1} \\[1em] = \dfrac{8 \pm \sqrt{64 + 36}}{2} \\[1em] = \dfrac{8 \pm \sqrt{100}}{2} \\[1em] = \dfrac{8 \pm 10}{2} \\[1em] = \dfrac{8 + 10}{2} \text{ or } \dfrac{8 - 10}{2} \\[1em] = \dfrac{18}{2} \text{ or } \dfrac{-2}{2} \\[1em] = 9 \text{ or } -1$

Hence, Option 2 is the correct option.

If x² - 2x - 3 = 0; values of x correct to two significant figures are :

1. 3.0 and 1.0

2. -1.0 and 3.0

3. 3.0 or -1.0

4. 3.00 or -1.00

Answer

Given,

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ x + 1 = 0 or x - 3 = 0

⇒ x = -1 or x = 3.

Rounding off to two significant figures, we get :

⇒ x = -1.0 or x = 3.0

Hence, Option 3 is the correct option.

The value (values) of x satisfying the equation x² - 6x - 16 = 0 is/are :

1. 8 or -2

2. -8 or 2

3. 8 and -2

4. -8 and 2

Answer

Comparing equation x² - 6x - 16 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -6 and c = -16.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times -16}}{2 \times 1} \\[1em] = \dfrac{6 \pm \sqrt{36 + 64}}{2} \\[1em] = \dfrac{6 \pm \sqrt{100}}{2} \\[1em] = \dfrac{6 \pm 10}{2} \\[1em] = \dfrac{6 + 10}{2} \text{ or } \dfrac{6 - 10}{2} \\[1em] = \dfrac{16}{2} \text{ or } \dfrac{-4}{2} \\[1em] = 8 \text{ or } -2.$

Substituting x = 8 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ 8² - 6(8) - 16

⇒ 64 - 48 - 16

⇒ 0.

Substituting x = -2 in L.H.S. of equation x² - 6x - 16 = 0, we get :

⇒ (-2)² - 6(-2) - 16

⇒ 4 + 12 - 16

⇒ 0.

Since, L.H.S. = R.H.S., thus x = -2 is a solution of the equation.

Thus 8 and -2 are the solution of the equation x² - 6x - 16 = 0.

Hence, Option 3 is the correct option.

Solve the following equation for x and give your answer correct to one decimal place :

x² - 8x + 5 = 0

Answer

Comparing x² - 8x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -8 and c = 5.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{8 \pm \sqrt{64 - 20}}{2} \\[1em] = \dfrac{8 \pm \sqrt{44}}{2} \\[1em] = \dfrac{8 \pm 2\sqrt{11}}{2} \\[1em] = 4 \pm \sqrt{11} \\[1em] = 4 + \sqrt{11} \text{ and } 4 - \sqrt{11} \\[1em] = 4 + 3.3 \text{ and } 4 - 3.3 \\[1em] = 7.3 \text{ and } 0.7$

Hence, x = 7.3 and 0.7

Solve the following equation for x and give your answer correct to one decimal place :

5x² + 10x - 3 = 0

Answer

Comparing 5x² + 10x - 3 = 0 with ax² + bx + c = 0 we get,

a = 5, b = 10 and c = -3.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(10) \pm \sqrt{(10)^2 - 4.(5).(-3)}}{2(5)} \\[1em] = \dfrac{-10 \pm \sqrt{100 + 60}}{10} \\[1em] = \dfrac{-10 \pm \sqrt{160}}{10} \\[1em] = \dfrac{-10 \pm 4\sqrt{10}}{10} \\[1em] = \dfrac{-10 \pm 12.8}{10} \\[1em] = \dfrac{-10 + 12.8}{10} \text{ and } \dfrac{-10 - 12.8}{10} \\[1em] = \dfrac{2.8}{10} \text{ and } \dfrac{-22.8}{10} \\[1em] = 0.28 \text{ and } -2.28 \\[1em] \approx 0.3 \text{ and } -2.3$

Hence, x = 0.3 and -2.3

Solve the following equation for x and give your answer correct to two decimal places :

2x² - 10x + 5 = 0

Answer

Comparing 2x² - 10x + 5 = 0 with ax² + bx + c = 0 we get,

a = 2, b = -10 and c = 5.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4.(2).(5)}}{2(2)} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{10 \pm 7.74}{4} \\[1em] = \dfrac{10 + 7.74}{4} \text{ and } \dfrac{10 - 7.74}{4} \\[1em] = \dfrac{17.74}{4} \text{ and } \dfrac{2.26}{4} \\[1em] = 4.44 \text{ and } 0.56$

Hence, x = 4.44 and 0.56

Solve the following equation for x and give your answer correct to two decimal places :

4x + $\dfrac{6}{x}$ + 13 = 0

Answer

Given,

$\Rightarrow 4x + \dfrac{6}{x} + 13 = 0 \\[1em] \Rightarrow \dfrac{4x^2 + 6 + 13x}{x} = 0 \\[1em] \Rightarrow 4x^2 + 13x + 6 = 0$

Comparing 4x² + 13x + 6 = 0 with ax² + bx + c = 0 we get,

a = 4, b = 13 and c = 6.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(13) \pm \sqrt{(-13)^2 - 4.(4).(6)}}{2(4)} \\[1em] = \dfrac{-13 \pm \sqrt{169 - 96}}{8} \\[1em] = \dfrac{-13 \pm \sqrt{73}}{8} \\[1em] = \dfrac{-13 \pm 8.54}{8} \\[1em] = \dfrac{-13 + 8.54}{8} \text{ and } \dfrac{-13 - 8.54}{8} \\[1em] = \dfrac{-4.46}{8} \text{ and } \dfrac{-21.54}{8} \\[1em] = -0.56 \text{ and } -2.69$

Hence, x = -0.56 and -2.69

Solve the following equation for x and give your answer correct to two decimal places :

4x² - 5x - 3 = 0

Answer

Comparing 4x² - 5x - 3 = 0 with ax² + bx + c = 0 we get,

a = 4, b = -5 and c = -3.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(4).(-3)}}{2(4)} \\[1em] = \dfrac{5 \pm \sqrt{25 + 48}}{8} \\[1em] = \dfrac{5 \pm \sqrt{73}}{8} \\[1em] = \dfrac{5 \pm 8.54}{8} \\[1em] = \dfrac{5 + 8.54}{8} \text{ and } \dfrac{5 - 8.54}{8} \\[1em] = \dfrac{13.54}{8} \text{ and } \dfrac{-3.54}{8} \\[1em] = 1.69 \text{ and } -0.44$

Hence, x = 1.69 and -0.44

Solve the following equation for x, giving your answer correct to 3 decimal places :

3x² - 12x - 1 = 0

Answer

Comparing 3x² - 12x - 1 = 0 with ax² + bx + c = 0 we get,

a = 3, b = -12 and c = -1.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-12) \pm \sqrt{(-12)^2 - 4.(3).(-1)}}{2(3)} \\[1em] = \dfrac{12 \pm \sqrt{144 + 12}}{6} \\[1em] = \dfrac{12 \pm \sqrt{156}}{6} \\[1em] = \dfrac{12 \pm 12.49}{6} \\[1em] = \dfrac{12 + 12.49}{6} \text{ or } \dfrac{12 - 12.49}{6} \\[1em] = \dfrac{24.49}{6} \text{ or } \dfrac{-0.49}{6} \\[1em] = 4.082 \text{ or } -0.082$

Hence, x = 4.082 or -0.082

Solve the following equation for x, giving your answer correct to 3 decimal places :

x² - 16x + 6 = 0

Answer

Comparing x² - 16x + 6 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -16 and c = 6.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-16) \pm \sqrt{(-16)^2 - 4.(1).(6)}}{2(1)} \\[1em] = \dfrac{16 \pm \sqrt{256 - 24}}{2} \\[1em] = \dfrac{16 \pm \sqrt{232}}{2} \\[1em] = \dfrac{16 \pm 15.232}{2} \\[1em] = \dfrac{16 + 15.232}{2} \text{ or } \dfrac{16 - 15.232}{2} \\[1em] = \dfrac{31.232}{2} \text{ or } \dfrac{0.768}{2} \\[1em] = 15.616 \text{ or } 0.384$

Hence, x = 15.616 or 0.384

Solve the following equation for x, giving your answer correct to 3 decimal places :

2x² + 11x + 4 = 0

Answer

Comparing 2x² + 11x + 4 = 0 with ax² + bx + c = 0 we get,

a = 2, b = 11 and c = 4.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(11) \pm \sqrt{(11)^2 - 4.(2).(4)}}{2(2)} \\[1em] = \dfrac{-11 \pm \sqrt{121 - 32}}{4} \\[1em] = \dfrac{-11 \pm \sqrt{89}}{4} \\[1em] = \dfrac{-11 \pm 9.434}{4} \\[1em] = \dfrac{-11 + 9.434}{4} \text{ or } \dfrac{-11 - 9.434}{4} \\[1em] = \dfrac{-1.566}{4} \text{ or } \dfrac{-20.434}{4} \\[1em] = -0.392 \text{ or } -5.109$

Hence, x = -0.392 or -5.109

Solve the following equation and give your answer correct to 3 significant figures :

5x² - 3x - 4 = 0

Answer

Comparing 5x² - 3x - 4 = 0 with ax² + bx + c = 0 we get,

a = 5, b = -3 and c = -4.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4.(5).(-4)}}{2(5)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 80}}{10} \\[1em] = \dfrac{3 \pm \sqrt{89}}{10} \\[1em] = \dfrac{3 \pm 9.434}{10} \\[1em] = \dfrac{3 + 9.434}{10} \text{ or } \dfrac{3 - 9.434}{10} \\[1em] = \dfrac{12.434}{10} \text{ or } \dfrac{-6.434}{10} \\[1em] = 1.2434 \text{ or } -0.6434 \\[1em] \approx 1.24 \text{ or } -0.643$

Hence, x = 1.24 or -0.643

Solve for x using the quadratic formula. Write your answer correct to two significant figures.

(x - 1)² - 3x + 4 = 0

Answer

Given,

⇒ (x - 1)² - 3x + 4 = 0

⇒ x² + 1 - 2x - 3x + 4 = 0

⇒ x² - 5x + 5 = 0

Comparing x² - 5x + 5 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -5 and c = 5.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4.(1).(5)}}{2(1)} \\[1em] = \dfrac{5 \pm \sqrt{25 - 20}}{2} \\[1em] = \dfrac{5 \pm \sqrt{5}}{2} \\[1em] = \dfrac{5 \pm 2.2}{2} \\[1em] = \dfrac{5 + 2.2}{2} \text{ or } \dfrac{5 - 2.2}{2} \\[1em] = \dfrac{7.2}{2} \text{ or } \dfrac{2.8}{2} \\[1em] = 3.6 \text{ or } 1.4$

Hence, x = 3.6 or 1.4

x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0, then other root is:

1. 1

2. 3

3. -3

4. -4

Answer

Given,

x = 3 is a solution of the quadratic equation (k + 2)x² - kx + 6 = 0.

Substituting x = 3 in equation (k + 2)x² - kx + 6 = 0,

⇒ (k + 2)(3)² - k(3) + 6 = 0

⇒ (k + 2)(9) - 3k + 6 = 0

⇒ 9k + 18 - 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = -24

⇒ k = $\dfrac{-24}{6}$

⇒ k = -4.

Substituting k = −4 in equation (k + 2)x² - kx + 6 = 0 :

⇒ (-4 + 2)x² - (-4)x + 6 = 0

⇒ -2x² + 4x + 6 = 0

⇒ -2(x² - 2x - 3) = 0

⇒ x² - 2x - 3 = 0

⇒ x² - 3x + x - 3 = 0

⇒ x(x - 3) + 1(x - 3) = 0

⇒ (x + 1)(x - 3) = 0

⇒ (x + 1) = 0 or (x - 3) = 0

⇒ x = -1 or x = 3.

Hence, option 4 is the correct option.