Quadratic Equations — Exercise 5.3

Solve the following equations by using formula:

2x² - 7x + 6 = 0

Answer

The given equation is 2x² - 7x + 6 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -7 , c = 6

By using formula,

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

we obtain:

$\Rightarrow x = \dfrac{-(-7) ± \sqrt{(-7^2) - 4\times 2 \times 6}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{49 - 48}}{4} \\[1em] \Rightarrow x = \dfrac{7 ± \sqrt{1}}{4} \\[1em] \Rightarrow x = \dfrac{7 + 1}{4} \text{ or } \dfrac{7 - 1}{4} \\[1em] \Rightarrow x = \dfrac{8}{4} \text { or } \dfrac{6}{4} \\[1em] x = 2 \text{ or } \dfrac{3}{2}$

Hence roots of the given equation are 2 , $\dfrac{3}{2}$.

Solve the following equations by using formula:

2x² - 6x + 3 = 0

Answer

The given equation is 2x² - 6x + 3 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -6 , c = 3

By using formula,

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

we obtain:

$\Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6^2) - 4\times 2 \times 3}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 24}}{4} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 + \sqrt{12}}{4} \text{ or } \dfrac{6 - \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 + 2\sqrt{3}}{4} \text { or } \dfrac{6 - 2\sqrt{3}}{4} \\[1em] x = \dfrac{3 + \sqrt{3}}{2} \text{ or } \dfrac{3 - \sqrt{3}}{2}$

Hence roots of the given equation are $\dfrac{3 + \sqrt{3}}{2} , \dfrac{3 - \sqrt{3}}{2}$.

Solve the following equations by using formula:

256x² - 32x + 1 = 0

Answer

The given equation is 256x² - 32x + 1 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 256 , b = -32 , c = 1

By using formula,

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

we obtain:

$\Rightarrow x = \dfrac{-(-32) ± \sqrt{(-32)^2 - 4\times 256 \times 1}}{2 \times 256} \\[1em] \Rightarrow x = \dfrac{32 ± \sqrt{1024 - 1024}}{512} \\[1em] \Rightarrow x = \dfrac{32 ± \sqrt{0}}{512} \\[1em] \Rightarrow x = \dfrac{32 + 0}{512} \text{ or } \dfrac{32 - 0}{512} \\[1em] \Rightarrow x = \dfrac{32}{512} \text{ or } \dfrac{32 - 0}{512} \\[1em] x = \dfrac{1}{16} \text{ or } \dfrac{1}{16}$

Hence roots of the given equation are $\dfrac{1}{16} , \dfrac{1}{16}$.

Solve the following equations by using formula:

25x² + 30x + 7 = 0

Answer

The given equation is 25x² + 30x + 7 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 25 , b = 30 , c = 7

By using formula,

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

we obtain:

$\Rightarrow x = \dfrac{-(30) ± \sqrt{(30^2) - 4\times 25 \times 7}}{2 \times 25} \\[1em] \Rightarrow x = \dfrac{-30 ± \sqrt{900 - 700}}{50} \\[1em] \Rightarrow x = \dfrac{-30 ± \sqrt{200}}{50} \\[1em] \Rightarrow x = \dfrac{-30 + \sqrt{200}}{50} \text{ or } \dfrac{-30 - \sqrt{200}}{50} \\[1em] \Rightarrow x = \dfrac{-30 + 10\sqrt{2}}{50} \text { or } \dfrac{-30 - 10\sqrt{2}}{50} \\[1em] x = \dfrac{-3 + \sqrt{2}}{5} \text{ or } \dfrac{-3 - \sqrt{2}}{5}$

Hence roots of the given equation are $\dfrac{-3 + \sqrt{2}}{5} , \dfrac{-3 - \sqrt{2}}{5}$.

Solve the following equations by using formula:

$2x^2 + \sqrt{5}x - 5 = 0$

Answer

The given equation is $2x^2 + \sqrt{5}x - 5 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = $\sqrt{5}$ , c = -5

By using formula,

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

we obtain:

$\Rightarrow x = \dfrac{-(\sqrt{5}) ± \sqrt{(\sqrt{5})^2 - 4\times 2 \times -5}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{5 + 40}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + \sqrt{45}}{4} \text{ or } \dfrac{-\sqrt{5} - \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + 3\sqrt{5}}{4} \text { or } \dfrac{-\sqrt{5} - 3\sqrt{5}}{4} \\[1em] \Rightarrow x = \dfrac{\sqrt{5}(-1 + 3)}{4} \text{ or } \dfrac{\sqrt{5}(-1 - 3)}{4} \\[1em] x = \dfrac{\sqrt{5}}{2} \text{ or } -\sqrt{5}$

Hence roots of the given equations are $\dfrac{\sqrt{5}}{2} , -\sqrt{5}$.

Solve the following equations by using formula:

$\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0$

Answer

The given equation is $\sqrt{3}x^2 + 10x - 8\sqrt{3} = 0$

Comparing it with ax² + bx + c = 0, we get
$a = \sqrt{3}, \space b =10, \space c = -8\sqrt{3}$

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(10) ± \sqrt{(10)^2 - 4\times \sqrt{3} \times -8\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 ± \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 ± \sqrt{196}}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{-10 + 14}{2\sqrt{3}} \text{ or } \dfrac{-10 - 14}{2\sqrt{3}}\\[1em] \Rightarrow x = \dfrac{4}{2\sqrt{3}} \text { or } \dfrac{-24}{2\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{2}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \text{ (Multiplying both roots by } \dfrac{\sqrt{3}}{\sqrt{3}}) \\[1em] \Rightarrow x = \dfrac{2\sqrt{3}}{3} \text{ or } -4\sqrt{3}.$

Hence roots of the given equations are $\dfrac{2\sqrt{3}}{3} , -4\sqrt{3}$.

Solve the following equations by using formula:

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4$

Answer

Given,

$\dfrac{x - 2}{x + 2} + \dfrac{x + 2}{x - 2} = 4 \\[1em] \Rightarrow \dfrac{(x - 2)^2 + (x + 2)^2}{(x - 2)(x + 2)} = 4 \\[1em] \Rightarrow \dfrac{x^2 + 4 - 4x + x^2 + 4 + 4x}{x^2 - 2x + 2x - 4} = 4 \\[1em] \Rightarrow \dfrac{2x^2 + 8}{x^2 - 4} = 4 \\[1em] \Rightarrow 2x^2 + 8 = 4(x^2 - 4) \\[1em] \Rightarrow 2x^2 + 8 = 4x^2 - 16 \\[1em] \Rightarrow 2x^2 - 4x^2 + 8 + 16 = 0 \\[1em] \Rightarrow -2x^2 + 24 = 0 \\[1em] 2x^2 - 24 = 0 \text{ (Multiplying equation by -1) }$

The equation is $2x^2 - 24 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 0 , c = -24

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-0 ± \sqrt{0^2 - 4 \times 2 \times (-24)}}{2 \times 2} \\[1em] \Rightarrow \dfrac{ ±\sqrt{192}}{4} \\[1em] \Rightarrow \dfrac{±8\sqrt{3}}{4} \\[1em] +2\sqrt{3} \text { or } -2\sqrt{3} \\[1em]$

Hence roots of the given equations are $2\sqrt{3} , -2\sqrt{3}$.

Solve the following equations by using formula:

$\dfrac{x + 1}{x + 3} = \dfrac{3x + 2}{2x + 3}$

Answer

Given,

$\dfrac{x + 1}{x + 3} = \dfrac{3x + 2}{2x + 3} \\[0.5em] \Rightarrow (x + 1)(2x + 3) = (3x + 2)(x + 3) \text{ On cross multiplication}\\[0.5em] \Rightarrow 2x^2 + 3x + 2x + 3 = 3x^2 + 9x + 2x + 6 \\[0.5em] \Rightarrow 2x^2 + 5x + 3 = 3x^2 + 11x + 6 \\[0.5em] \Rightarrow 2x^2 - 3x^2 + 5x - 11x + 3 - 6 = 0 \\[0.5em] \Rightarrow -x^2 - 6x - 3 = 0 \\[0.5em] \Rightarrow x^2 + 6x + 3 = 0 \text{ (Multiplying equation by -1) } \\[0.5em]$

The equation is x² + 6x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 6 , c = 3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-6 ± \sqrt{6^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-6 ±\sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + \sqrt{24}}{2} \text { or } \dfrac{-6 - \sqrt{24}}{2} \\[1em] \Rightarrow \dfrac{-6 + 2\sqrt{6}}{2} \text{ or } \dfrac{-6 - 2\sqrt{6}}{2} \\[1em] -3 + \sqrt{6} \text{ or } -3 - \sqrt{6}$

Hence roots of the given equations are $-3 + \sqrt{6} , -3 - \sqrt{6}$.

Solve the following equations by using formula:

a(x² + 1) = (a² + 1)x , a ≠ 0

Answer

Given,

a(x² + 1) = (a² + 1)x

First converting the equation in the form ax² + bx + c = 0.

$\Rightarrow ax^2 + a = a^2x + x \\[0.5em] \Rightarrow ax^2 - a^2x - x + a = 0 \\[0.5em] \Rightarrow ax^2 -(a^2 + 1)x + a = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = a , b = -(a² + 1) , c = a

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-(a^2 + 1)) ± \sqrt{(-(a^2 + 1))^2 - 4 \times a \times a}}{2 \times a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 + 1)^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 + 2a^2 - 4a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{a^4 + 1 - 2a^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± \sqrt{(a^2 - 1)^2}}{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 ± (a^2 - 1) }{2a} \\[1em] \Rightarrow \dfrac{a^2 + 1 + a^2 - 1}{2a} \text{ or } \dfrac{a^2 + 1 - a^2 + 1}{2a} \\[1em] \Rightarrow \dfrac{2a^2}{2a} \text{ or } \dfrac{2}{2a} \\[1em] \Rightarrow a \text{ or } \dfrac{1}{a}$

Hence roots of the given equations are a , $\dfrac{1}{a}$.

Solve the following equations by using formula:

4x² - 4ax + (a² - b²) = 0

Answer

The given equation is 4x² - 4ax + (a² - b²) = 0.

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -4a , c = a² - b²

By using formula,

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

we obtain:

$\Rightarrow \dfrac{-(-4a) ± \sqrt{(-4a)^2 - 4 \times 4 \times (a^2 - b^2)}}{2 \times 4} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16a^2 - 16(a^2 - b^2)}}{8} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16a^2 - 16a^2 + 16b^2}}{8} \\[1em] \Rightarrow \dfrac{4a ± \sqrt{16b^2}}{8} \\[1em] \Rightarrow \dfrac{4a ± 4b}{8} \\[1em] \Rightarrow \dfrac{4a + 4b}{8} \text{ or } \dfrac{4a - 4b}{8} \\[1em] \Rightarrow \dfrac{a + b}{2} \text{ or } \dfrac{a - b}{2} \\[1em]$

Hence roots of the given equations are $\dfrac{a + b}{2} , \dfrac{a - b}{2}$.

Solve the following equations by using formula:

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

Answer

Given,

$x - \dfrac{1}{x} = 3 , x$ ≠ 0

$\Rightarrow \dfrac{x^2 - 1}{x} = 3 \text{ (By taking L.C.M) } \\[0.5em] \Rightarrow x^2 - 1 = 3x \\[0.5em] \Rightarrow x^2 - 3x - 1 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -3 , c = -1

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-3) ± \sqrt{(-3)^2 - 4 \times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{9 - (-4)}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{13}}{2} \\[1em] \Rightarrow \dfrac{3 + \sqrt{13}}{2} \text{ or } \dfrac{3 - \sqrt{13}}{2} \\[1em]$

Hence roots of the given equations are $\dfrac{3 + \sqrt{13}}{2} , \dfrac{3 - \sqrt{13}}{2}$.

Solve the following equations by using formula:

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 , x$ ≠ $0, 2.$

Answer

Given,

$\dfrac{1}{x} + \dfrac{1}{x - 2} = 3 \\[1em] \Rightarrow \dfrac{x - 2 + x }{x(x - 2)} = 3 \\[1em] \Rightarrow \dfrac{2x - 2}{x^2 - 2x} = 3 \\[1em] \Rightarrow 2x - 2 = 3(x^2 - 2x) \\[1em] \Rightarrow 2x - 2 = 3x^2 - 6x \\[1em] \Rightarrow 3x^2 - 6x - 2x + 2 = 0 \\[1em] \Rightarrow 3x^2 - 8x + 2 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -8 , c = 2

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \\[1em] \Rightarrow \dfrac{-(-8) ± \sqrt{(-8)^2 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{64 - 24}}{6} \\[1em] \Rightarrow \dfrac{8 ± \sqrt{40}}{6} \\[1em] \Rightarrow \dfrac{8 ± 2\sqrt{10}}{6} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{10}}{3} \\[1em] \Rightarrow \dfrac{4 + \sqrt{10}}{3} \text{ or } \dfrac{4 - \sqrt{10}}{3}$

Hence roots of the given equations are $\dfrac{4 + \sqrt{10}}{3} , \dfrac{4 - \sqrt{10}}{3}$.

Solve for x :

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5, x$ ≠ $-3, \dfrac{1}{2}$

Answer

Given,

$2\Big(\dfrac{2x - 1}{x + 3}\Big) - 3\Big(\dfrac{x + 3}{2x - 1}\Big) = 5 \\[1em] \Rightarrow \text{ Taking } y = \dfrac{2x - 1}{x + 3} \text{ the equation becomes} \\[1em] \Rightarrow 2y - \dfrac{3}{y} = 5 \\[1em] \Rightarrow 2y^2 - 3 = 5y \\[1em] \Rightarrow 2y^2 - 5y - 3 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -5 , c = -3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow \dfrac{-(-5) ± \sqrt{(-5)^2 - 4 \times 2 \times -3}}{2 \times 2} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{25 + 24}}{4} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{5 ± 7}{4} \\[1em] \Rightarrow \dfrac{5 + 7}{4} \text{ or } \dfrac{5 - 7}{4}\\[1em] \Rightarrow \dfrac{12}{4} \text{ or } \dfrac{-2}{4} \\[1em] \Rightarrow 3 \text{ or } -\dfrac{1}{2} \\[1em]$

But,

$y = \dfrac{2x - 1}{x + 3} \\[1em] \therefore 3 = \dfrac{2x - 1}{x + 3} \text{ or } -\dfrac{1}{2} = \dfrac{2x - 1}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 2x - 1 \text{ or } -(x + 3) = 2(2x - 1) \\[1em] \Rightarrow 3x + 9 = 2x - 1 \text{ or } -x - 3 = 4x - 2 \\[1em] \Rightarrow 3x - 2x = -1 - 9 \text{ or } -x - 4x = -2 + 3 \\[1em] \Rightarrow x = -10 \text{ or } -5x = 1 \\[1em] \Rightarrow x = -10 \text{ or } x = -\dfrac{1}{5} \\[1em]$

Hence roots of the given equations are -10 , $-\dfrac{1}{5}$.

Solve the following quadratic equations for x and give your answer correct to 2 decimal places :

(i) x² - 5x - 10 = 0

(ii) x² + 7x = 7

Answer

(i) The given equation is x² - 5x - 10 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -5 , c = -10

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-5) ± \sqrt{(-5)^2 - 4\times 1 \times -10}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{25 + 40}}{2} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{65}}{2} \\[1em] \Rightarrow x = \dfrac{5 + \sqrt{65}}{2} \text{ or } \dfrac{5 - \sqrt{65}}{2} \\[1em] \text{ Also } \sqrt{65} = 8.062 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{5 + 8.062}{2} \text { or } \dfrac{5 - 8.062}{2} \\[1em] \Rightarrow x = \dfrac{13.062}{2} \text{ or } \dfrac{-3.062}{2} \\[1em] \Rightarrow x = 6.531 \text{ or } -1.531 \\[1em] x = 6.53 \text{ or } -1.53 \text{ (correct to two decimal places) }$

Hence roots of the given equations are 6.53 , -1.53.

(ii) Given, x² + 7x = 7

or , x² + 7x - 7 = 0

The given equation is x² + 7x - 7 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = 7 , c = -7

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(7) ± \sqrt{(7)^2 - 4\times 1 \times -7}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{-7 ± \sqrt{49 + 28}}{2} \\[1em] \Rightarrow x = \dfrac{-7 ± \sqrt{77}}{2} \\[1em] \Rightarrow x = \dfrac{-7 + \sqrt{77}}{2} \text{ or } \dfrac{-7 - \sqrt{77}}{2} \\[1em] \text{ Also } \sqrt{77} = 8.775 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{-7 + 8.775}{2} \text { or } \dfrac{-7 - 8.775}{2} \\[1em] \Rightarrow x = \dfrac{1.775}{2} \text{ or } \dfrac{-15.775}{2} \\[1em] \Rightarrow x = 0.885 \text{ or } -7.885 \\[1em] \Rightarrow x = 0.89\text{ or } -7.89 \text{ (correct to two decimal places) }$

Hence roots of the given equations are 0.89 , -7.89.

Solve the following equations by using quadratic formula and give answer in correct to 2 decimal places :

(i) 4x² - 5x - 3 = 0

(ii) x² - 7x + 3 = 0

Answer

(i) The given equation is 4x² - 5x - 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = 4 , b = -5 , c = -3

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-5) ± \sqrt{(-5)^2 - 4\times 4 \times -3}}{2 \times 4} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{25 + 48}}{8} \\[1em] \Rightarrow x = \dfrac{5 ± \sqrt{73}}{8} \\[1em] \Rightarrow x = \dfrac{5 + \sqrt{73}}{8} \text{ or } \dfrac{5 - \sqrt{73}}{8} \\[1em] \text{ Also } \sqrt{73} = 8.54 \text{ (From tables)} \\[1em] \Rightarrow x = \dfrac{5 + 8.54}{8} \text { or } \dfrac{5 - 8.54}{8} \\[1em] \Rightarrow x = \dfrac{13.54}{8} \text{ or } \dfrac{-3.54}{8} \\[1em] \Rightarrow x = 1.69 \text{ or } -0.44$

Hence roots of the given equations are 1.69, -0.44.

(ii) For a quadratic equation in the form :

ax² + bx + c = 0

The solutions are :

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

Comparing equation x² - 7x + 3 = 0, with ax² + bx + c = 0, we get :

a = 1, b = -7 and c = 3

$\Rightarrow x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 1 \times 3}}{2 \times 1} \\[1em] = \dfrac{7 \pm \sqrt{49 - 12}}{2} \\[1em] = \dfrac{7 \pm \sqrt{37}}{2} \\[1em] = \dfrac{7 \pm 6.08}{2} \\[1em] = \dfrac{13.08}{2} \text{ or } \dfrac{0.92}{2}\\[1em] = 6.54 \text{ or } 0.46$

Hence, the value of x = 6.54 or 0.46

Solve the following quadratic equations and give your answer correct to two significant figures :

(i) x² - 4x - 8 = 0

(ii) $x -\dfrac{18}{x} = 6$

Answer

(i) The given equation is x² - 4x - 8 = 0

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -4 , c = -8

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-4) ± \sqrt{(-4)^2 - 4\times 1 \times -8}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{16 + 32}}{2} \\[1em] \Rightarrow x = \dfrac{4 ± \sqrt{48}}{2} \\[1em] \Rightarrow x = \dfrac{4 + \sqrt{48}}{2} \text{ or } \dfrac{4 - \sqrt{48}}{2} \\[1em] \text{ Also } \sqrt{48} = 6.928 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{4 + 6.928}{2} \text { or } \dfrac{4 - 6.928}{2} \\[1em] \Rightarrow x = \dfrac{10.928}{2} \text{ or } \dfrac{-2.928}{2} \\[1em] \Rightarrow x = 5.464 \text{ or } -1.464 \\[1em] x = 5.5 \text{ or } -1.5 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 5.5 , -1.5.

(ii) Given,

$\Rightarrow x - \dfrac{18}{x} = 6 \\[1em] \Rightarrow \dfrac{x^2 - 18}{x} = 6 \\[1em] \Rightarrow x^2 - 18 = 6x \\[1em] \Rightarrow x^2 - 6x - 18 = 0 \\[1em]$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = -6 , c = -18

By using formula, $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times -18}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 + 72}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{108}}{2} \\[1em] \Rightarrow x = \dfrac{6 + \sqrt{108}}{2} \text{ or } \dfrac{6 - \sqrt{108}}{2} \\[1em] \text{ Also } \sqrt{108} = 10.392 (\text{From tables}) \\[1em] \Rightarrow x = \dfrac{6 + 10.392}{2} \text { or } \dfrac{6 - 10.392}{2} \\[1em] \Rightarrow x = \dfrac{16.392}{2} \text{ or } \dfrac{-4.392}{2} \\[1em] \Rightarrow x = 8.196 \text{ or } -2.195 \\[1em] \Rightarrow x = 8.2 \text{ or } -2.2 \text{ (correct to two significant figures) }$

Hence roots of the given equations are 8.2 , -2.2.

Solve the equation 2x² - 10x + 5 = 0 and give your answer correct to 3 significant figures.

Answer

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

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

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2} \\[1em] = \dfrac{10 \pm \sqrt{100 - 40}}{4} \\[1em] = \dfrac{10 \pm \sqrt{60}}{4} \\[1em] = \dfrac{10 \pm 2\sqrt{15}}{4} \\[1em] = \dfrac{2(5 \pm \sqrt{15})}{4} \\[1em] = \dfrac{5 \pm \sqrt{15}}{2} \\[1em] = \dfrac{5 \pm 3.87}{2} \\[1em] = \dfrac{5 + 3.87}{2}, \dfrac{5 - 3.87}{2} \\[1em] = \dfrac{8.87}{2}, \dfrac{1.13}{2} \\[1em] = 4.44, 0.565$

Hence, x = 4.44, 0.565