Question

Solve the given questions by both the methods, first by method of completing the squares, and then by using quadratic formula:

a^{2} x^{2} + 4 a b x + 4 b^{2} = 0

Accepted Answer

Solution of the given quadratic equation by method of completing the squares:

We have,

$a^{2} x^{2} + 4 a b x + 4 b^{2} = 0 \Rightarrow x^{2} + \frac{4 a b}{a^{2}} x + \frac{4 b^{2}}{a^{2}} = 0 \Rightarrow x^{2} + \frac{4 b}{a} x + \frac{4 b^{2}}{a^{2}} = 0 \Rightarrow {(x)}^{2} + 2 \times \frac{2 b}{a} \times x + {(\frac{2 b}{a})}^{2} = 0 \Rightarrow {(x + \frac{2 b}{a})}^{2} = 0 \Rightarrow (x + \frac{2 b}{a}) (x + \frac{2 b}{a}) = 0 \Rightarrow x + \frac{2 b}{a} = 0 or x + \frac{2 b}{a} = 0 \Rightarrow x = - \frac{2 b}{a} o r x = - \frac{2 b}{a}$

Therefore, the required solution is $- \frac{2 b}{a}, - \frac{2 b}{a}$ .

Now, solution of the given quadratic equation by method of quadratic formula:

Comparing the equation $a^{2} x^{2} + 4 a b x + 4 b^{2} = 0$ with $A x^{2} + B x + C = 0, (A \neq 0)$ , we get,

$A = a^{2}, B = 4 a b, C = 4 b^{2} .$

Now, we know,

$x = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A}$

$\Rightarrow x = \frac{- 4 a b \pm \sqrt{{(4 a b)}^{2} - 4 \times a^{2} \times 4 b^{2}}}{2 \times a^{2}} \Rightarrow x = \frac{- 4 a b \pm \sqrt{16 a^{2} b^{2} - 16 a^{2} b^{2}}}{2 a^{2}} \Rightarrow x = \frac{- 4 a b \pm 0}{2 a^{2}} \Rightarrow x = \frac{- 4 a b + 0}{2 a^{2}} or x = \frac{- 4 a b - 0}{2 a^{2}} \Rightarrow x = \frac{- 4 a b}{2 a^{2}} or x = \frac{- 4 a b}{2 a^{2}} \Rightarrow x = - \frac{2 b}{a} or x = - \frac{2 b}{a}$

Therefore, the required solution is $- \frac{2 b}{a}, - \frac{2 b}{a} .$

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Quadratic Equations, Exercise 3: EXERCISE

H K Dass Mathematics Solutions for Exercise - H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Quadratic Equations, Exercise 3: EXERCISE

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