Question

Q1.find roots of ax²+ (a+b)x+b=0 by quadratic formula​

Answer 1

Step-by-step explanation:

Given:

$\begin{gathered}a {x}^{2} + (a + b)x + b = 0 \\ \end{gathered}$

To find: Roots of equation using quadratic formula.

Solution:

Quadratic formula:

$\begin{gathered}\boxed{\bold{x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a}}} \\ \end{gathered}$

Here

$\begin{gathered}a = a \\ b = a + b \\ c = b \\ \end{gathered}$

on comparison with standard quadratic equation

ax²+bx+c,a≠0

$\begin{gathered}x_{1,2} = \frac{ -(a + b) ± \sqrt{ {(a + b)}^{2} - 4ab } }{2a} \\ \\ x_{1,2} = \frac{ -(a + b) ± \sqrt{ { {a}^{2} + b }^{2} + 2ab- 4ab } }{2a} \\ \\\\ x_{1,2} = \frac{ -(a + b) ± \sqrt{ { {a}^{2} + b }^{2} - 2ab } }{2a} \\ \\x_{1,2} =\frac{ -(a + b) ± \sqrt{ { {(a - b)}^{2}}}}{2a} \\ \\ x_{1,2} =\frac{ -(a + b) ± a - b}{2a} \end{gathered}$

Case 1:Take +ve sign

$\begin{gathered}x_1 = \frac{ - a - b + a - b}{2a} \\ \\ x_1 = \frac{ - 2b}{2a} \\ \\ \bold{\red{x_1 = \frac{ - b}{a}}} \\ \\ \end{gathered}$

Case 2: Take (-) sign

$\begin{gathered}x_2= \frac{ - a - b - a + b}{2a} \\ \\ x_2 = \frac{ - 2a}{2a} \\ \\ \bold{\green{x_2 = - 1}} \\ \\ \end{gathered}$

Final answer:

$</p><p>\begin{gathered}\bold{\red{x = \frac{ - b}{a} }} \\ \\\bold{\green{ x = - 1}} \\ \end{gathered}$

Hope it helps you.

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