Question

Q1.Show that the sum of roots of a quadratic equation ax²+ bx+c=0(a = 0) is -b/a​

Answer 1

Step-by-step explanation:

GIVEN :-

A quadratic equation ax² + bx + c = 0.

TO SHOW :-

Sum of product of the quadratic equation id -b/a [ α + β = -b/a ]

PROOF :-

Let us considered that a polynomial ax² + bx + c = 0 has descriminant ( D = b² - 4ac ) and the roots α and β i.e,

$\begin{gathered} \\ : \implies \displaystyle \sf \: \alpha = \frac{ - b + \sqrt{b ^{2} - 4ac } }{2a} \\ \\ \\ \end{gathered}$

$\begin{gathered}: \implies \displaystyle \sf \: \beta = \frac{ - b - \sqrt{b ^{2} - 4ac} }{2a} \\ \end{gathered}$

Now according to the question we have to find the value of α + β,

$\begin{gathered} \\ : \implies \displaystyle \sf \: \alpha + \beta = \frac{ - b + \sqrt{b ^{2} - 4ac } }{2a} + \frac{ - b - \sqrt{b ^{2} - 4ac } }{2a} \\ \\ \\ \end{gathered}$

$\begin{gathered} : \implies \displaystyle \sf \: \alpha + \beta = \frac{ - b + \sqrt{b ^{2} - 4ac} + ( - b) - \sqrt{b ^{2} - 4ac } }{2a} \\ \\ \\ \end{gathered}$

$</p><p>\begin{gathered}: \implies \displaystyle \sf \: \alpha + \beta = \frac{ - b + \sqrt{b ^{2} - 4ac} - b- \sqrt{b ^{2} - 4ac } }{2a} \\ \\ \\ \end{gathered}$

$\begin{gathered}: \implies \displaystyle \sf \: \alpha + \beta = \frac{ - b - b}{2a} \\ \\ \\ \end{gathered}$

$\begin{gathered}: \implies \displaystyle \sf \: \alpha + \beta = \frac{ - 2b}{2a} \\ \\ \\ \end{gathered}$

$\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \: \alpha + \beta = \frac{ - b}{a} }} \\ \end{gathered}$

Hence Proved.