Question

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

Answer 1

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,

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

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

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

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

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

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

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

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

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

Hence Proved.

Answer 2

Given:-

• Quadratic eq. = ax² + bx + c = 0

Show That:-

• Sum of its roots is -b/a

Solution:-

Let, α and β be zeroes of quadratic eq. ax² + bx +c = 0

Thus, by using quadratic formula:-

$\huge{ \underline{\boxed{\sf\dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}}}}$

$\implies\sf \alpha = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a}$

$\implies\sf \beta = \dfrac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

Now, sum of zeroes:-

$\sf \dashrightarrow \alpha + \beta = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} + \dfrac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

✪$Taking\:L.C.M$

$\\ \sf \dashrightarrow \alpha + \beta = \dfrac{ - b + \sqrt{ {b}^{2} - 4ac} + - b - \sqrt{ {b}^{2} - 4ac} }{2a}$

$\sf \dashrightarrow \alpha + \beta = \dfrac{ - b \cancel{ + \sqrt{ {b}^{2} - 4ac}} + - b \cancel{- \sqrt{ {b}^{2} - 4ac}} }{2a}$

$\sf \dashrightarrow \alpha + \beta = \dfrac{ - 2b }{2a}$

$\sf \dashrightarrow \alpha + \beta = \dfrac{ - \not{2}b }{ \not{2}a}$

$\sf \dashrightarrow \alpha + \beta = \dfrac{ - b }{a}$

$\small{\sf \therefore \underline{\alpha + \beta = \dfrac{ - b }{a}}}$

Hence, Sum of roots of quadratic eq. ax² + bx + c = 0 is -b/a