If roots of the equation ax 2 -bx-c-0 are a a-1 and a-1 a then prove that a-b-c 2 - b 2 -4ac

We have, #10; #10;Given that, #10; #10; ax^2+bx+c=0 #10; #10;Roots are aa 1 and a+1a #10; #10;Then, #10; #10;We know that, ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If roots of t...

Question

If roots of the equation ${a x}^{2} + b x + c = 0 a r e \frac{a}{a - 1} a n d \frac{a + 1}{a}$ then prove that $(a + b + c)^{2} - b^{2} - 4 a c$

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Solution

We have,

Given that,

$a x^{2} + b x + c = 0$

Roots are $\frac{a}{a - 1}$ and $\frac{a + 1}{a}$

Then,

We know that,

Sum of roots

$\frac{a}{a - 1} + \frac{a + 1}{a} = - \frac{b}{a}$

$\Rightarrow \frac{a^{2} + (a + 1) (a - 1)}{a (a - 1)} = \frac{- b}{a}$

$\Rightarrow \frac{a^{2} + a^{2} - 1}{a (a - 1)} = \frac{- b}{a}$

$\Rightarrow \frac{2 a^{2} - 1}{a (a - 1)} = - \frac{b}{a}$

$\Rightarrow \frac{2 a^{2} - 1}{(a - 1)} = - \frac{b}{1}$

$\Rightarrow 2 a^{2} - 1 = - b (a - 1) . . . . . . (1)$

Product of roots

$\frac{a}{a - 1} \times \frac{a + 1}{a} = \frac{a + 1}{a - 1} = \frac{c}{a}$

$\Rightarrow a (a + 1) = c (a - 1)$

$\Rightarrow (a - 1) = \frac{a (a + 1)}{c} . . . . . . (2)$

From equation (1) and (2) to and we get,

$\begin{align}

$\frac{2 a^{2} - 1}{- b} = \frac{a (a + 1)}{c}$

$\Rightarrow 2 a^{2} c - c = - a^{2} b - a b$

$\Rightarrow 2 a^{2} c + a^{2} b = c - a b$

LHS

${(a + b + c)}^{2} - b^{2} - 4 a c$

$\Rightarrow a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a - b^{2} - 4 a c$

$\Rightarrow a^{2} + c^{2} + 2 a b + 2 b c - 2 c a$

$\Rightarrow {(a - c)}^{2} + 2 c (b - a)$

Hence proved.

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Similar questions

Q. If roots of the equation

a x^{2} + b x + c = 0 a r e . \frac{α}{α - 1} . a n d . \frac{α + 1}{α}

, then prove that

(a + b + c)^{2} = b^{2} - 4 a c

Q. Prove that root of equation

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

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Q. Prove that the roots of

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

are

\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Q. Prove the root of equation

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

\Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

The roots of a quadratic equation $a x^{2} + b x + c = 0$ are given by $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , provided $b^{2} - 4 a c \geq 0$ .

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If roots of the equation ax2+bx+c=0 are aa−1and a+1a then prove that (a+b+c)2−b2−4ac

If roots of the equation ${a x}^{2} + b x + c = 0 a r e \frac{a}{a - 1} a n d \frac{a + 1}{a}$ then prove that $(a + b + c)^{2} - b^{2} - 4 a c$