Question

if the roots of a^2 x^2 - ab x + c = 0 are consecutive integers then b^2 - a^2​

Answer 1

Solution:

Let m, m+1 are two roots of the equation,

Compare given Quadratic equation a²x²-abx+c=0 with

Ax²+Bx+C = 0, we get

A = a² , B = -ab, C = c

i) Sum of the roots = -B/A

=> m+m+1 = -(-ab)/a²

=> 2m = b/a-1

=> m = (b-a)/2a ---(1)

ii) Product of the roots = C/A

=> m(m+1) = c/a²

=> [(b-a)/2a][(b-a)/2a+1]=c/a²

=> [(b-a)/2a][(b-a+2a)/2a]=c/a²

=> [(b-a)(b+a)]/4a² = c/a²

=> (b²-a²)/(4a²) = c/a²

=> b²-a² = [c(4a²)]/a²

=> b²-a² = 4c

••••

Answer 2

let α, β are the roots of the equation

Here, the roots are consecutive so, the roots will be taken as α, α+1

according to the equation a²x²- (ab)x + c =0

a = a², b = -ab, c = c

$\alpha + \beta = \frac{b}{a} \\ \alpha = \frac{b - a}{2a} \: \: (eq \: 1)$

$\alpha \beta = \frac{c}{a {}^{2} } \\ \alpha {}^{2} + \alpha = \frac{c}{a {}^{2} } \: \: (eq \: 2)$

from equation (1) and (2)

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

$\frac{b {}^{2} + a {}^{2} - 2ba + 2ab - 2a {}^{2} }{4a {}^{2} } = \frac{c}{a {}^{2} } \\ \frac{b {}^{2} - a {}^{2} }{4a {}^{2} } = \frac{c}{a {}^{2} } \\ b {}^{2} - a {}^{2} = 4c$

hope it's help ful to all