Solve the equation $x^{2} - 2 b x + (b^{2} - a^{2}) = 0$

Open in App

Solution

$x^{2} - 2 b x + (b^{2} - a^{2}) = 0$

$⟹ x^{2} - (b + a) x - (b - a) x + (b + a) (b - a) = 0$

$⟹ x [x - (b + a)] - (b - a) [x + (b + a)] = 0$

$⟹ [x - (b + a)] [x - (b - a)] = 0$

$∴ x = (b + a)$ or $x = (b - a)$

Suggest Corrections

Similar questions

Q. Solve the quadratic equation by factorisation method

a^{2} b^{2} x^{2} - (a^{2} + b^{2}) x + 1 = 0

Q. Solve the following quadratic equation by factorisation method:

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

Q. If A.M of the roots of

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

is equal to the of the roots of the equation

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

. They are in :

Q. Solve

x^{2} - 4 x + (a^{2} - b^{2}) = 0

Suppose $a, b ϵ R a n d a \neq 0, b \neq 0$ . Let $α, β$ be the roots of $x^{2} + a x + b = 0$ . Find the equation whose roots are $α^{2}$ , $β^{2}$ .

Join BYJU'S Learning Program

Solve the equation x2−2bx+(b2−a2)=0

x2−2bx+(b2−a2)=0⟹x2−(b+a)x−(b−a)x+(b+a)(b−a)=0⟹x[x−(b+a)]−(b−a)[x+(b+a)]=0⟹[x−(b+a)][x−(b−a)]=0∴x=(b+a) or x=(b−a)

Solve the equation $x^{2} - 2 b x + (b^{2} - a^{2}) = 0$

$x^{2} - 2 b x + (b^{2} - a^{2}) = 0$

$⟹ x^{2} - (b + a) x - (b - a) x + (b + a) (b - a) = 0$

$⟹ x [x - (b + a)] - (b - a) [x + (b + a)] = 0$

$⟹ [x - (b + a)] [x - (b - a)] = 0$

$∴ x = (b + a)$ or $x = (b - a)$