Question

if the quadratic equation (a-1)x^2-(a+1)x+a+1 =0 has real and equal roots find the value of a ?

Answer 1

D=b^2-4ac.

(a+1)^2-4(a-1)(a+1)=0

D=0 because it has real and equal roots..

a^2+1+2a-4a^2+4 =0

-3a^2+5+2a=0

Using factorisation method.

(3a-5)(a+1)

a=-1 and a=5\3

Answer 2

$HEYA \: \\ \\ for \: equal \: and \: real \: roots \: descrimnant \: \\ must \: be \: = 0 \\ \\ descrimnant \: of \: given \: quadratic \: \\ equation \: is \: \\ \\ d = (- (a + 1)) {}^{2} - 4(a - 1) \times (a + 1) \\ \\ d = a {}^{2} + 1 + 2a - 4a {}^{2} + 4 \\ \\ d = 0 \: \: \: for \: equal \: roots \\ \\ - 3a {}^{2} + 2a + 5 = 0 \\ \\ 3a {}^{2} - 2a - 5 = 0 \\ \\ 3a {}^{2} - 5a + 3a - 5 = 0 \\ \\ a(3a - 5) + 1(3a - 5) = 0 \\ \\ (a + 1) = 0 \: \: \: \: or \: \: \: \: (3a - 5) = 0 \\ \\ a = - 1 \: \: \: or \: \: \: a = 5 \div 3$