Question

Find the values of $a$ for which the roots of the equation $(2a-5)x^2-2(a-1)x+3=0$ are equal.

Answer 1

If roots are equal, then $D=0$

Thereofore $b^2-4ac=0$

$(-2(a-1){)}^2-4\times (2a-5)\times 3$

$4(a^2+1-2a)-12(2a-5)=0$

$a^2+1-2a-6a+15=0$

$a^2-8a+16=0$

$(a-4)(a-4)=0$

$\Rightarrow a=4$