Nature of The Roots of The Quadratic Equation $(a-1)x^2+(4a-2)x+4a+1=0$

Question

For which values of the real parameter $a$ are the roots of the quadratic equation: $$ (a-1)x^2+(4a-2)x+4a+1=0 $$ a) Real

b) Positive

Answer 1

$(a−1)x^2+(4a−2)x+4a+1=0$ = 0

Given $ax^2 + bx + c = 0$

if $b^2 - 4ac < 0$ the roots are complex.

if $b^2 - 4ac = 0$ the roots are real and equal

if $b^2 - 4ac > 0$ the roots are real and distinct.

For real roots, Equating with the quadratic formula and substituting into if $b^2 - 4ac $ We have $(4a-2)^2-4(a-2)(4a+1) \geq 0$ which gives $-4a+8 \geq 0$ , which means $a \leq 2$

For positive roots you must solve $(-(4a-2) + \sqrt{-4a+8} ) / (2a-2)> 0 $ and $(-(4a-2) - \sqrt{-4a+8} )/ (2a-2) > 0 $ and the answer would be the intervals that overlap.

Answer 2

The discriminant is $$\Delta=(4a-2)^2-4(a-1)(4a+1)$$

a) For which values of $a$ is $\Delta$ non-negative?

b) For which values of $a$ is $\displaystyle \frac{2-4a-\sqrt{\Delta}}{a-1}>0$?