Question

Find the value of a for which roots of the equation $(a-3)x^2-2ax+5a=0$ are positive.

Answer 1

$(a-3)x^2-2ax+5a=0;FirstD\ge 0\Rightarrow (2a{)}^2-4(5a\left)\right(a-3)\ge 0\Rightarrow a^2-5a^2+15a\ge 0\Rightarrow -4a^2+15a\ge 0\Rightarrow a(4a-15)\le 0[Multiplybothsidesby-1]$

$\Rightarrow aϵ[0,\frac{15}{4}]-\left(A\right)$
Now consider 2 cases, (a) or (b)
(a)

$f\left(0\right)>0\Rightarrow 0-0+5a>0\Rightarrow a>0\left(i\right)$
Coefficient of $x^2$ less than 0
$\Rightarrow a-3<0\Rightarrow a<3\left(ii\right)$
From (i) and (ii), $aϵ(0,3)$ (B)

(b)

$f\left(0\right)>0\Rightarrow 0-0+5a>0$
$\Rightarrow a>0\left(iii\right)$
Coefficient of $x^2$ greater than 0
$\Rightarrow a-3>0\Rightarrow a>3\left(iii\right)$
From (iii) and (iv), $aϵ(3,\infty )$ (C)
Hence, solution is A and B or C i.e.,
$(0,3)\cup (3,\frac{15}{4})$