Question

Find greatest integer value of $m$ for which the equation $\left(2m-3\right)x^2-4x+\left(2m-3\right)=0$ has both negative roots.

Answer 1

As we know that if all terms in a quadratic equation are greater than $0$ and $D>0$, then both roots of the quadratic equation will be negative.

$(2m-3)x^2-4x+(2m-3)=0$

$(3-2m)x^2+4x+(3-2m)=0$

For both roots to be negative,

(i) $3-2m>0$

$m<\frac{3}{2}.....\left(1\right)$

(ii) $D>0$

$4^2-4{(3-2m)}^2>0$

$16-36-16m^2+48m>0$

$4m^2-12m+5<0$

$(2m-5)(2m-1)<0$

$m\in (\frac{1}{2},\frac{5}{2}).....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$m\in (\frac{1}{2},\frac{3}{2})$

Hence the value of $m$ will lie in the interval $(\frac{1}{2},\frac{3}{2})$.