Question

If the roots of the equation $(a+1)x^2-3ax+4a=0$ $(a\ne -1)$ are greater than unity, then the values of $a$ are.

• A. $[-\frac{16}{7},-1)$
• B. $[0,-1]$
• C. $[-\frac{16}{7},1]$
• D. $[0,1]$

Answer 1

Answer: (A)

$[-\frac{16}{7},-1)$

$(a+1)x^2-3ax+4a=0$

Roots greater than unity

$\therefore △\ge 0$ and $-\frac{b}{2a}>1$

$9a^2-16a(a+1)\ge 0$ and $\frac{3a}{2(a+1)}>1$

$-7a^2-16a\ge 0$ $\frac{3a}{a+1}-2>0$

$7a^2+16a\le 0$ $\frac{3a-2a-2}{a+1}>0$

$a(7a+16)\le 0$ $\frac{a-2}{a+1}>0$

$a\in [-\frac{16}{7},0)$ $a\in (-\infty ,-1)\cup (2,\infty )$

$\therefore a\in [-\frac{16}{7},-1)$