Question

If both the roots of the equation $x^2+(3-2k)x-6k=0$ belong to the interval $(-6,10)$, then the largest value of the k is.

• A. $-1$
• B. $3$
• C. $5$
• D. Not defined

Answer 1

The correct option is C $5$

Given: equation $x^2+(3-2k)x-6k=0$, its roots belong to the interval $(-6,10)$

To find the largest value of the $k$

Sol: The given equation is of the form $a{x}^2+bx+c=0$.

Hence $a=1,b=(3-2k),c=-6k$

The roots are $x=\frac{-(3-2k)\pm \sqrt{(3-2k{)}^2-4(1\left)\right(-6k)}}{2\left(1\right)}⟹x=\frac{-3+2k\pm \sqrt{9+4k^2-12k+24k}}{2}⟹x=\frac{-3+2k\pm \sqrt{4k^2+12k+9}}{2}⟹x=\frac{-3+2k\pm \sqrt{(2k+3{)}^2}}{2}⟹x=\frac{-3+2k\pm (2k+3)}{2}$

i.e., $x_1=\frac{-3+2k+2k+3}{2},x_2=\frac{-3+2k-2k-3}{2}⟹x_1=\frac{4k}{2},x_2=\frac{-6}{2}⟹x_1=2k,x_2=-3$

Hence the roots to be in the interval $(-6,10)$ the largest value of $k$ can be $2k=10⟹k=5$