Question

The set of values of k for which the given quadratic equation has real roots
$4x^2$ - 3kx + 1 = 0 is k $\le -\frac{4}{3}$ or k $\ge \frac{4}{3}$

• A. True
• B. False

Answer 1

The correct option is A True

$\Rightarrow$ The given equation is $4x^2-3kx+1=0$, comparing it with $a{x}^2+bx+c=0$.

$\Rightarrow$ We get, $a=4,b=-3k,c=1$

$\Rightarrow$ It is given that roots are real.

$\therefore$ $b^2-4ac\ge 0$

$\Rightarrow$ $(-3k{)}^2-4(4\left)\right(1)\ge 0$

$\Rightarrow$ $9k^2-16\ge 0$

$\Rightarrow$ $(3k{)}^2-(4{)}^2\ge 0$

$\Rightarrow$ $(3k+4)(3k-4)\ge 0$

$\therefore$ $k\le \frac{-4}{3}$ or $k\ge \frac{4}{3}$

$\Rightarrow$ It can be seen that values of $k$ given in question are correct.