Question

Find the values of k for which the given quadratic equation has real and distinct roots:

(i) $k{x}^2+6x+1=0$

(ii) $x^2-kx+9=0$

(iii) $9x^2+3kx+4=0$

(iv) $5x^2-kx+1=0$

Answer 1

1)
K$x^2+6x+1=0$

The given equation is K$x^2+6x+1=0$

Here, a= k, b= 6, c= 1

The discriminant (D) =$b^2–4ac\ge 0$

$\to 36-4k\ge 0$

$\to 4k\le 36$

$\to k\le 9$

The value of $k\le 9$

for which the quadratic equation is having real and equal roots.

2) $x^2-kx+9=0$

The given equation is $x^2-kx+9=0$

Here, a= 1, b= -k, c= 9

Given that the equation is having real and distinct roots.

Hence, the discriminant(D) =$b^2–4ac\ge 0$

$\to k^2-4\left(1\right)\left(9\right)\ge 0$

$k^2-36\ge 0$

$\to k\ge 6$ and k $\le 6$

The value of k lies between -6 and 6 respectively to have the real and distinct roots.

(iii) Given equation is $9x^2+3kx+4=0$

a=9, b = 3k, c=4

the discriminant(D) =$b^2–4ac\ge 0$

$\to (3k{)}^2–4\times 9\times 4\ge 0$

$\to 9k^2–144\ge 0$

$\to k^2–16\ge 0$

For values $k\ge 4$ and $k\le -4$ the equation will have real and distinct roots

(iv) Given equation is $5x^2-kx+1=0$

a=5, b = -k, c=1

the discriminant(D) =$b^2–4ac\ge 0$

$\to (-k{)}^2–4\times 5\times 1\ge 0$

$\to k^2–20\ge 0$

$\to k^2\ge 20$

For values $k\ge 2\sqrt{5}$ and $k\le -2\sqrt{5}$ the equation will have real and distinct roots