3. In the following, determine the set of values of k for which the given quadratic equation has real roots:
3. In the following, determine the set of values of k for which the given quadratic equation has real roots:

Quadratic is that type of problem which deals with a variable multiplied by itself – an operation known also as squaring.

(iii) 2{{x}^{2}}-5x-k=0

Solution:

Given,

2{{x}^{2}}-5x-k=0

It’s of the form of a{{x}^{2}}+bx+c=0

Where, a=2,b=-5,c=-k

For the given quadratic equation to have real roots D={{b}^{2}}-4ac\ge 0

D={{\left( -5 \right)}^{2}}-4\left( 2 \right)\left( -k \right)\ge 0

25+8k\ge 0

k\ge -25/8

The value of k should be lesser than -25/8 to have real roots.

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

Solution:

Given,

k{{x}^{2}}+6x+1=0

It’s of the form of a{{x}^{2}}+bx+c=0

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

For the given quadratic equation to have real roots D={{b}^{2}}-4ac\ge 0

D={{6}^{2}}-4\left( k \right)\left( 1 \right)\ge 0

36-4k\ge 0

The given equation will have real roots if,

36-4k\ge 0

36\ge 4k

36/4\ge k

9\ge k

⇒ so, k\le 9

The value of k should not exceed 9 to have real roots.