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Home » 2. Find the values of for which the roots are real and equal in each of the following equations:
2. Find the values of k for which the roots are real and equal in each of the following equations:
Class 10 | ICSE | Maths | Quadratic Equations | RD Sharma
2. Find the values of k for which the roots are real and equal in each of the following equations:

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

(xv) \left( 4-k \right){{x}^{2}}+\left( 2k+4 \right)x+\left( 8k+1 \right)=0

Solution:

The given equation \left( 4-k \right){{x}^{2}}+\left( 2k+4 \right)x+\left( 8k+1 \right)=0 is in the form of a{{x}^{2}}+bx+c=0

Where a=\left( 4-k \right),b=\left( 2k+4 \right),c=\left( 8k+1 \right)

For the equation to have real and equal roots, the condition is

D={{b}^{2}}-4ac=0

⇒ {{\left( 2k+4 \right)}^{2}}-4\left( 4-k \right)\left( 8k+1 \right)=0

⇒ 4{{k}^{2}}+16k+16-4\left( -8{{k}^{2}}+32k+4-k \right)=0

⇒ 4{{k}^{2}}+16k+16+32{{k}^{2}}-124k-16=0

⇒ 36{{k}^{2}}-108k=0

Taking common,

⇒ 9k\left( k-3 \right)=0

Now, either 9k=0 ⇒ k=0 or k-3=0 ⇒ k=3,

So, the value of k can either be 0 or 3.

(xvi) \left( 2k+1 \right){{x}^{2}}+2\left( k+3 \right)x+\left( k+5 \right)=0

Solution:

The given equation \left( 2k+1 \right){{x}^{2}}+2\left( k+3 \right)x+\left( k+5 \right)=0 is in the form of a{{x}^{2}}+bx+c=0

Where a=\left( 2k+1 \right),b=2\left( k+3 \right),c=\left( k+5 \right)

For the equation to have real and equal roots, the condition is

D={{b}^{2}}-4ac=0

⇒ {{\left( 2\left( k+3 \right) \right)}^{2}}-4\left( 2k+1 \right)\left( k+5 \right)=0

⇒ 4{{\left( k+3 \right)}^{2}}-4\left( 2{{k}^{2}}+11k+5 \right)=0

⇒ {{\left( k+3 \right)}^{2}}-\left( 2{{k}^{2}}+11k+5 \right)=0 [dividing by 4 both sides]

⇒ {{k}^{2}}+5k-4=0

Now, solving for k by completing the square we have

⇒ {{k}^{2}}+2\times \left( 5/2 \right)\times k+{{\left( 5/2 \right)}^{2}}=4+{{\left( 5/2 \right)}^{2}}

⇒ {{\left( k+5/2 \right)}^{2}}=4+25/4=\sqrt{41}/4

⇒ k+\left( 5/2 \right)=\pm \sqrt{41}/2

⇒ k=\left( \sqrt{41}-5 \right)/2 or -\left( \sqrt{41}+5 \right)/2

So, the value of k can either be \left( \sqrt{41}-5 \right)/2 or -\left( \sqrt{41}+5 \right)/2

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