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9. Find the values of k for which the quadratic equation \left( 3k+1 \right){{x}^{2}}+2\left( k+1 \right)x+1=0 has equal roots. Also, find the roots.
Class 10 | ICSE | Maths | Quadratic Equations | RD Sharma
9. Find the values of k for which the quadratic equation \left( 3k+1 \right){{x}^{2}}+2\left( k+1 \right)x+1=0 has equal roots. Also, find the roots.

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

Solution:

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

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

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

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

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

⇒ {{\left( k+1 \right)}^{2}}-\left( 3k+1 \right)=0 [After dividing by 4 both sides]

⇒ {{k}^{2}}+2k+1-3k-1=0

⇒ {{k}^{2}}-k=0

⇒ k\left( k-1 \right)=0

Either k=0 Or, k-3=0 ⇒ k=1,

So, the value of k can either be 0 or 1

Now, using k = 0 in the given quadratic equation we get

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

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

⇒ {{\left( x+1 \right)}^{2}}=0

Thus, x=-1 is the root of the given quadratic equation.

Next, on using k=1 in the given quadratic equation we get

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

4{{x}^{2}}+4x+1=0

⇒ {{\left( 2x+1 \right)}^{2}}=0

Thus, 2x=-1 ⇒ x=-1/2 is the root of the given quadratic equation.

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