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Home » 10. Find the values of p for which the quadratic equation has equal roots. Also, find the roots.
10. Find the values of p for which the quadratic equation \left( 2p+1 \right){{x}^{2}}-\left( 7p+2 \right)x+\left( 7p-3 \right)=0 has equal roots. Also, find the roots.
Class 10 | ICSE | Maths | Quadratic Equations | RD Sharma
10. Find the values of p for which the quadratic equation \left( 2p+1 \right){{x}^{2}}-\left( 7p+2 \right)x+\left( 7p-3 \right)=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( 2p+1 \right){{x}^{2}}-\left( 7p+2 \right)x+\left( 7p-3 \right)=0 is in the form of a{{x}^{2}}+bx+c=0

Where a=\left( 2p+1 \right),b=-\left( 7p+2 \right),c=\left( 7p-3 \right)

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

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

⇒ {{\left( -\left( 7p+2 \right) \right)}^{2}}-4\left( 2p+1 \right)\left( 7p-3 \right)=0

⇒ {{\left( 7p+2 \right)}^{2}}-4\left( 14{{p}^{2}}+p-3 \right)=0

⇒ 49{{p}^{2}}+28p+4-56{{p}^{2}}-4p+12=0

⇒ -7{{p}^{2}}+24p+16=0

Solving for p by factorization,

⇒ -7{{p}^{2}}+28p-4p+16=0

⇒ -7p\left( p-4 \right)-4\left( p-4 \right)=0

⇒ \left( p-4 \right)\left( -7p-4 \right)=0

Either p-4=0 ⇒ p=4 Or, 7p+4=0 ⇒ p=-4/7

So, the value of k can either be 4 or -4/7

Now, using k=4 in the given quadratic equation we get

\left( 2\left( 4 \right)+1 \right){{x}^{2}}-\left( 7\left( 4 \right)+2 \right)x+\left( 7\left( 4 \right)-3 \right)=0

9{{x}^{2}}-30x+25=0

⇒ {{\left( 3x-5 \right)}^{2}}=0

Thus, x=5/3 is the root of the given quadratic equation.

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

\left( 2\left( -4/7 \right)+1 \right){{x}^{2}}-\left( 7\left( -4/7 \right)+2 \right)x+\left( 7\left( -4/7 \right)-3 \right)=0

{{x}^{2}}-14x+49=0

⇒ {{\left( x-7 \right)}^{2}}=0

Thus, x-7=0 ⇒ x=7 is the root of the given quadratic equation.

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