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# Find the values of p for which the quadratic equation ( 2 p + 1 ) x 2 − ( 7 p + 2 ) x + ( 7 p − 3 ) = 0 has equal roots. Also, find these roots. - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Factorization

#### Question

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 these roots.

#### Solution

The given quadric equation is $\left( 2p + 1 \right) x^2 - \left( 7p + 2 \right)x + \left( 7p - 3 \right) = 0$ and roots are real and equal.

Then, find the value of p.

Here,

$a = 2p + 1, b = - 7p - 2 \text { and } c = 7p - 3$
As we know that $D = b^2 - 4ac$
Putting the values of  $a = 2p + 1, b = - 7p - 2 \text { and } c = 7p - 3$.

$D = \left[ - \left( 7p + 2 \right) \right]^2 - 4\left( 2p + 1 \right)\left( 7p - 3 \right)$

$= (49 p^2 + 28p + 4) - 4\left( 14 p^2 - 6p + 7p - 3 \right)$

$= 49 p^2 + 28p + 4 - 56 p^2 - 4p + 12$

$= - 7 p^2 + 24p + 16$

The given equation will have real and equal roots, if D = 0

Thus,

$- 7 p^2 + 24p + 16 = 0$

$\Rightarrow 7 p^2 - 24p - 16 = 0$

$\Rightarrow 7 p^2 - 28p + 4p - 16 = 0$

$\Rightarrow 7p(p - 4) + 4(p - 4) = 0$

$\Rightarrow (7p + 4)(p - 4) = 0$

$\Rightarrow 7p + 4 = 0 \text { or } p - 4 = 0$

$\Rightarrow p = - \frac{4}{7} \text { or } p = 4$

Therefore, the value of p is 4 or $- \frac{4}{7}$.

Now, for p = 4, the equation becomes

$9 x^2 - 30x + 25 = 0$

$\Rightarrow 9 x^2 - 15x - 15x + 25 = 0$

$\Rightarrow 3x(3x - 5) - 5(3x - 5) = 0$

$\Rightarrow (3x - 5 )^2 = 0$

$\Rightarrow x = \frac{5}{3}, \frac{5}{3}$

for p = $- \frac{4}{7}$ the equation becomes

$\left( - \frac{8}{7} + 1 \right) x^2 - \left( - 4 + 2 \right)x + \left( - 4 - 3 \right) = 0$

$\Rightarrow \left( \frac{- 8 + 7}{7} \right) x^2 + 2x - 7 = 0$

$\Rightarrow - \frac{1}{7} x^2 + 2x - 7 = 0$

$\Rightarrow - x^2 + 14x - 49 = 0$

$\Rightarrow x^2 - 14x + 49 = 0$

$\Rightarrow x^2 - 7x - 7x + 49 = 0$

$\Rightarrow x(x - 7) - 7(x - 7) = 0$

$\Rightarrow (x - 7 )^2 = 0$

$\Rightarrow x = 7, 7$

Hence, the roots of the equation are $\frac{5}{3} \text{ and } 7$.

Is there an error in this question or solution?

#### APPEARS IN

Solution Find the values of p for which the quadratic equation ( 2 p + 1 ) x 2 − ( 7 p + 2 ) x + ( 7 p − 3 ) = 0 has equal roots. Also, find these roots. Concept: Solutions of Quadratic Equations by Factorization.
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