Find the Value of P for Which the Quadratic Equation `(2p+1)X^2-(7p+2)X+(7p-3)=0` Has Real and Equal Roots. - Mathematics

Find the value of p for which the quadratic equation $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ has real and equal roots.

The given equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$

This is of the form $a x^{2} + b x + c = 0$ where a= $2 p + 1, b = - (7 p - 2) and c = 7 p - 3$

∴ $D = b^{2} - 4 a c$

= $- {[- 7 p + 2]}^{2} - 4 \times (2 p + 1) \times (7 p - 3)$

= $(49 p^{2} + 28 p + 4) - 4 (14 p^{2} + p - 3)$

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

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

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

∴ $- 7 p^{2} + 24 p + 16 = 0$

⇒ $7 p^{2} - 24 p - 16 = 0$

⇒ $7 p^{2} - 28 p + 4 p - 16 = 0$

⇒ $7 p (p - 4) + 4 (p - 4) = 0$

⇒ $(p - 4) (7 p + 4) = 0$

⇒ $p - 4 = 0 or 7 p + 4 = 0$

⇒ $p = 4 or p = - \frac{4}{7}$

Hence, 4 and $- \frac{4}{7}$ are the required values of p.

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Relationship Between Discriminant and Nature of Roots

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Chapter 10: Quadratic Equations - Exercises 4

Chapter 10 Quadratic Equations
Exercises 4 | Q 9

Find the Value of P for Which the Quadratic Equation $(2 p + 1) X^{2} - (7 p + 2) X + (7 p - 3) = 0$ Has Real and Equal Roots. - Mathematics