#### Question

Without solving the following quadratic equation, find the value of ‘p' for which the given equation has real and equal roots:

x^{2} + (p – 3)x + p = 0.

#### Solution

x^{2} + (p – 3)x + p = 0

Comparing with ax^{2} + bx + c = 0, we have

a = 1, b = (p – 3), c = p

Since, the roots are real and equal, D = 0

`⇒ b^2 - 4ac = 0`

`=> (p - 3)^2 - 4(1)(p) = 0`

`=> p^2 + 9 - 6p - 4p = 0`

`=> p^2 - 10p + 9 = 0`

`=> (p - 1)(p - 9) = 0`

`=> p = 1 or p = 9`

Is there an error in this question or solution?

#### APPEARS IN

Solution Without Solving the Following Quadratic Equation, Find the Value of ‘P' for Which the Given Equation Has Real and Equal Roots: X2 + (P – 3)X + P = 0. Concept: Quadratic Equations.