Question

Find the values of P for which the equation below has equal roots x^2 - (p-2)x + 2p + 1 = 0

Answer 1

Answer:

p = 0, p = 12

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 then the nature of its roots is given by the discriminant

Δ = b² - 4ac

For equal roots b² - 4ac = 0

Given

x² -(p - 2)x + 2p + 1 = 0 ← in standard form

with a = 1, b = - (p - 2), c = 2p + 1, then

[- (p - 2)]² - (4 × 1 × 2p + 1) = 0

(p - 2)² - 4(2p + 1) = 0

p² - 4p + 4 - 8p - 4 = 0

p² - 12p = 0 ← factor out p from each term

p(p - 12) = 0

Equate each factor to zero and solve for x

p = 0

p - 12 = 0 ⇒ p = 12

Answer 2

Answer:

$\huge\boxed{\sf p = 0 \ \ \ \ OR \ \ \ \ p = 12}$

Step-by-step explanation:

Given the equation:

$\sf x^2 - (p-2)x+2p+1= 0$.

Comparing it with $\sf ax^2 + bx+c = 0$, we get:

$\sf a = 1 , b = -(p-2), \ c = 2p+1$

Equal Roots mean Discriminant = 0

So,

$\sf b^2 -4ac = 0\\\\Given\ that\ a = 1 , b = -(p-2) \ and \ c = 2p+1 \\\\\ [-(p-2)\ ]^2 - 4(1)(2p+1) = 0\\\\(p-2)^2 -4(2p+1)=0\\\\p^2 - 4p+4 -8p-4 = 0 \\\\p^2 -12p = 0\\\\Taking \ p \ common\\\\p(p-12) = 0 \\\\Either, \\\\p = 0 \ \ \ \ \bold{OR} \ \ \ \ p-12 = 0\\\\p = 0 \ \ \ \ \bold{OR} \ \ \ \ p = 12$