Question

Find the value of p for which the quadratic equation (p + 4)x2 + px + x + 1 = 0 has equal
roots.​

Answer 1

Answer:

px² - px + 1 = 0

The quadratic equation has equal roots

⇒ The discriminant is equal to zero

Find p:

b² - 4ac = 0

Sub a = p , b = -p , c = 1 into the equation:

(-p)² - 4(p)(1) = 0

Evaluate each term:

p² - 4p = 0

Take out common factor p:

p( p - 4) = 0

Apply zero product property:

p = 4 or p = 0 (rejected as it will make the equation meaningless)

Answer::::- 4...

Answer 2

Answer

• p = 5 , - 3

Given

• Quadratic equation (p+4)x² + (p+1)x + 1 = 0 has equal roots

To Find

• Value of p

Solution

For a quadratic equation ax² + bx + c = 0  has equal roots .

Then , Value of discriminant , b² - 4ac = 0 .

Compare given equation (p + 4)x² + (p+1)x + 1 with ax² + bx + c , we get ,

• a = p + 4 , b = p + 1 , c = 1

Since ,

b² - 4ac = 0

⇒ (p+1)² - 4(p+4)(1) = 0

⇒ p² + 1 + 2p - 4p - 16 = 0

⇒ p² - 2p - 15 = 0

⇒ p² - 5p + 3p - 15 = 0

⇒ p ( p - 5 ) + 3 ( p - 5 ) = 0

⇒ ( p - 5 ) ( p + 3 ) = 0

⇒ p = 5 or p = - 3