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 *​

Answer 1

Step-by-step explanation:

$x^{2} + (p-3)x + p = 0\\A=1 , B=(p-3) , C=p ,D=0 \\\\B^{2} - 4ac =0\\(p-3)^{2} -4(1) (p)=0\\p^2 +9-6p-4p=0\\p^2-10p+9=0\\p^2-9p-p+9=0\\p(p-9) -1(p-9)=0\\p=9 or p=1$

Answer 2

Answer:

• p = 9 or 1

Step-by-step explanation:

Given

• x² + (p - 3)x + p = 0

To find

• Value of p which has real and equal roots.

Solution

x² + (p - 3)x + p = 0

Comparing to ax² + bx + c = 0

We get:

• a = 1
• b = p - 3
• c = p

∴ Discriminant (D) = b² - 4ac

• (p - 3)² - 4 × 1 × p
• (p - 3)² - 4p

For real and equal roots, D = 0

So:

• (p - 3)² - 4p = 0
• p² - 10p + 9 = 0
• p² - 9p - p + 9 = 0
• p(p - 9) - (p - 9) = 0
• (p - 9) (p - 1) = 0
• p = 9 or p = 1

∴ p = 9 or 1