Question

For what value of p the equation
9x^2–3 (p-1)x+(p-1)=o has real and equal roots.​

Answer 1

Step-by-step explanation:

Given quadratic equation is 9x^2 - 3 (p-1)x + (p-1) = 0

From equation we get, a = 9

b = - 3(p-1)

c = (p-1)

Since, the roots are real and equal.

Discriminant = 0

${b}^{2} - 4ac = 0$

$= > ( { - 3(p - 1))}^{2} - 4 \times 9 \times (p - 1) = 0$

$= > 9( {p }^{2} - 2p + 1) - 36p + 36 = 0$

$= > 9 {p}^{2} - 18p + 9 - 36p + 36 =0$

$= > 9 {p}^{2} - 54p + 45 = 0$

$= > 9( {p}^{2} - 6 p+ 5) = 0$

$= > {p}^{2} - 6p + 5 = 0$

$= > {p}^{2} - ( 5 + 1) p + 5 = 0$

$= > {p}^{2} - 5p - p + 5 = 0$

$= > p(p - 5) - 1(p - 5) = 0$

$= > (p - 5)(p - 1) = 0$

Either, p-5 =0 , =>p= 5

Or, p-1 =0 , => p = 1

The two values of p are 5 and 1.