Question

Find the value of p for which the equation px2 - px +1=0 has equal roots

Answer 1

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: p = 4

Answer 2

Your question a needs correction.


Correct question : Find the value of p for which the equation px2 - px +1=0 has equal roots and real ( positive ) roots



Given Equation : px² - px + 1 = 0


On comparing the given equation with a²x+ bx + c = 0 we get that the value of a is p , value of b is - p and the value of c is 1 .


Discriminant = b² - 4ac

In the question discriminant will be 0 as the equation has real & equal roots.


b² - 4ac = 0

( - p )² - 4( 1 × p ) = 0

p² - 4 p = 0

p( p - 4 ) = 0



By Zero Product Rule, value of p is 0 or 4. As p is also with x² , it can't be 0 because if it is 0 , given equation can't be a quadratic equation.



Therefore, value of p is 4.

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