find the values p for which 9x^2+px +1=0 has equal roots.
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Answered by
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Answer:
b^2-4ac=0
(px)^2-4.9.1=0
(px)^2-36=0
(px)^2=36
px=√36
px=6
hope it is helpful to you.
Answered by
1
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.
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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