Find the values of p for which the equation (p - 12) x² + 2 (p-12) x + 2 = 0 has two equal roots.
Answers
Step-by-step explanation:
Given:-
The quadratic equation (p - 12) x² + 2 (p - 12) x + 2 = 0 has two equal roots.
To find:-
Find the values of p ?
Solution:-
Given quadratic equation is (p - 12) x² + 2 (p - 12) x + 2 = 0
On comparing with the standard quadratic
equation ax² + bx + c = 0
We have,
a = (p - 12)
b = 2 (p - 12)
c = 2
Since it has two equal roots then it's discriminant must be zero.
=> b² - 4ac = 0
=> [ 2 (p - 12) ]² - 4 (p - 12)(2) = 0
=> 4(p - 12)² - 8 (p - 12) = 0
=> 4[(p - 12)² - 2(p - 12)] = 0
=> (p - 12)² - 2(p - 12) = 0/4
=> (p - 12)² - 2(p - 12) = 0
=>(p - 12)[(p - 12) - 2] = 0
=>(p - 12)(p - 14) = 0
=>p - 12 = 0 or p - 14 = 0
=>p = 12 or p = 14
If p = 12 then the given quadratic equation does not exist.
Therefore, P = 14
Answer:-
The value of p for the given problem = 14
Used formulae:-
If the discriminant of the quadratic equation is zero then it has two equal roots.
ax²+bx+c=0 is the quadratic equation then it's
discriminant is D=b²-4ac.
If D=0 then it has real and two equal roots.
If D>0 then it has two distinct and real roots.
If D<0 then it has no real roots.
Answer
(p−12)x
2
−2(p−12)x+2=0
For quadratic equation to has equal roots
b
2
=4ac
4(p−12)
2
=4(p−12)(2)
p−12=2
p=2+12
p=14