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01 Feb 2021, 08:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:22) correct
32%
(02:27)
wrong
based on 140
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of P for which the quadratic equation \((P + 1)x^2 – 6(P + 1)x + 3(P + 3) = 0\), where \(P ≠ - 1\) has equal roots?
A. -3
B. -2
C. 0
D. 3
E. 5
A. -3
B. -2
C. 0
D. 3
E. 5
01 Feb 2021, 10:03
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For a quadratic equation to have equal roots, its discriminant (D=b^2-4ac) should equal to zero. In our equation it must be true that:
36(P+1)^2-4(P+1)*3(P+3)=0
(P+1)(36P+36-12P-36)=0
24P(P+1)=0
We are given that P is not equal to -1. So P must be equal to 0.
If P=0, then the equation x^2-6x+9=0 has a root of 3.
But there is no zero among answer choices.
36(P+1)^2-4(P+1)*3(P+3)=0
(P+1)(36P+36-12P-36)=0
24P(P+1)=0
We are given that P is not equal to -1. So P must be equal to 0.
If P=0, then the equation x^2-6x+9=0 has a root of 3.
But there is no zero among answer choices.
09 Feb 2021, 10:56
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Bunuel
For a general quadratic eqn. such as \(ax^2 +bx +c=0\) if roots are equal then
\(b^2 -4ac=0 \)
Comparing with the eqn. in the question we notice:
\(a= (P + 1)\)
\(b=– 6(P + 1)\)
\(c=3(P + 3)\)
Using \(b^2 -4ac=0\) and substituting the values of a, b and c respectively
we finally get
\(24p^2 + 24p=0 \)
\(24p(p+1)=0 \)
so either \(p=-1\) or \(p=0 \)
Given \( p \neq\) 1 hence \(p= 0\)
Ans \(p=0 \)
Bunuel, unless I have made a mistake somewhere there should be 0 among the answer choices. Hope you will look into this. Thank you.
