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What is the value of P for which the quadratic equation (P + 1)x^2 – 6

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Bunuel
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What is the value of P for which the quadratic equation (P + 1)x^2 – 6 [#permalink] New post   01 Feb 2021, 08:45
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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
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stne
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Re: What is the value of P for which the quadratic equation (P + 1)x^2 – 6 [#permalink] New post   09 Feb 2021, 10:56
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Bunuel wrote:
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. 1
D. 3
E. 5


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.
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Re: What is the value of P for which the quadratic equation (P + 1)x^2 – 6 [#permalink] New post   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.
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Re: What is the value of P for which the quadratic equation (P + 1)x^2 – 6 [#permalink] New post   09 Feb 2021, 11:00
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stne wrote:
Bunuel wrote:
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. 1
D. 3
E. 5


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.

__________________
Fixed. Thank you.
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Re: What is the value of P for which the quadratic equation (P + 1)x^2 – 6 [#permalink]
09 Feb 2021, 11:00
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