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Let p and q be the roots of the quadratic equation

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Let p and q be the roots of the quadratic equation [#permalink]

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New post 09 Jun 2018, 06:35
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Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12

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Let p and q be the roots of the quadratic equation [#permalink]

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New post 09 Jun 2018, 07:10
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12




\(x^{2}-(a-2)x-a-1=0\) = \(x^{2}-(a-2)x+(-a-1)=0\) equivalent to \(x^{2}-(p+q)x+pq=0\)
Hence \(p+q=a-2\) & \(pq=-a-1\)
\(p^{2} + q^{2}\)= \((p+q)^2-2pq\) = \((a-2)^2-2(-a-1)\) = \(a^2-4a+4+2a+2\) = \(a^2-2a+6\)

* When a =0 or 2 , \(p^{2} + q^{2}\)=6 but this is not the min value.

Now , \(a^2-2a+6\) is min when \(a^2-2a\) is minimum i.e, a=1
Hence \(p^{2} + q^{2}\)= 6-1=5.......................................................Hence I would go for option D.
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Let p and q be the roots of the quadratic equation [#permalink]

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New post 09 Jun 2018, 07:20
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12


Lets equate the co-efficient s of given quadratic equation with the standard form \(ax^2+bx+c=0\)

a=1,b=-(a-2),c=-(a+1)

Sum of roots, p+q=\(-\frac{b}{a}\)=-(-(a-2))=a-2

product of roots, pq=\(\frac{c}{a}\)=-(a+1)

Now, \(p^2+q^2\)=\((p+q)^2-2pq\)=\((a-2)^2+2(a+1)\)=\(a^2-2a+6\)=\((a-1)^2+5\)

Now, min(\(p^2+q^2\))=min(\((a-1)^2\))+5)=0+5=5 (Since minimum value of any square expression is zero.)

Ans. D
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Let p and q be the roots of the quadratic equation [#permalink]

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New post 09 Jun 2018, 07:47
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12


Quadratic equation \(x^{2} - (a-2)x -a-1 = 0\) can be re-written as \(x^{2} - (a-2)x -(a+1) = 0\)

For a quadratic equation \(ax^2 + bx + x\). The sum of the roots is \(\frac{-b}{a}\) and product of the roots is \(\frac{c}{a}\)

Sum of the roots(p+q) = \((a-2)\) | Product of the roots(pq) = \(-(a+1)\)

\(p^{2} + q^{2} = (p+q)^2 - 2*pq\)

Here, \(p^{2} + q^{2} = (a-2)^2 - 2(-a-1) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6\)

Since we need the minimum value, a^2 - 2a = -1 when a = 1. Therefore, \(p^{2} + q^{2} = 5\) (Option C)
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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 14:18
How do you figure out that p^2+q^2 is equal to (p+q)^2-2pq?
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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 14:27
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dmod14 wrote:
How do you figure out that p^2+q^2 is equal to (p+q)^2-2pq?


Hi dmod14

You must be aware of the property: \((p+q)^2 = p^2 + 2pq + q^2\)

Re-arranging this we get \(p^2+q^2 = (p+q)^2 - 2pq\)

Hope this helps you!
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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 15:05
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5
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Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 16:39
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1
tarunanandani wrote:
Let p and q be the roots of the quadratic equation \(x^{2}-(a-2)x-a-1=0\).
What is the minimum possible value of \(p^{2} + q^{2}\)?

A. 0
B. 3
C. 4
D. 5
E. 12

For any quadratic in the form \(x^2 + bx + c=0\):
Product of the roots = c.
Sum of the roots = -b.

Converting \(x^{2}-(a-2)x-a-1=0\) into the form \(x^2 + bx + c=0\), we get:
\(x^{2} + (-a+2)x + (-a-1)=0\).

In the quadratic above, c = -a-1 and -b = a-2.
Since the roots are p and q, we get:
Product of the roots = pq = -a-1.
Sum of the roots = p+q = a-2.

Squaring the blue equation, we get:
\((p+q)^2\) = \((a-2)^2\)
\(p^2 + q^2 + 2pq\) = \(a^2 + 4 - 4a\)

Substituting pq = -a-1 into the red equation, we get:
\(p^2 + q^2 + 2(-a-1)\) = \(a^2 + 4 - 4a\)
\(p^2 + q^2 -2a - 2\) = \(a^2 + 4 - 4a\)
\(p^2 + q^2\) = \(a^2 - 2a + 6\)

To minimize the value of \(p^2 + q^2\), we must minimize the value of \(a^2 - 2a + 6\).
Test easy values in \(a^2 - 2a + 6\):
a = 0 --> \(a^2 - 2a + 6 = 6\)
a = 1 --> \(a^2 - 2a + 6 = 5\)
a = 2 --> \(a^2 - 2a + 6 = 6\)
The resulting values on the right -- 6, 5, 6 -- indicate that \(y = a^2 - 2a + 6\) is a U-shaped parabola with a minimum value of 5.
Thus:
The minimum value of \(p^2 + q^2 = 5\).


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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 19:33
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dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


a^2-2a+6 = p^2+q^2, and you are asked to find min of p^2+q^2, so you have to find the minimum of a^2-2a+6 - parabola
The minimum of parabola is point x= -b/2a=-(-2)/2*1=1, so y will be 1^2-2*1+6=5. Hence min p^2+q^2=5
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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 11 Jun 2018, 22:57
1
dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


Hey dmod14

If you observe the solution, you will arrive at \(p^2+q^2 = a^2-2a+6\)

We have been asked to find the minimum value of the expression \(p^2+q^2\).
This will happen when the expression \(a^2-2a\) takes the minimum value.
For that to happen, we test values and find that when a = 1 , \(a^2-2a\) = -1.
Now, the value of the overall expression \(a^2-2a +6\) now becomes \(-1+6 = 5\)

Hope this helps you!
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Re: Let p and q be the roots of the quadratic equation [#permalink]

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New post 12 Jun 2018, 04:32
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dmod14 wrote:
Thanks pushpitkc! That makes a lot of sense.

Can you explain how you got the minimum value, specifically how you went from a^2-2a+6 to p^2+q^2=5


The equation \(p^{2}+q^{2}=a^{2}−2a+6\) can be re-written as \(p^{2}+q^{2}=(a^{2}−2a+1^{2})+5\).
Now the last equation can be simplified as \(p^{2}+q^{2}=(a-1)^{2}+5\) and for \(p^{2}+q^{2}\) to be minimum the term \((a-1)^{2}\) has to be minimum, i.e., 0 as square of any term can't be negative. Therefore the minimum value of \(p^{2}+q^{2}=0+5=5\).
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Re: Let p and q be the roots of the quadratic equation   [#permalink] 12 Jun 2018, 04:32
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