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# Let p and q be the roots of the quadratic equation

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Manager
Joined: 01 Apr 2018
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Let p and q be the roots of the quadratic equation  [#permalink]

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09 Jun 2018, 05:35
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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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09 Jun 2018, 06:10
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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09 Jun 2018, 06:20
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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09 Jun 2018, 06:47
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

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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11 Jun 2018, 13:18
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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11 Jun 2018, 13:27
1
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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11 Jun 2018, 14:05
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
Senior Manager
Joined: 04 Aug 2010
Posts: 322
Schools: Dartmouth College
Let p and q be the roots of the quadratic equation  [#permalink]

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11 Jun 2018, 15:39
1
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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11 Jun 2018, 18:33
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

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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11 Jun 2018, 21:57
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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12 Jun 2018, 03:32
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

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 &nbs [#permalink] 12 Jun 2018, 03:32
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# Let p and q be the roots of the quadratic equation

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