Let p and q be the roots of the quadratic equation

Question

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

u1983 · Answer

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 .

PKN · Answer

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

pushpitkc · Answer

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)

dmod14 · Answer

How do you figure out that p^2+q^2 is equal to (p+q)^2-2pq?

pushpitkc · Answer

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!

dmod14 · Answer

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

GMATGuruNY · Answer

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 .

D

Hero8888 · Answer

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

pushpitkc · Answer

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!

tarunanandani · Answer

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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Let p and q be the roots of the quadratic equation

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