Let p and q be the roots of the quadratic equation

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)

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!

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

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 \(p^2 + q^2 = 5\).

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

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!

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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