If-p-and-q-are-the-roots-for-the-x-2-a-1-x-a-5-2-0-The-minimum-value-of-p-2-q-2-is-

Question Number 10933 by Joel576 last updated on 02/Mar/17

If p and q are the roots for the

x^{2} - (a + 1) x + (- a - \frac{5}{2}) = 0

The minimum value of p^{2} + q^{2} is \dots

Answered by ridwan balatif last updated on 02/Mar/17

there 2 solution to solve this question

x^{2} - (a + 1) x + (- a - \frac{5}{2}) = 0

p + q = a + 1 & p \times q = - (a + \frac{5}{2})

let p^{2} + q^{2} = k, then

k = {(p + q)}^{2} - 2 p \times q

k = {(a + 1)}^{2} - 2 \times {- (a + \frac{5}{2})}

k = a^{2} + 2 a + 1 + (2 a + 5)

k = a^{2} + 4 a + 6

* first solution

k will minimum if k^{'} = 0

k^{'} = 0

2 a + 4 = 0 \to a = - 2, so

k = {(- 2)}^{2} + 4 \times (- 2) + 6 = 2

* * second solution

k = a^{2} + 4 a + 6

k = {(a + 2)}^{2} + 2

if we want k is minimum, {(a + 2)}^{2} must be zero

so k = 0 + 2 = 2

Commented by Joel576 last updated on 02/Mar/17

t h a n k y o u v e r y m u c h

Commented by Joel576 last updated on 02/Mar/17

t h a n k y o u v e r y m u c h

Answered by bahmanfeshki last updated on 02/Mar/17

p^{2} + q^{2} = S^{2} - 2 P = {(a + 1)}^{2} + 2 a + 5 = a^{2} + 4 a + 6

m i n i m u m v a l u e = \frac{4 \times 1 \times 6 - 4^{2}}{4 \times 1} = 2