SOLUTION: Fill in the blanks to make a quadratic whose roots are -5 and 5. x^2 + ___ x + ___

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Question 1209216: Fill in the blanks to make a quadratic whose roots are -5 and 5.
x^2 + ___ x + ___
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.


Notice that the leading coefficient at x^2 is 1.


Apply the Vieta's theorem, which says that the sum of the roots 
of such quadratic equation is the coefficient at x with the opposite sign,
while the product of the roots is the constant terms.


It gives you the coefficient at x  of  -((-5)+5) = -0 = 0

and the constant term of (-5)*5 = -25.


Therefore, the restored equation is  x^2 + 0*x - 25 = 0,  or simply  x^2 - 25 = 0.


ANSWER.  First blank (the coefficient at x) is  0  (zero).

         Second blank (the constant term) is  -25.

Solved, with explanations.



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

If -5 and 5 are the roots of this monic polynomial, then (x+5) and (x-5) are factors

(x+5)(x-5) = x^2 - 25 due to the difference of squares rule.

Therefore the answer is:
x^2 +  0 x +  -25

--------------------------------------------------------------------------

Another approach.

Let's say the blanks are b and c for now.

We have x^2+bx+c

If p,q are the roots of this quadratic, then (x-p) and (x-q) are factors.

(x-p)(x-q) = x^2-qx-px+pq
(x-p)(x-q) = x^2-(p+q)x+pq

We have
x^2+bx+c = x^2-(p+q)x+pq

Comparing terms shows that
-(p+q)x = bx which leads to b = -(p+q)
and,
c = pq

In short,
b = -(p+q)
c = pq
Which are part of Vieta's Formulas.

In this case the roots are p = -5 and q = 5
b = -(p+q) = -(-5+5) = 0
c = p*q = -5*5 = -25

We arrive at the answer 
x^2 +  0 x +  -25


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