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(52779)   (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
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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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