Find a quadratic equation that has the numbers below as solutions.

Question

1. 1/2,-10
2. 2/3,-5/7

How do I even do these problems? Those solutions are so hard to find.

Answer 1

if a polynomial is factored and set to zero, as in

(x-a)(x-b)(x-c) = 0
Thane any one of the factors can be zero. In fact, at least one of the factors must be zero.

So, f(x) = 0 when x = a or b or c

So, your quadratics can be factored as

(x - 1/2)(x-(-10) = 0
(x - 1/2)(x+10) = 0
You can clear the fraction by multiplying both sides of the equation by 2, to get

(2x-1)(x+10) = 0

Now just expand that if you want it in the usual quadratic form.

Do the other similarly. What do you get?

Answer 2

2/3,-5/7

(x-2/3)(x+5/7) = 0

x^2 + 5/7x - 2/3x - 10/21

x^2 + 1/21x - 10/21

Is this right?

Answer 3

ok, but I'd clear the fractions to make it look nicer:

21x^2 + x - 10 = 0

Answer 4

Thank you so much! I never spotted that I could clear the fractions like that :).