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

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

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

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?

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?

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

21x^2 + x - 10 = 0

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

To find a quadratic equation with given solutions, you can use the fact that the solutions of a quadratic equation in the form ax² + bx + c = 0 are the values of x that make the equation equal to zero.

To start, let's take a look at the first set of solutions: 1/2 and -10.

1. Solution: 1/2
If 1/2 is one of the solutions, it means that when x = 1/2, the equation is equal to zero. Let's substitute x = 1/2 into the equation to see if it satisfies the condition:
a(1/2)² + b(1/2) + c = 0
(1/4)a + (1/2)b + c = 0

2. Solution: -10
Similarly, if -10 is another solution, it means that when x = -10, the equation is equal to zero. Substituting x = -10 into the equation gives us:
a(-10)² + b(-10) + c = 0
100a - 10b + c = 0

Now we have two equations:
(1/4)a + (1/2)b + c = 0 (equation 1)
100a - 10b + c = 0 (equation 2)

To find the quadratic equation, we need to eliminate one of the variables. Let's solve equation 1 for c and substitute it into equation 2.

From equation 1: c = -(1/4)a - (1/2)b

Substituting this into equation 2:
100a - 10b + (-(1/4)a - (1/2)b) = 0
Simplifying:
(100 - 1/4)a - (10 + 1/2)b = 0
(400 - 1)a - (40 + 2)b = 0
399a - 42b = 0

So, the quadratic equation with the given solutions of 1/2 and -10 is:
399x² - 42x = 0

The same process can be applied to the second set of solutions (2/3 and -5/7) to find the quadratic equation with these solutions.