i am stuck on this problem! help!
Write a quadratic equation that has two solution, 4 and -4. Leave the polynomial in reduced factored form.
If you know how to solve a quadratic by factoring, then you know how to do this.
Just reverse your steps
if x=4
then x-4 is a factor
if x = -4
then x+4 is the second factor
so (x-4)(x+4 = 0
expanded: x^2 - 16 = 0
To write a quadratic equation that has two solutions, 4 and -4, we know that the equation can be factored into two binomial expressions. We also know that the solutions of a quadratic equation in the form of ax^2 + bx + c = 0 can be found using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Since 4 and -4 are the solutions, we can set up two equations using the quadratic formula:
For x = 4:
4 = (-b + √(b^2 - 4ac)) / (2a)
For x = -4:
-4 = (-b - √(b^2 - 4ac)) / (2a)
Let's simplify these equations further. Since we have two equations with unknowns a, b, and c, we can solve them simultaneously using an algebraic method or substitution.
First, let's solve the equation for x = 4:
4 = (-b + √(b^2 - 4ac)) / (2a)
Cross-multiplying, we have:
4 * 2a = -b + √(b^2 - 4ac)
8a = -b + √(b^2 - 4ac)
Rearranging and squaring both sides, we get:
64a^2 = b^2 - 4ac
Now, let's solve the equation for x = -4:
-4 = (-b - √(b^2 - 4ac)) / (2a)
Cross-multiplying, we have:
-4 * 2a = -b - √(b^2 - 4ac)
-8a = -b - √(b^2 - 4ac)
Rearranging and squaring both sides, we get:
64a^2 = b^2 - 4ac
Notice that the right-hand side of both equations is the same, which means that b^2 - 4ac = 64a^2.
Now, let's proceed with factoring the polynomial:
b^2 - 4ac = 64a^2
b^2 - 64a^2 = 4ac
We can rewrite this as:
(b - 8a)(b + 8a) = 4ac
Now we have a factored form of the quadratic polynomial. To write it in reduced factored form, we can express it as a product of binomials:
(b - 8a)(b + 8a) = 4ac
Thus, the quadratic equation that has two solutions, 4 and -4, in reduced factored form is:
(a)(x - 4)(x + 4) = 0
or simply:
(x - 4)(x + 4) = 0