I don't understand how to do this, I would appreciate if someone could help explain.
Two of the factors of ax^2+bx+c are (2x-2) and (x+5). Which is one of x-intercepts of y=ax^2+bx+c?
Choices:
-2
-1
5
1
x=1, and x=-5, so in the choices, only x=1 is there
Thank you! :) How would you do this problem though?
(2x-2)(x+5)=0 so either
2x-2 is zero, or
x+5 is zero.
so set each to zero, and solve for x.
To find the x-intercepts of a quadratic equation, you need to set the equation equal to zero and solve for x. In this case, the given quadratic equation is ax^2 + bx + c.
First, let's find the value of a, b, and c based on the given factors (2x-2) and (x+5).
We can do this by expanding the factors:
(2x-2)(x+5) = 2x^2 + 10x - 2x - 10 = 2x^2 + 8x - 10
Comparing this to the given equation ax^2 + bx + c, we can match coefficients to find the values of a, b, and c:
a = 2, b = 8, c = -10
Next, we will set the equation equal to zero:
ax^2 + bx + c = 0
2x^2 + 8x - 10 = 0
To solve this quadratic equation, we can use the factoring method, quadratic formula, or completing the square method. Let's use the factoring method.
Factor the quadratic equation by finding two numbers that multiply to the product of the coefficient of x^2 (a) and the constant term (c), which is 2*(-10) = -20, and also add up to the coefficient of x (b), which is 8.
The two numbers are 10 and -2:
2x^2 + 10x - 2x - 10 = 0
(2x^2 + 10x) + (-2x - 10) = 0
2x(x + 5) - 2(x + 5) = 0
(2x - 2)(x + 5) = 0
Now, we have factored the quadratic equation into two binomial factors:
(2x - 2)(x + 5) = 0
To find the x-intercepts, we set each factor equal to zero and solve for x:
Setting 2x - 2 = 0:
2x = 2
x = 1
Setting x + 5 = 0:
x = -5
So, the possible x-intercepts of the equation ax^2 + bx + c are x = 1 and x = -5.
However, neither of these choices -2, -1, or 5 is equal to either of the possible x-intercepts. Therefore, none of the given choices is one of the x-intercepts of the equation.