Which of the following equations would have the roots x=6, x=−1? (options in description as well as my thoughts on it)

Question

1. x²-7x+6=0
2. x²-7x-6=0
3. x²+7x-6=0
4. x²-5x+6=0

so basically youd have to plug in all the x's and solve the quadradic equation for each of them so like for 1. lets plug it in (6)²-7(-1)+6=0 so id first remove parenthesis and simplify 6*2 and 7*-1 to get 12+7+6=0 then come out with 25=0 which means no solution. Is that how im supposed to be doing this? I think its not my mathmatical abilities that is stumping me over its that im not sure what the question wants from me so i dont really need you to solve it if you see my thought process and tell me its right, thank you for taking your time to answer! :D

Answer 1

Which of the following equations would have the roots x=6, x=−1? (options in description as well as my thoughts on it)

1. x²-7x+6=0
2. x²-7x-6=0
3. x²+7x-6=0
4. x²-5x+6=0

You are right that a root of an equation makes it true. Thus, if an equation has two roots, they both make the equation true.

These equations have zero on the right side. That makes factoring easy.

The equation (x-6)(x+1) = 0 has roots x=6 and x=-1. Expanding that (use F-O-I-L) gives:

x² + x - 6x - 6 = 0

x² - 5x - 6 = 0

There are other, even quicker, methods for easy quadratic equations. Part of the goals of this problem is to get practice in choosing a method.

Answer 2

A root of a polynomial function is a value of x that makes the value of the function equal to zero.

By the Factor Theorem, if c is a root then x - c is a factor.

So, since 6 and -1 are both roots, x-6 and x-(-1) = x+1 are both factors.

(x - 6)(x + 1) = x² - 5x - 6 = 0 (Is there a typo in choice 4??)