x2-7x+6=0x2-7x-6=0x2+7x-6=0x2-5x+6=0

so usually youd need to plug in every the x"s and also solve the quadradic equation because that each of lock so prefer for 1. Lets plug it in (6)2-7(-1)+6=0 so id very first remove parenthesis and simplify 6*2 and also 7*-1 to get 12+7+6=0 then come out through 25=0 which means no solution. Is that how im claimed to be doing this? i think its not my mathmatical abilities that is stumping me end its the im not sure what the inquiry wants native me so ns dont really require you to fix it if you watch my thought procedure and call me its right, thank you for acquisition your time to answer! :D


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Which of the following equations would have actually the roots x=6, x=−1? (options in description as well as my thoughts on it)x2-7x+6=0x2-7x-6=0x2+7x-6=0x2-5x+6=0

You are right that a source of an equation renders it true. Thus, if an equation has two roots, castle both make the equation true.

You are watching: Which equation has x = –6 as the solution?

These equations have zero ~ above the appropriate side. The makes factoring easy.

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

x2 + x - 6x - 6 = 0

x2 - 5x - 6 = 0

There space other, also quicker, approaches for simple quadratic equations. Part of the goals of this difficulty is to obtain practice in choosing a method.


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A root of a polynomial role is a worth of x that provides the value of the role equal come zero.

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

So, due to the fact that 6 and also -1 are both roots, x-6 and x-(-1) = x+1 space both factors.

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


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Carsyn F.

no sir
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mark M.


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then none of the four choices is correct.
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