How do you solve #x²+6x+2=9#?

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Answer 1 · 2018-06-11T16:55:38

#x = 1 or x = -7#

#x^2 + 6x + 2 = 9#

Collecting like terms;

#x^2 + 6x +2 - 9 = 0#

#x^2 + 6x - 7 = 0#

Using Factorization Method..

Factors: #1 and 7#

#6x = 7x - 1x#

#-7 = 7 xx -1#

Therefore;

#x^2 - x + 7x - 7 = 0#

Grouping;

#(x^2 -x)+(7x - 7) = 0#

Factorizing;

#x(x - 1) +7(x - 1) = 0#

#(x - 1) (x + 7) = 0#

#x - 1 or x + 7 = 0#

#x = 1 or x = -7#

Answer 2 · 2018-06-11T16:59:30

#x=1# and #x=-7#

We're dealing with a quadratic, so we want to set it equal to zero to find its zeroes. We now have

#x^2+6x-7=0#

To factor this business, let's do a little thought experiment:

What two numbers sum up to #6# and have a product of #-7#? After some trial and error, we arrive at #-1# and #7#. Now, we can factor this as

#(x-1)(x+7)=0#

Setting both factors equal to zero, we get

#x=1# and #x=-7#

Hope this helps!