How do you factor #x^2+14x=-49#?

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Answer 1 · 2018-04-07T21:39:32

#(x+7)^2=0#

#x^2+14x=-49|+49#
#x^2+14x+49=0#

#"Complete the perfect square"#

#(x+14/2)^2+49-49=0#
#(x+7)^2=0#

Answer 2 · 2018-04-07T21:44:10

#(x+7)^2#

Here's how I factored it:

#x^2 + 14x = -49#

To factor this, we first have to make one side equal to zero. We do this by adding #49# to both sides of the equation:
#x^2 + 14x + 49#

This is a perfect square trinomial:
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We know that:

Therefore, the expression #x^2 + 14x + 49# is equivalent to:
#(x+7)^2#

#----------------#

To check our answer, let's see if #(x+7)^2# equals to the original expression:
#(x+7)(x+7)#
#x * x = x^2#

#x * 7 = 7x#

#7 * x = 7x#

#7 * 7 = 49#

And when we combine this we get:
#x^2 + 7x + 7x + 49#

We can still combine like terms (#7x + 7x#):
#x^2 + 14x + 49#

Therefore, we know that #(x+7)^2# is correct.

Hope this helps!