How do you find the x intercepts of # 36x^2 + 84x + 49 = 0#?

2 Answers
Alan P.
Aug 2, 2017

#y=36x^2+84x+49# has a single #x#-intercept at #x=-7/6#

Explanation:

Technically an equation in one variable does not have intercepts; it has a (or multiple) solution(s).

#36x^2+84x+49#
can be factored as #(6x+7)^2 or (6x+7)*(6x+7)#

So if #36x^2+84x+49=0#
then #(6x+7)*(6x+7)=0#
which implies #(6x+7)=0#
#color(white)("xxxxxxxx")x=-7/6#

smendyka
Aug 2, 2017

See a solution process below:

Explanation:

We can factor the left side of the equation as:

#(6x + 7)^2 = 0#

Or

#(6x + 7)(6x + 7) = 0#

Because both terms on the left are the same there will be only one #x# intercept. We can solve one of the terms for #0#:

#6x + 7 = 0#

#6x + 7 - color(red)(7) = 0 - color(red)(7)#

#6x + 0 = -7#

#6x = -7#

#(6x)/color(red)(6) = -7/color(red)(6)#

#(color(red)(cancel(color(black)(6)))x)/cancel(color(red)(6)) = -7/color(red)(6)#

#x = -7/6#

As you can see from the graph the parabola touches the x-axis at just one point: #-7/6# or #(-7/6, 0)#

graph{(y - 36x^2 - 84x - 49)((x+7/6)^2+(y)^2-0.0005) = 0 [-2, 0, -0.5, 0.51]}

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