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# binomials

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If $$(w+13)^2=(3w+7)(2w+4)$$ , find $$w^2$$ . Express your answer as a decimal.

ant101  Dec 29, 2017
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#1
+2

Solve for w:
(w + 13)^2 = (2 w + 4) (3 w + 7)

Write the quadratic polynomial on the right-hand side in standard form.

Expand out terms of the right-hand side:
(w + 13)^2 = 6 w^2 + 26 w + 28

Move everything to the left-hand side.
Subtract 6 w^2 + 26 w + 28 from both sides:
-28 - 26 w - 6 w^2 + (w + 13)^2 = 0

Write the quadratic polynomial on the left-hand side in standard form.
Expand out terms of the left-hand side:
141 - 5 w^2 = 0

Isolate terms with w to the left-hand side.
Subtract 141 from both sides:
-5 w^2 = -141

Divide both sides by a constant to simplify the equation.
Divide both sides by -5:
w^2 =± 141/5

Guest Dec 29, 2017
#2
+6943
+4

(w + 13)2  =  (3w + 7)(2w + 4)

First multiply out the parenthesees on both sides.

(w + 13)(w + 13)   =   (3w + 7)(2w + 4)

w2 + 13w + 13w + 169   =   6w2 + 12w + 14w + 28

w2 + 26w + 169   =   6w2 + 26w + 28

Subtract  26w  from both sides.

w2 + 169  =  6w2 + 28

Subtract  w2  from both sides.

169  =  5w2 + 28

Subtract  28  from both sides.

141  =  5w2

Divide both sides by  5 .

141/5  =  w2

w2  =  28.2

hectictar  Dec 29, 2017
#3
+111
+4

We expand both sides to find

\begin{align*} (w+13)(w+13)&=(3w+7)(2w+4)\\ w^2+26w+169&=3w(2w+4)+7(2w+4)\\ w^2+26w+169&=6w^2+12w+14w+28\\ w^2+26w+169&=6w^2+26w+28\\ w^2+169&=6w^2+28\\ 141&=5w^2\\ \frac{141}{5}&=w^2.\\ \end{align*}

So, expressed as a decimal, our answer is $$\frac{141}{5}=\boxed{28.2}$$ .

azsun  Dec 29, 2017
#4
+92217
0

Hi Azsun...

I am seriously wondering why your LaTex is displaying properly when no one elses is ???

How strange...

Melody  Jan 8, 2018

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