+885 If \((w+13)^2=(3w+7)(2w+4)\) , find \(w^2\) . Express your answer as a decimal.
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
+9479 (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
+199 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 \(\frac{141}{5}=\boxed{28.2}\) .
