Hi All,
What are the solutions of the equation, (5x-1)^{2} = (2x+7)^{2} ?
My answer:8/3. Are there any other solutions?
Thanks.
There are two answers of quadratic.
Factors of the quadratic
$$(3x-8)(7x+6)=0$$
Therefore,$$\begin{equation}\begin{split}3x-8&=0 \\3x&=8 \\x&=\dfrac{8}{3} \end{split} \end{equation}$$
and$$\begin{equation}\begin{split}7x+6&=0\\7x&=-6 \\ x&=\dfrac{-6}{7} \end{split} \end{equation}$$
Hence, $x=\dfrac{8}{3},\;\dfrac{-6}{7}$
It good to know multiple ways of solving this type of equation.
There is relatively more work involved in the method in post # 2, because 1) it requires expanding the square of two binomials,
2) collecting like terms, and 3) factoring a higher degree polynomial, a quadratic, whose first and last coefficients have a
relatively large total number of ways of breaking them down into usable factors.
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$\displaystyle (5x - 1)^2 \ = \ (2x + 7)^2$
$\displaystyle 5x - 1 \ = \ \pm\sqrt{2x + 7}$
$\displaystyle 5x - 1 \ = \ -2x - 7 \ \ \ \ or \ \ \ \ 5x - 1 \ = \ 2x + 7 $
$\displaystyle 7x \ = \ -6 \ \ \ \ or \ \ \ \ 3x \ = \ 8$
$\displaystyle x \ = \ \dfrac{-6}{7} \ \ \ \ or \ \ \ \ x \ = \ \dfrac{8}{3}$