1. ## Solutions to quadratic equation

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.

2. ## Re: Solutions to quadratic equation

$(5x-1)^2 = (2x+7)^2$

$(5x-1)^2 -(2x+7)^2 = 0$

$21 x^2-38 x-48=0$

$(3x-8)(7x+6) = 0$

$x = \dfrac 8 3,~-\dfrac 6 7$

3. ## Re: Solutions to quadratic equation

Originally Posted by allo95

What are the solutions of the equation, (5x-1)2 = (2x+7)2 ?
$(5x-1)^2 - (2x+7)^2 = 0$

$[(5x-1)-(2x+7)] \cdot [(5x-1)+(2x+7)] = 0$

$(3x-8)(7x+6) = 0$

4. ## Re: Solutions to quadratic equation

Yes, there is also a negative answer.

5. ## Re: Solutions to quadratic equation

$$(3x-8)(7x+6)=0$$
Therefore,$$$$\begin{split}3x-8&=0 \\3x&=8 \\x&=\dfrac{8}{3} \end{split}$$$$
and$$$$\begin{split}7x+6&=0\\7x&=-6 \\ x&=\dfrac{-6}{7} \end{split}$$$$
Hence, $x=\dfrac{8}{3},\;\dfrac{-6}{7}$

6. ## Re: Solutions to quadratic equation

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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$\displaystyle (5x - 1)^2 \ = \ (2x + 7)^2$

$\displaystyle 5x - 1 \ = \ \pm\sqrt{2x + 7}$

$\displaystyle 5x - 1 \ = \ -2x - 7 \ \ \ \ or \ \ \ \ 5x - 1 \ = \ 2x + 7$

$\displaystyle 7x \ = \ -6 \ \ \ \ or \ \ \ \ 3x \ = \ 8$

$\displaystyle x \ = \ \dfrac{-6}{7} \ \ \ \ or \ \ \ \ x \ = \ \dfrac{8}{3}$