How do you solve $5 x^{2} - 10 x = 23$ $5x^2-10x=23$ by completing the square?

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How do you solve $5 x^{2} - 10 x = 23$ $5x^2-10x=23$ by completing the square?

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Answer 1 · 2018-05-21T10:31:37

$x = 1 \pm \sqrt{\frac{28}{5}}$ $x=1+-sqrt(28/5)$

Explanation:

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$∙ the coefficient of the x^{2} term must be 1$ $• " the coefficient of the "x^2" term must be 1"$

$factor out 5$ $"factor out 5"$

$5 (x^{2} - 2 x) = 23$ $5(x^2-2x)=23$

$∙ add to both sides {(\frac{1}{2} coefficient of x-term)}^{2}$ $• "add to both sides "(1/2"coefficient of x-term")^2$

$5 (x^{2} + (- 1) x + 1) = 23 + 5 \leftarrow note$ $5(x^2+(-1)xcolor(red)(+1))=23color(red)(+5)larr"note"$

$\Rightarrow 5 {(x - 1)}^{2} = 28$ $rArr5(x-1)^2=28$

$\Rightarrow {(x - 1)}^{2} = \frac{28}{5}$ $rArr(x-1)^2=28/5$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x - 1)}^{2}} = \pm \sqrt{\frac{28}{5}} \leftarrow note plus or minus$ $sqrt((x-1)^2)=+-sqrt(28/5)larrcolor(blue)"note plus or minus"$

$\Rightarrow x - 1 = \pm \sqrt{\frac{28}{5}}$ $rArrx-1=+-sqrt(28/5)$

$add 1 to both sides$ $"add 1 to both sides"$

$\Rightarrow x = 1 \pm \sqrt{\frac{28}{5}} \leftarrow exact solutions$ $rArrx=1+-sqrt(28/5)larrcolor(red)"exact solutions"$

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How do you solve $5 x^{2} - 10 x = 23$ $5x^2-10x=23$ by completing the square?

1 Answer

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How do you solve 5x2−10x=235x^2-10x=23 by completing the square?

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How do you solve $5 x^{2} - 10 x = 23$ $5x^2-10x=23$ by completing the square?