How do you solve $x^{2} - 18 x + 74 = 0$ $x^2 - 18x + 74 = 0$ by completing the square?

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How do you solve $x^{2} - 18 x + 74 = 0$ $x^2 - 18x + 74 = 0$ by completing the square?

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Answer 1 · 2016-06-13T06:21:25

$x = 9 \pm \sqrt{7}$ $x=9+-sqrt(7)$

Explanation:

Complete the square and use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

with $a = (x - 9)$ $a=(x-9)$ and $b = \sqrt{7}$ $b=sqrt(7)$ as follows:

$0 = x^{2} - 18 x + 74$ $0 = x^2-18x+74$

$= x^{2} - 18 x + 81 - 7$ $= x^2-18x+81-7$

$= {(x - 9)}^{2} - {(\sqrt{7})}^{2}$ $= (x-9)^2-(sqrt(7))^2$

$= ((x - 9) - \sqrt{7}) ((x - 9) + \sqrt{7})$ $= ((x-9)-sqrt(7))((x-9)+sqrt(7))$

$= (x - 9 - \sqrt{7}) (x - 9 + \sqrt{7})$ $= (x-9-sqrt(7))(x-9+sqrt(7))$

Hence the zeros are:

$x = 9 \pm \sqrt{7}$ $x = 9 +-sqrt(7)$

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How do you solve $x^{2} - 18 x + 74 = 0$ $x^2 - 18x + 74 = 0$ by completing the square?

1 Answer

Explanation:

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How do you solve $x^{2} - 18 x + 74 = 0$ $x^2 - 18x + 74 = 0$ by completing the square?