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iren [92.7K]
2 years ago
11

How do I solve for x : 2(x-5)^2+13=31

Mathematics
1 answer:
Setler [38]2 years ago
3 0
                         2(x - 5)² + 13 = 31
                 2(x - 5)(x - 5) + 13 = 21
        2(x(x - 5) - 5(x - 5)) + 13 = 21
2(x(x) - x(5) - 5(x) + 5(5)) + 13 = 21
        2(x² - 5x - 5x + 25) + 13 = 21
             2(x² - 10x + 25) + 13 = 21
     2(x²) - 2(10x) + 2(25) + 13 = 21
               2x² - 20x + 50 + 13 = 21
                       2x² - 20x + 63 = 21
                       <u>                - 21  - 21</u>
                       2x² - 20x + 42 = 0
             2(x²) - 2(20x) + 2(21) = 0
                     <u>2(x² - 20x + 21)</u> = <u>0</u>
                                 2               2
                         x² - 20x + 21 = 0
                                            x = <u>-(-20) +/- √((-20)² - 4(1)(21))</u>
                                                                      2(1)
                                            x = <u>20 +/- √(400 - 84)</u>
                                                               2
                                            x = <u>20 +/- √(316)
</u>                                                            2
                                            x = <u>20 +/- 2√(79)</u>
                                                            2
                                            x = 10 <u>+</u> √(79)
                                            x = 10 + √(79)  U  x = 10 - √(79)
<u />
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