How do you solve using completing the square method $x^{2} - 7 x + 8 = 0$ $x^2-7x+8=0$ ?

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Answer 1 · 2017-07-12T12:28:21

Solution: $x = \frac{1}{2} (7 + \sqrt{17}) or x = \frac{1}{2} (7 - \sqrt{17})$ $x = 1/2(7+ sqrt17) or x = 1/2(7- sqrt17)$

$x^{2} - 7 x + 8 = 0 or x^{2} - 7 x + {(\frac{7}{2})}^{2} - \frac{49}{4} + 8 = 0$ $x^2-7x+8=0 or x^2-7x +(7/2)^2 -49/4+8=0$ or

${(x - \frac{7}{2})}^{2} - \frac{17}{4} = 0 or {(x - \frac{7}{2})}^{2} = \frac{17}{4}$ $(x-7/2)^2 -17/4=0 or (x-7/2)^2 =17/4$ or

$(x - \frac{7}{2}) = \pm \sqrt{\frac{17}{4}} or x = \frac{7}{2} \pm \sqrt{\frac{17}{4}}$ $(x-7/2) =+- sqrt (17/4) or x = 7/2+- sqrt (17/4)$ or

$x = \frac{7}{2} \pm \frac{\sqrt{17}}{2} or x = \frac{1}{2} (7 \pm \sqrt{17})$ $x = 7/2+- sqrt 17/2 or x = 1/2(7+- sqrt17)$

Solution: $x = \frac{1}{2} (7 + \sqrt{17}) or x = \frac{1}{2} (7 - \sqrt{17})$ $x = 1/2(7+ sqrt17) or x = 1/2(7- sqrt17)$ [Ans]

How do you solve using completing the square method x^2-7x+8=0x2−7x+8=0x^2-7x+8=0?