How do you solve $x^{2} - 7 x + 5 = 0$ $x^2-7x+5=0$ by completing the square?

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Answer 1 · 2016-08-07T04:14:22

$x = \frac{7 \pm \sqrt{29}}{2}$ $x=(7+-sqrt29)/2$

$x^{2} - 7 x + 5 = 0$ $x^2-7x+5=0$

$\Rightarrow x^{2} - 2 \cdot x \cdot \frac{7}{2} + {(\frac{7}{2})}^{2} - {(\frac{7}{2})}^{2} + 5 = 0$ $=>x^2-2*x*7/2+(7/2)^2-(7/2)^2+5=0$

$\Rightarrow {(x - \frac{7}{2})}^{2} = \frac{49}{4} - 5 = \frac{29}{4}$ $=>(x-7/2)^2=49/4-5=29/4$

$\Rightarrow (x - \frac{7}{2}) = \pm \frac{\sqrt{29}}{2}$ $=>(x-7/2)=+-sqrt29/2$

$\Rightarrow x = \frac{7 \pm \sqrt{29}}{2}$ $=>x=(7+-sqrt29)/2$

How do you solve x2−7x+5=0x^2-7x+5=0 by completing the square?