How do you solve $x^{2} - 5 x = - 4$ $x^2-5x=-4$ by completing the square?

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How do you solve $x^{2} - 5 x = - 4$ $x^2-5x=-4$ by completing the square?

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Answer 1 · 2015-04-30T14:50:49

In this way:

$x^{2} - 5 x = - 4 \Rightarrow x^{2} - 5 x + \frac{25}{4} - \frac{25}{4} = - 4 \Rightarrow$ $x^2-5x=-4rArrx^2-5x+25/4-25/4=-4rArr$

${(x - \frac{5}{2})}^{2} = \frac{25}{4} - 4 \Rightarrow {(x - \frac{5}{2})}^{2} = \frac{25 - 16}{4} \Rightarrow$ $(x-5/2)^2=25/4-4rArr(x-5/2)^2=(25-16)/4rArr$

${(x - \frac{5}{2})}^{2} = \frac{9}{4} \Rightarrow x - \frac{5}{2} = \pm \frac{3}{2} \Rightarrow x = \frac{5}{2} \pm \frac{3}{2} \Rightarrow$ $(x-5/2)^2=9/4rArrx-5/2=+-3/2rArrx=5/2+-3/2rArr$

$x_{1} = \frac{5}{2} - \frac{3}{2} = 1$ $x_1=5/2-3/2=1$

$x_{2} = \frac{5}{2} + \frac{3}{2} = 4$ $x_2=5/2+3/2=4$ .

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How do you solve $x^{2} - 5 x = - 4$ $x^2-5x=-4$ by completing the square?

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How do you solve x2−5x=−4 x^2-5x=-4 by completing the square?

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How do you solve $x^{2} - 5 x = - 4$ $x^2-5x=-4$ by completing the square?