How do you solve using the completing the square method $x^{2} + 5 x + 4 = 0$ $x^2 +5x +4=0$ ?

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How do you solve using the completing the square method $x^{2} + 5 x + 4 = 0$ $x^2 +5x +4=0$ ?

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Answer 1 · 2017-05-03T14:35:25

$x = - 4 or x = - 1$ $x=-4" or "x=-1$

Explanation:

$To solve by completing the square$ $"To solve by "color(blue)"completing the square"$

add ${(\frac{1}{2} coefficient of the x-term)}^{2} to both sides$ $(1/2"coefficient of the x-term")^2" to both sides"$

$that is {(\frac{5}{2})}^{2} = \frac{25}{4}$ $"that is " (5/2)^2=25/4$

$\Rightarrow (x^{2} + 5 x + \frac{25}{4}) + 4 = 0 + \frac{25}{4}$ $rArr(x^2+5xcolor(red)(+25/4))+4=0color(red)(+25/4)$

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

$\Rightarrow {(x + \frac{5}{2})}^{2} = \frac{25}{4} - 4 = \frac{9}{4}$ $rArr(x+5/2)^2=25/4-4=9/4$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x + \frac{5}{2})}^{2}} = \pm \sqrt{\frac{9}{4}}$ $sqrt((x+5/2)^2)=+-sqrt(9/4)$

$\Rightarrow x + \frac{5}{2} = \pm \frac{3}{2}$ $rArrx+5/2=+-3/2$

$\Rightarrow x = - \frac{5}{2} \pm \frac{3}{2}$ $rArrx=-5/2+-3/2$

$\Rightarrow x_{1} = - \frac{5}{2} - \frac{3}{2} = - 4 and x_{2} = - \frac{5}{2} + \frac{3}{2} = - 1$ $rArrx_1=-5/2-3/2=-4" and " x_2=-5/2+3/2=-1$

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How do you solve using the completing the square method $x^{2} + 5 x + 4 = 0$ $x^2 +5x +4=0$ ?

1 Answer

Explanation:

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How do you solve using the completing the square method x2+5x+4=0x^2 +5x +4=0?

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How do you solve using the completing the square method $x^{2} + 5 x + 4 = 0$ $x^2 +5x +4=0$ ?