How do you solve the equation $x^{2} - 3 x - 7 = 0$ $x^2-3x-7=0$ by completing the square?

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How do you solve the equation $x^{2} - 3 x - 7 = 0$ $x^2-3x-7=0$ by completing the square?

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Answer 1 · 2018-03-03T08:01:50

$x = 4.54, - 1.54$ $color(blue)(x = 4.54, -1.54$

Explanation:

$x^{2} - 3 x - 7 = 0$ $x^2 - 3x - 7 = 0$

$x^{2} - (2 \cdot (\frac{3}{2}) \cdot x) = 7$ $x^2 - (2*(3/2)*x) = 7$

Adding ${(\frac{3}{2})}^{2}$ $(3/2)^2$ to both sides,

$x^{2} - (2 \cdot \frac{3}{2}) \cdot x) + {(\frac{3}{2})}^{2} = 7 + {(\frac{3}{2})}^{2} = \frac{37}{4}$ $x^2 - (2 * 3/2) * x) + (3/2)^2 = 7 + (3/2)^2 = 37/4$

${(x - (\frac{3}{2}))}^{2} = {(\sqrt{\frac{37}{4}})}^{2}$ $(x - (3/2))^2 = (sqrt(37/4))^2$

$x = \pm \sqrt{\frac{37}{4}} + (\frac{3}{2})$ $x =+- sqrt(37/4)+ (3/2)$

$x = 4.54, - 1.54$ $color(blue)(x= 4.54, -1.54$

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How do you solve the equation $x^{2} - 3 x - 7 = 0$ $x^2-3x-7=0$ by completing the square?

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