How do you solve using the completing the square method $a^{2} + 3 a - 40 = 0$ $a^2+3a-40=0$ ?

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How do you solve using the completing the square method $a^{2} + 3 a - 40 = 0$ $a^2+3a-40=0$ ?

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Answer 1 · 2016-08-04T08:05:12

$a = 5$ $a=5$
$a = - 8$ $a=-8$

Explanation:

Given -

$a^{2} + 3 a - 40 = 0$ $a^2+3a-40=0$

Take the constant term to the right

$a^{2} + 3 a = 40$ $a^2+3a=40$

Divide the coefficient of $x$ $x$ by 2. Square it and add the same to both sides

$a^{2} + \frac{3}{2} x + \frac{9}{4} = 40 + \frac{9}{4}$ $a^2+3/2x+9/4=40+9/4$

${(a + \frac{3}{2})}^{2} = \frac{160 + 9}{4} = \frac{169}{4}$ $(a+3/2)^2=(160+9)/4=169/4$

$a + \frac{3}{2} = \pm \sqrt{\frac{169}{4}} = \pm \frac{13}{2}$ $a+3/2=+-sqrt(169/4)=+-13/2$

$a = \frac{13}{2} - \frac{3}{2} = \frac{10}{2} = 5$ $a=13/2-3/2=10/2=5$

$a = - \frac{13}{2} - \frac{3}{2} = - \frac{16}{2} = - 8$ $a=-13/2-3/2=-16/2=-8$

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How do you solve using the completing the square method $a^{2} + 3 a - 40 = 0$ $a^2+3a-40=0$ ?

1 Answer

Explanation:

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How do you solve using the completing the square method a^2+3a-40=0a2+3a−40=0a^2+3a-40=0?

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How do you solve using the completing the square method $a^{2} + 3 a - 40 = 0$ $a^2+3a-40=0$ ?