What are the zero(s) of: x^2 = 6x + 6 = 0?

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What are the zero(s) of: x^2 = 6x + 6 = 0?

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Answer 1 · 2015-11-09T22:09:14

$3 + \sqrt{15}, 3 - \sqrt{15}$ $3+ sqrt(15) , 3- sqrt(15)$

Explanation:

We can use the quadratic formula to find the zeros. We are given:

$x^{2} = 6 x + 6$ $x^2=6x+6$

We can arrange this into a quadratic equation:

$x^{2} - 6 x - 6 = 0$ $x^2-6x-6=0$

The quadratic formula:

$x = \frac{- b (\frac{+}{-}) \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b(+/-) sqrt(b^2-4ac))/(2a)$

If:
$a = 1, b = - 6, c = - 6$ $a=1, b=-6, c=-6$
Then:

$x = \frac{- (- 6) (\frac{+}{-}) \sqrt{{(- 6)}^{2} - 4 (1) (- 6)}}{2 (1)} = \frac{6 (\frac{+}{-}) \sqrt{36 + 24}}{2}$ $x=(-(-6)(+/-) sqrt((-6)^2-4(1)(-6)))/(2(1))= (6(+/-) sqrt(36+24))/2$
$x = \frac{6 (\frac{+}{-}) \sqrt{60}}{2} = \frac{6 (\frac{+}{-}) 2 \sqrt{15}}{2} = 3 (\frac{+}{-}) \sqrt{15}$ $x=(6(+/-)sqrt(60))/2=(6(+/-) 2sqrt(15))/2= 3(+/-) sqrt(15)$

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What are the zero(s) of: x^2 = 6x + 6 = 0?

1 Answer

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