How do you solve $\sqrt{6 a - 6} = a + 1$ $sqrt(6a-6)=a+1$ and check your solution?

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How do you solve $\sqrt{6 a - 6} = a + 1$ $sqrt(6a-6)=a+1$ and check your solution?

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Answer 1 · 2017-09-30T16:34:49

No solutions

or using imaginary numbers
$a = 2 \pm \sqrt{3} i$ $a=2+-sqrt(3)i$

Explanation:

To start we would square both sides to get rid of the square root.

${(\sqrt{6 a - 6})}^{2} = {(a + 1)}^{2}$ $(sqrt(6a-6))^2=(a+1)^2$
$6 a - 6 = {(a + 1)}^{2}$ $6a-6=(a+1)^2$

We can now expand the brackets.

$6 a - 6 = a^{2} + 2 a + 1$ $6a-6=a^2+2a+1$

Move everything to one side to get the equation equal to 0.

$0 = a^{2} - 4 a + 7$ $0=a^2-4a+7$

To solve this we will use the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
$a = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (7)}}{2 (1)}$ $a=(-(-4)+-sqrt((-4)^2-4(1)(7)))/(2(1))$
$a = \frac{4 \pm \sqrt{16 - 28}}{2}$ $a=(4+-sqrt(16-28))/2$
$a = \frac{4 \pm \sqrt{- 12}}{2}$ $a=(4+-sqrt(-12))/2$

However we are left with the square root of a negative, so there are no solutions.

Whereas if you are using imaginary number we can continue. If you are not using imaginary numbers stop here.

We can split $\sqrt{- 12}$ $sqrt(-12)$ into $\sqrt{- 1} \sqrt{12}$ $sqrt(-1)sqrt(12)$

$a = \frac{4 \pm \sqrt{- 12}}{2}$ $a=(4+-sqrt(-12))/2$
$a = \frac{4 \pm \sqrt{- 1} \sqrt{12}}{2}$ $a=(4+-sqrt(-1)sqrt(12))/2$

Put $i$ $i$ in as $\sqrt{- 1}$ $sqrt(-1)$ and simplify $\sqrt{12}$ $sqrt12$ .

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

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How do you solve $\sqrt{6 a - 6} = a + 1$ $sqrt(6a-6)=a+1$ and check your solution?

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How do you solve √6a−6=a+1sqrt(6a-6)=a+1 and check your solution?

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Explanation:

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How do you solve $\sqrt{6 a - 6} = a + 1$ $sqrt(6a-6)=a+1$ and check your solution?