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

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Answer 1 · 2017-09-30T16:34:49

No solutions

or using imaginary numbers
$a=2+-sqrt(3)i$

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

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

We can now expand the brackets.

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

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

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

To solve this we will use the quadratic formula.

$x=(-b+-sqrt(b^2-4ac))/(2a)$
$a=(-(-4)+-sqrt((-4)^2-4(1)(7)))/(2(1))$
$a=(4+-sqrt(16-28))/2$
$a=(4+-sqrt(-12))/2$

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 $sqrt(-12)$ into $sqrt(-1)sqrt(12)$

$a=(4+-sqrt(-12))/2$
$a=(4+-sqrt(-1)sqrt(12))/2$

Put $i$ in as $sqrt(-1)$ and simplify $sqrt12$ .

$a=(4+-2sqrt(3)i)/2$
$a=2+-sqrt(3)i$

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