1. Simplify: sqrt6(sqrt2/2+sqrt3)-sqrt8
2. Simplify: sqrt(72x^3)-5x sqrt(2x)
3. Simplify: (2sqrt6-sqrt3)^2
4. Simplify: 6/3+2sqrt3
5. Solve: 2-y=sqrt(y+4)
I need all of the steps.
Thanks
Sure! I will provide step-by-step explanations for each question.
1. Simplify: sqrt6(sqrt2/2+sqrt3)-sqrt8
To simplify this expression, we'll start by expanding the square root and combining like terms:
sqrt6(sqrt2/2+sqrt3)-sqrt8
= sqrt6(sqrt2)/sqrt2 + sqrt6(sqrt3) - sqrt8
Next, we'll rationalize the denominators:
= sqrt6(sqrt2)/sqrt2 + sqrt6(sqrt3) - sqrt(8) * sqrt(2)/sqrt(2)
= sqrt6(sqrt2)/sqrt2 + sqrt6(sqrt3) - sqrt16
Now, simplify the square roots:
= sqrt3(sqrt2)/2 + sqrt(6sqrt3) - 4
2. Simplify: sqrt(72x^3) - 5x sqrt(2x)
We begin by factoring and taking out common factors inside the square root:
sqrt(72x^3) - 5x sqrt(2x)
= sqrt(36x^2 * 2x) - 5x sqrt(2x)
Now, simplify the square roots separately:
= 6x sqrt(2x) - 5x sqrt(2x)
Combine the like terms:
= (6x - 5x) sqrt(2x)
= x sqrt(2x)
3. Simplify: (2sqrt6 - sqrt3)^2
Since we have a binomial squared, we can use the formula (a - b)^2 = a^2 - 2ab + b^2:
(2sqrt6 - sqrt3)^2
= (2sqrt6)^2 - 2(2sqrt6)(sqrt3) + (sqrt3)^2
= 4 * 6 - 2 * 2 * sqrt6 * sqrt3 + 3
= 24 - 4 * sqrt18 + 3
= 27 - 4 * sqrt(9 * 2)
= 27 - 4 * sqrt9 * sqrt2
= 27 - 4 * 3 * sqrt2
= 27 - 12 * sqrt2
= 27 - 12sqrt2
4. Simplify: 6/3 + 2sqrt3
First, divide 6 by 3:
= 2 + 2sqrt3
There are no like terms to combine, so the simplified form is:
= 2 + 2sqrt3
5. Solve: 2 - y = sqrt(y + 4)
To solve this equation, we'll need to isolate the variable y.
Start by moving the square root term to the other side of the equation:
2 - sqrt(y + 4) = y
Next, square both sides of the equation to eliminate the square root:
(2 - sqrt(y + 4))^2 = y^2
Simplify the left side using the formula (a - b)^2 = a^2 - 2ab + b^2:
4 - 4sqrt(y + 4) + (sqrt(y + 4))^2 = y^2
Combine like terms and square the square root term:
4 - 4sqrt(y + 4) + (y + 4) = y^2
Simplify further:
8 - 4sqrt(y + 4) + y = y^2
Rearrange the equation:
y^2 - y - 8 + 4sqrt(y + 4) = 0
This quadratic equation is ready for further factoring or use of the quadratic formula to solve for y if needed.