Square roots. Woohoo. Want to check some work I did.
1. Perform indicated operations
3sqrt[3]+2sqrt[27]-sqrt[12]
3sqrt[3]+2sqrt[3*9]-sqrt[2*6]
3sqrt[3]+3*2sqrt[3]-2sqrt[3]
3sqrt+6sqrt[3]-2sqrt[3]
= 7sqrt[3]
2.Simplify
sqrt[49x^12y^4z^8]
= 7x^6y^2z^4
3.Multply
(8sqrt[6]+3sqrt[2])(4sqrt[6]-5sqrt[2]
32sqrt[36]-40sqrt[12]+12sqrt[12]-15sqrt[4]
32*6-40sqrt[12]+12sqrt[12]-15*2
192 - 28sqrt[12] -30
162 - 28sqrt[12]
or am i missing a step?
192-40sqrt[4*3]+12sqrt[4*3]-30
162-40*4sqrt[3]+12*4sqrt[3]
162-160sqrt[3]+48sqrt[3]
162- 128sqrt[3] which I think turns into 162 -56sqrt[3]
1 last square root :
Rationalize the denominator
5/sqrt[3]+sqrt[5]=
5*sqrt[3]-sqrt[5]/ sqrt[3]sqrt5[5]*sqrt[3]-sqrt[5]=
5sqrt[3]-5sqrt[5]/sqrt[9]-sqrt[15]+sqrt[15]- sqrt[25]=
5sqrt[3]-5sqrt[5]/-16 =
5sqrt[3]-5sqrt[5]/ -2
first two ok
3.Multply
(8sqrt[6]+3sqrt[2])(4sqrt[6]-5sqrt[2]
32sqrt[36]-40sqrt[12]+12sqrt[12]-15sqrt[4]
32*6-40sqrt[12]+12sqrt[12]-15*2
192 - 28sqrt[12] -30
162 - 28sqrt[12] then
162-28sqrt (4*3)
162-28*2sqrt3 and then multiply the 28*2
1 last square root :
Rationalize the denominator
5/sqrt[3]+sqrt[5]=
5*sqrt[3]-sqrt[5]/ sqrt[3]sqrt5[5]*sqrt[3]-sqrt[5]=
5sqrt[3]-5sqrt[5]/sqrt[9]-sqrt[15]+sqrt[15]- sqrt[25]=
5sqrt[3]-5sqrt[5]/-16 = No. the demoniator is (3-5)=-2
I have no idea where you got this line in your work, it does not make sense
5sqrt[3]-5sqrt[5]/ -2
Oh I see. So 1 more step on the first one and I had it.
I see my big huge mistake on that last one. I went wild. I get it now.
Thankyou!
Let's go through the steps and check your work:
1. Perform indicated operations
\[3\sqrt{3}+2\sqrt{27}-\sqrt{12}\]
\[=3\sqrt{3}+2\sqrt{3\cdot 9}-\sqrt{2\cdot 6}\]
\[=3\sqrt{3}+3\cdot 2\sqrt{3}-2\sqrt{3}\]
\[=3\sqrt{3}+6\sqrt{3}-2\sqrt{3}\]
\[=7\sqrt{3}\]
Your answer is correct: \(7\sqrt{3}\).
2. Simplify
\[\sqrt{49x^{12}y^4z^8}\]
\[=7x^6y^2z^4\]
Your answer is correct: \(7x^6y^2z^4\).
3. Multiply
\((8\sqrt{6}+3\sqrt{2})(4\sqrt{6}-5\sqrt{2})\)
\[=32\sqrt{6}-40\sqrt{12}+12\sqrt{12}-15\sqrt{4}\]
\[=32\cdot 6-40\sqrt{12}+12\sqrt{12}-15\cdot 2\]
\[=192-28\sqrt{12}-30\]
\[=162-28\sqrt{12}\]
Your answer is correct: \(162-28\sqrt{12}\).
4. Rationalize the denominator
\(\frac{5}{\sqrt{3}+\sqrt{5}}\)
\[\frac{5\sqrt{3}-\sqrt{5}}{\sqrt{3}\sqrt{3}+\sqrt{3}\sqrt{5}}\]
\[\frac{5\sqrt{3}-\sqrt{5}}{3+5}\]
\[\frac{5\sqrt{3}-\sqrt{5}}{8}\]
Your answer is correct: \(\frac{5\sqrt{3}-\sqrt{5}}{8}\).
Overall, your work is correct! Well done.
1. Perform indicated operations:
To simplify the expression 3√3 + 2√27 - √12, you correctly broke down the radicals into their simplest form:
3√3 + 2√(3*9) - √(2*6)
= 3√3 + 2√(3^2) - √(2^2*3)
= 3√3 + 2*3√3 - 2√3
= 3√3 + 6√3 - 2√3
= 7√3
So your final answer is 7√3.
2. Simplify:
To simplify the expression √(49x^12y^4z^8), you correctly applied the property of radicals that says √(ab) = √a * √b:
√(49x^12y^4z^8) = √(7^2 * x^12 * y^4 * z^8)
= 7 * x^6 * y^2 * z^4
So your final answer is 7x^6y^2z^4.
3. Multiply:
In your first attempt to multiply (8√6 + 3√2)(4√6 - 5√2), you made a mistake in distributing √36 and √4 square roots. Here's the correct step-by-step process:
(8√6 + 3√2)(4√6 - 5√2)
= 8√6 * 4√6 + 8√6 * (-5√2) + 3√2 * 4√6 + 3√2 * (-5√2)
= 32√36 - 40√12 + 12√12 - 15√4
Now, simplify each radical:
32√36 = 32√(6^2) = 32 * 6 = 192
40√12 = 40√(2^2 * 3) = 40 * 2√3 = 80√3
12√12 = 12√(2^2 * 3) = 12 * 2√3 = 24√3
15√4 = 15√(2^2) = 15 * 2 = 30
Combining these terms, you get:
192 - 40√3 + 24√3 - 30 = 162 - 16√3
So your final answer is 162 - 16√3.
Rationalizing the Denominator:
To rationalize the expression 5/(√3 + √5), you correctly multiplied both the numerator and the denominator by the conjugate of the denominator, which is √3 - √5:
5(√3 - √5)/(√3 + √5)(√3 - √5)
= 5(√3 - √5)/(3 - 5)
= 5(√3 - √5)/(-2)
So your final answer is (5√3 - 5√5)/(-2).