You need to show us what you've tried for each of them...
I can't figure these out....
1)square root of 252x^14y^31
2) 3 square root of 10 times square root of 490
3) 7 cubed root of 40 (minus) cubed root of 625 (plus) 7 cubed root of 5.
4) for the square root of negative 25...is the answer 5i ?
any help would be awesome because I'm stuck
These problems ask you to apply certain fundamental principles of roots and exponents to simplify expressions.
$\sqrt{ab} \equiv \sqrt{a} * \sqrt{b}.$ This means you can analyze a product under a square root sign in terms of the square roots of the individual factors. So, for the first problem
$\sqrt{252x^{14}y^{31}} = \sqrt{252} * \sqrt{x^{14}} * \sqrt{y^{31}}.$ With me to here?
Factor 252 into primes giving $252 = 2^2 * 3^2 * 7 = (2 * 3)^2 * 7 = 6^2 * 7.$ So: $\sqrt{252} = \sqrt{6^2 * 7} = \sqrt{6^2} * \sqrt{7} = 6\sqrt{7}.$ Make sense?
Use the laws of exponents to help you with the rest of the factors.
$\sqrt{x^{14}} = \sqrt{x^{7 * 2}} = \sqrt{(x^7)^2} = x^7.$ Any problems here?
The next one is a tiny bit trickier.
$\sqrt{y^{31}} = \sqrt{y^{30 + 1}} = \sqrt{y^{30} * y^1} = \sqrt{y^{30}} * \sqrt{y} = \sqrt{y^{15 * 2}} * \sqrt{y} = \sqrt{(y^{15})^2} * \sqrt{y} =y^{15}\sqrt{y}.$
Now put it all together
$\sqrt{252x^{14}y^{31}} = \sqrt{252} * \sqrt{x^{14}} * \sqrt{y^{31}} = 6\sqrt{7} * x^7 * y^{15}\sqrt{y} = 6x^7y^{15}\sqrt{7y}.$
Now you try #2 and 3 and show us your work so we can point out if you have made a slip