The product of two radicals is equal to the radicand. The square root deletes itself.
So the result is y^3
Answer:
no
Step-by-step explanation:
Answer:
Step-by-step explanation:
1 In general, given a{x}^{2}+bx+cax
2
+bx+c, the factored form is:
a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a
2a
−b+√
b
2
−4ac
)(x−
2a
−b−√
b
2
−4ac
)
2 In this case, a=1a=1, b=-2b=−2 and c=-2c=−2.
(x-\frac{2+\sqrt{{(-2)}^{2}-4\times -2}}{2})(x-\frac{2-\sqrt{{(-2)}^{2}-4\times -2}}{2})(x−
2
2+√
(−2)
2
−4×−2
)(x−
2
2−√
(−2)
2
−4×−2
)
3 Simplify.
(x-\frac{2+2\sqrt{3}}{2})(x-\frac{2-2\sqrt{3}}{2})(x−
2
2+2√
3
)(x−
2
2−2√
3
)
4 Factor out the common term 22.
(x-\frac{2(1+\sqrt{3})}{2})(x-\frac{2-2\sqrt{3}}{2})(x−
2
2(1+√
3
)
)(x−
2
2−2√
3
)
5 Cancel 22.
(x-(1+\sqrt{3}))(x-\frac{2-2\sqrt{3}}{2})(x−(1+√
3
))(x−
2
2−2√
3
)
6 Simplify brackets.
(x-1-\sqrt{3})(x-\frac{2-2\sqrt{3}}{2})(x−1−√
3
)(x−
2
2−2√
3
)
7 Factor out the common term 22.
(x-1-\sqrt{3})(x-\frac{2(1-\sqrt{3})}{2})(x−1−√
3
)(x−
2
2(1−√
3
)
)
8 Cancel 22.
(x-1-\sqrt{3})(x-(1-\sqrt{3}))(x−1−√
3
)(x−(1−√
3
))
9 Simplify brackets.
(x-1-\sqrt{3})(x-1+\sqrt{3})(x−1−√
3
)(x−1+√
3
)
Answer:
x=18
Step-by-step explanation:
am=by which means you can make the equations equal each other
4x-22=2x+14
2x=36
x=18
The answer is A. :). Have a great day!
