Question

Algebra 1A help please thank you!!

Answer 1

`\large\text{$√(50)=\boxed{5√(2)}$}`

`\large\text{$√(50)=\boxed{5\cdot 2^{(1)/(2)}}$}`

`\large\text{$\sqrt[10]{2^5}=\boxed{2^{(1)/(2)}}$}`

Explanation:

Given expression:

`\large\text{$\sqrt[10]{2^5}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{√(50)}{2^{(1)/(2)}}$}`

To simplify the given expression, begin by simplifying and rewriting the radicals.

To simplify the radical √(50) using perfect square factors, we can rewrite 50 as the product of 5² · 2:

`\large\text{$√(50)=√(5^2 \cdot 2)$}`

`\textsf{Apply the radical rule:} \quad √(ab)=\sqrt{\vphantom{b}a}√(b)`

`\large\text{$√(50)=√(5^2) √( 2)$}`

`\textsf{Apply the radical rule:} \quad √(a^2)=a, \quad a \geq 0`

`\large\text{$√(50)=5√(2)$}`

The square root can be represented as a rational exponent of 1/2. Therefore:

`\large\text{$√(50)=5\cdot 2^{(1)/(2)}$}`

To rewrite
`\sqrt[10]{2^5}`, we can apply the exponent rule
`\sqrt[n]{a^m}=a^{(m)/(n)}`:

`\large\text{$\sqrt[10]{2^5}=2^{(5)/(10)}=2^{(1)/(2)}$}`

Therefore, the expression simplified by rewriting the radicals is:

`\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{5\cdot 2^{(1)/(2)}}{2^{(1)/(2)}}$}`

To simplify further, cancel the common factor
`2^{(1)/(2)}` of the rational:

`\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-\frac{5\cdot \diagup\!\!\!\!\! 2^{(1)/(2)}}{\diagup\!\!\!\!\!2^{(1)/(2)}}$}`

`\large\text{$2^{(1)/(2)}\cdot 5 \cdot 2^{-(1)/(2)}-5$}`

`\textsf{Apply the exponent rule:} \quad a^b \cdot a^c=a^(b+c)`

`\large\text{$2^{(1)/(2)}\cdot 2^{-(1)/(2)}\cdot 5 -5$}`

`\large\text{$2^{(1)/(2)-(1)/(2)}\cdot 5 -5$}`

`\large\text{$2^(0)\cdot 5 -5$}`

As any number to the power of zero is always 1:

`\large\text{$1\cdot 5 -5$}`

`\large\text{$5 -5$}`

`\large\text{$0$}`

Therefore, the expression is equal to zero.