Question

PLEASE PLEASE PLEASE HELP ME THIS IS OVER IN LESS THAN A DAY LOOK AT THE PICTURE BELOW

Answer 1

`\textsf{1)}\quad 2x^2y^3 z^4\:\sqrt[4]{8xz}`

`\textsf{2)}\quad 7√(2) \;i`

`\textsf{3)} \quad 26-7i`

Explanation:

Question 1

Given expression:

`\sqrt[4]{128x^9y^(12)z^(17)}`

First, rewrite 128 as a product of its prime factors: 128 = 2⁷

`\sqrt[4]{2^7x^9y^(12)z^(17)}`

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

`\sqrt[4]{2^7}\;\sqrt[4]{x^9}\;\sqrt[4]{y^(12)}\;\sqrt[4]{z^(17)}`

Rewrite the exponents of each term so that the sums of the exponents include multiples of 4:

`\sqrt[4]{2^(4+3)}\;\sqrt[4]{x^(8+1)}\;\sqrt[4]{y^(12)}\;\sqrt[4]{z^(16+1)}`

Apply the product exponent rule:

`\sqrt[4]{2^(4)\cdot 2^(3)}\;\sqrt[4]{x^8\cdot x}\;\sqrt[4]{y^(12)}\;\sqrt[4]{z^(16)\cdot z}`

Apply the power of a product exponent rule to x⁸, y¹² and z¹⁶ so that their outer exponents are 4:

`\sqrt[4]{2^(4)\cdot 2^(3)}\;\sqrt[4]{(x^2)^4\cdot x}\;\sqrt[4]{(y^3)^4}\;\sqrt[4]{(z^4)^4\cdot z}`

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

`\sqrt[4]{2^(4)}\;\sqrt[4]{2^(3)}\;\sqrt[4]{(x^2)^4}\;\sqrt[4]{x}\;\sqrt[4]{(y^3)^4}\;\sqrt[4]{(z^4)^4}\;\sqrt[4]{z}`

`\textsf{Apply the exponent rule:} \quad \sqrt[n]{a^n}=a`

`2 \cdot \sqrt[4]{2^(3)}\cdot x^2 \cdot\sqrt[4]{x}\cdot y^3\cdot z^4\cdot\sqrt[4]{z}`

`2 \cdot \sqrt[4]{8}\cdot x^2 \cdot\sqrt[4]{x}\cdot y^3\cdot z^4\cdot\sqrt[4]{z}`

Rearrange to collect like terms:

`2x^2y^3 z^4\;\sqrt[4]{8}\;\sqrt[4]{x}\;\sqrt[4]{z}`

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

`2x^2y^3 z^4\;\sqrt[4]{8xz}`

`\hrulefill`

Question 2

Given expression:

`√(-98)`

First, rewrite 98 as 98 × (-1):

`√(98\cdot (-1))`

Rewrite 98 as a product of its prime factors: 98 = 7² × 2

`√(7^2 \cdot 2\cdot (-1))`

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

`√(7^2)\;√(2) \;√(-1)`

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

`7√(2) \;√(-1)`

`\textsf{Apply the imaginary number rule:} \quad √(-1)=i`

`7√(2) \;i`

`\hrulefill`

Question 3

Given expression:

`(5+2i)(4-3i)`

`\textsf{Apply the complex arithmetic rule:}\quad (a+bi)(c+di)=(ac-bd)+(ad+bc)i`

In this case:

• a = 5
• b = 2
• c = 4
• d = -3

Therefore:

`\begin{aligned}(5+2i)(4-3i)&=(5 \cdot 4-2 \cdot (-3))+(5 \cdot (-3)+2 \cdot 4)i\\\\&=(20+6)+(-15+8)i\\\\&=26-7i\end{aligned}`