(a) Simplify $\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}$, giving your answer in the form $x^{a} y^{b}$, where $a$ and $b$ are integers. [2]

(b) (i) Show that $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}$ can be written in the form $(t-2)^{p}(q t+r)$, where $p, q$ and $r$ are constants to be found.

(ii) Hence solve the equation $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}=0$.

Let's solve the problem step by step.

### Part (a) Simplify $\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}$

We need to simplify the expression $\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}$.

#### Step 1: Simplify each radical expression

1. $\sqrt{x^8 y^{10}}$:
   \[
   \sqrt{x^8 y^{10}} = x^{\frac{8}{2}} y^{\frac{10}{2}} = x^4 y^5
   \]

2. $\sqrt[3]{x^3 y^{-6}}$:
   \[
   \sqrt[3]{x^3 y^{-6}} = x^{\frac{3}{3}} y^{\frac{-6}{3}} = x^1 y^{-2}
   \]

#### Step 2: Divide the simplified expressions
\[
\frac{x^4 y^5}{x^1 y^{-2}} = x^{4-1} y^{5 - (-2)} = x^3 y^{7}
\]

Thus, the simplified form is $x^3 y^7$.

### Part (b)(i) Show that $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}$ can be written in the form $(t-2)^{p}(qt+r)$.

We want to express $4(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}}$ in the form $(t-2)^p(qt + r)$.

#### Step 1: Factor out the common term $(t-2)^{\frac{1}{2}}$:
\[
4(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} = (t-2)^{\frac{1}{2}} \left(4 + 5(t-2)\right)
\]

#### Step 2: Simplify the expression inside the parentheses:
\[
4 + 5(t-2) = 4 + 5t - 10 = 5t - 6
\]

Thus, the expression can be written as:
\[
4(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} = (t-2)^{\frac{1}{2}}(5t - 6)
\]

Here, $p = \frac{1}{2}$, $q = 5$, and $r = -6$.

### Part (b)(ii) Solve the equation $4(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} = 0$.

From part (b)(i), we can rewrite the equation as:
\[
(t-2)^{\frac{1}{2}}(5t - 6) = 0
\]

This equation is a product of two factors, so each factor can be set to zero.

#### Case 1: $(t-2)^{\frac{1}{2}} = 0$
\[
(t-2)^{\frac{1}{2}} = 0 \implies t-2 = 0 \implies t = 2
\]

#### Case 2: $5t - 6 = 0$
\[
5t - 6 = 0 \implies 5t = 6 \implies t = \frac{6}{5}
\]

### Final Answer:
The solutions to the equation are $t = 2$ and $t = \frac{6}{5}$.

---

Do you want further details or have any questions? 

Here are some related questions for practice:

1. Simplify $\sqrt{x^{12} y^{8}} \div \sqrt[4]{x^{8} y^{-12}}$.
2. Express $3(t-1)^{\frac{1}{2}} + 2(t-1)^{\frac{3}{2}}$ in the form $(t-1)^p (qt + r)$.
3. Solve the equation $3(t-1)^{\frac{1}{2}} + 2(t-1)^{\frac{3}{2}} = 0$.
4. Simplify $ \sqrt{a^4b^6} \div \sqrt[3]{a^9b^{-3}} $.
5. Write $7(t-3)^{\frac{1}{2}} - 2(t-3)^{\frac{3}{2}}$ in the form $(t-3)^p(kt + m)$.

**Tip:** When simplifying expressions involving radicals, always look to express them in their simplest forms by factoring out common terms.

Simplify $\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}$, giving your answer in the form $x^{a} y^{b}$, where $a$ and $b$ are integers. (b) (i) Show that $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}$ can be written in the form $(t-2)^{p}(q t+r)$, where $p$, $q$ and $r$ are constants to be found. (ii) Hence solve the equation $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}=0$.

Explore the process of simplifying radicals and solving equations involving exponents with step-by-step solutions. This problem covers simplification of radicals, factoring, and solving quadratic-type equations. Ideal for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation