Math Problem Statement

(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 $4(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}}=0$.

Solution

Let's solve the problem step by step.

Part (a) Simplify x8y10÷x3y63\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}

We need to simplify the expression x8y10÷x3y63\sqrt{x^{8} y^{10}} \div \sqrt[3]{x^{3} y^{-6}}.

Step 1: Simplify each radical expression

  1. x8y10\sqrt{x^8 y^{10}}: x8y10=x82y102=x4y5\sqrt{x^8 y^{10}} = x^{\frac{8}{2}} y^{\frac{10}{2}} = x^4 y^5

  2. x3y63\sqrt[3]{x^3 y^{-6}}: x3y63=x33y63=x1y2\sqrt[3]{x^3 y^{-6}} = x^{\frac{3}{3}} y^{\frac{-6}{3}} = x^1 y^{-2}

Step 2: Divide the simplified expressions

x4y5x1y2=x41y5(2)=x3y7\frac{x^4 y^5}{x^1 y^{-2}} = x^{4-1} y^{5 - (-2)} = x^3 y^{7}

Thus, the simplified form is x3y7x^3 y^7.

Part (b)(i) Show that 4(t2)12+5(t2)324(t-2)^{\frac{1}{2}}+5(t-2)^{\frac{3}{2}} can be written in the form (t2)p(qt+r)(t-2)^{p}(qt+r).

We want to express 4(t2)12+5(t2)324(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} in the form (t2)p(qt+r)(t-2)^p(qt + r).

Step 1: Factor out the common term (t2)12(t-2)^{\frac{1}{2}}:

4(t2)12+5(t2)32=(t2)12(4+5(t2))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(t2)=4+5t10=5t64 + 5(t-2) = 4 + 5t - 10 = 5t - 6

Thus, the expression can be written as: 4(t2)12+5(t2)32=(t2)12(5t6)4(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} = (t-2)^{\frac{1}{2}}(5t - 6)

Here, p=12p = \frac{1}{2}, q=5q = 5, and r=6r = -6.

Part (b)(ii) Solve the equation 4(t2)12+5(t2)32=04(t-2)^{\frac{1}{2}} + 5(t-2)^{\frac{3}{2}} = 0.

From part (b)(i), we can rewrite the equation as: (t2)12(5t6)=0(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: (t2)12=0(t-2)^{\frac{1}{2}} = 0

(t2)12=0    t2=0    t=2(t-2)^{\frac{1}{2}} = 0 \implies t-2 = 0 \implies t = 2

Case 2: 5t6=05t - 6 = 0

5t6=0    5t=6    t=655t - 6 = 0 \implies 5t = 6 \implies t = \frac{6}{5}

Final Answer:

The solutions to the equation are t=2t = 2 and t=65t = \frac{6}{5}.


Do you want further details or have any questions?

Here are some related questions for practice:

  1. Simplify x12y8÷x8y124\sqrt{x^{12} y^{8}} \div \sqrt[4]{x^{8} y^{-12}}.
  2. Express 3(t1)12+2(t1)323(t-1)^{\frac{1}{2}} + 2(t-1)^{\frac{3}{2}} in the form (t1)p(qt+r)(t-1)^p (qt + r).
  3. Solve the equation 3(t1)12+2(t1)32=03(t-1)^{\frac{1}{2}} + 2(t-1)^{\frac{3}{2}} = 0.
  4. Simplify a4b6÷a9b33\sqrt{a^4b^6} \div \sqrt[3]{a^9b^{-3}}.
  5. Write 7(t3)122(t3)327(t-3)^{\frac{1}{2}} - 2(t-3)^{\frac{3}{2}} in the form (t3)p(kt+m)(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.

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Math Problem Analysis

Mathematical Concepts

Simplification of Radicals
Exponents and Radicals
Factoring
Solving Equations

Formulas

Simplifying \(\sqrt{x^a y^b}\) as \(x^{\frac{a}{2}} y^{\frac{b}{2}}\)
Simplifying \(\sqrt[3]{x^a y^b}\) as \(x^{\frac{a}{3}} y^{\frac{b}{3}}\)

Theorems

Zero Product Property

Suitable Grade Level

Grades 10-12