Question

1. simplify: \\(\sqrt{40x^{10}y^8}\\)

\\(2x^5y^4\sqrt{10}\\)

1. solve the radical equation: \\(\sqrt{5x - 11} = 7\\)

1. solve: \\(\sqrt3{x} - 4 = 3\\) (2 pts)

Explanation:

Problem 5: Simplify \(\sqrt{40x^{10}y^6}\)

Step 1: Factor the radicand

Factor \(40\) into \(4 \times 10\), \(x^{10}\) as \((x^5)^2\), and \(y^6\) as \((y^3)^2\). So we have \(\sqrt{4\times10\times(x^5)^2\times(y^3)^2}\).

Step 2: Use the square - root property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) and \(\sqrt{a^2}=a\) (\(a\geq0\))

We know that \(\sqrt{4\times10\times(x^5)^2\times(y^3)^2}=\sqrt{4}\times\sqrt{10}\times\sqrt{(x^5)^2}\times\sqrt{(y^3)^2}\). Since \(\sqrt{4} = 2\), \(\sqrt{(x^5)^2}=x^5\) and \(\sqrt{(y^3)^2}=y^3\), we get \(2x^5y^3\sqrt{10}\).

Step 1: Eliminate the square root

Square both sides of the equation \(\sqrt{5x - 11}=7\) to get rid of the square root. \((\sqrt{5x - 11})^2=7^2\), which simplifies to \(5x-11 = 49\).

Step 2: Solve for \(x\)

Add \(11\) to both sides of the equation \(5x-11 = 49\): \(5x-11 + 11=49 + 11\), so \(5x=60\). Then divide both sides by \(5\): \(\frac{5x}{5}=\frac{60}{5}\), which gives \(x = 12\).

Step 3: Check the solution

Substitute \(x = 12\) into the original equation: \(\sqrt{5\times12-11}=\sqrt{60 - 11}=\sqrt{49}=7\), which is equal to the right - hand side. So \(x = 12\) is a valid solution.

Step 1: Isolate the cube root

Add \(4\) to both sides of the equation \(\sqrt[3]{x}-4 = 3\): \(\sqrt[3]{x}-4 + 4=3 + 4\), so \(\sqrt[3]{x}=7\).

Step 2: Eliminate the cube root

Cube both sides of the equation \(\sqrt[3]{x}=7\). \((\sqrt[3]{x})^3=7^3\), which gives \(x = 343\).

Step 3: Check the solution

Substitute \(x = 343\) into the original equation: \(\sqrt[3]{343}-4=7 - 4 = 3\), which is equal to the right - hand side. So \(x = 343\) is a valid solution.

Answer:

\(2x^{5}y^{3}\sqrt{10}\)

Problem 6: Solve the Radical Equation \(\sqrt{5x - 11}=7\)