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Published byPeregrine Stone Modified over 4 years ago

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Remember! For a square root, the index of the radical is 2.

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**Example 1A: Solving Equations Containing One Radical**

Solve each equation. Check Subtract 5. Simplify. Square both sides. Simplify. Solve for x.

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**Example 1B: Solving Equations Containing One Radical**

Solve each equation. Check 3 7 7 5x = Divide by 7. 7 Simplify. Cube both sides. Simplify. Solve for x.

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**Example 4A: Solving Equations with Rational Exponents**

Solve each equation. 1 3 (5x + 7) = 3 Write in radical form. Cube both sides. 5x + 7 = 27 Simplify. 5x = 20 Factor. x = 4 Solve for x.

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**Example 2: Solving Equations Containing Two Radicals**

Solve Square both sides. 7x + 2 = 9(3x – 2) Simplify. 7x + 2 = 27x – 18 Distribute. 20 = 20x Solve for x. 1 = x

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Example 3 Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. –3x + 33 = 25 – 10x + x2 Simplify. 0 = x2 – 7x – 8 Write in standard form. 0 = (x – 8)(x + 1) Factor. x – 8 = 0 or x + 1 = 0 Solve for x. x = 8 or x = –1

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Example 3 Continued Method 2 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. 3 –3 x Because x = 8 is extraneous, the only solution is x = –1.

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**Check It Out! Example 3a Continued**

Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. 2x + 14 = x2 + 6x + 9 0 = x2 + 4x – 5 Write in standard form. Factor. 0 = (x + 5)(x – 1) x + 5 = 0 or x – 1 = 0 Solve for x. x = –5 or x = 1

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**Check It Out! Example 3a Continued**

Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. –2 x Because x = –5 is extraneous, the only solution is x = 1.

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HW pg. 633 #’s 27 – 41

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