Solving for x in Radical Equations
In solving for x in a radical equation, first, remove the radical sign by raising both sides of the equation with the given index. Simplify. Next, if x has a numerical coefficient, divide both sides of the equation with the numerical coefficient of x. Again simplify. Check if the equation is balanced.
Learning Task 1:
Answers:
- x = 81
- x = 72
- x = [tex]\frac{9}{4}[/tex]
- x = 25
- x = 27
- x = 9
- x = 32
- x = 189
- x = 1
- x = 64
- x = 8
- x = ±12
- x = 3, x = -1
- x = -1/4, x = 1
- x = 0
Solutions:
1. [tex]\sqrt{x}[/tex] = 9 2. [tex]\sqrt[3]{3x}[/tex] = 6 3. 4[tex]\sqrt{x}[/tex] = 6
([tex]\sqrt{x}[/tex])² = 9² ([tex]\sqrt[3]{3x}[/tex])³ = 6³ 4²([tex]\sqrt{x}[/tex])² = 6²
x = 81 3x = 216 16x = 36
[tex]\frac{3x}{3}[/tex] = [tex]\frac{216}{3}[/tex] [tex]\frac{16x}{16}[/tex] = [tex]\frac{36}{16}[/tex]
x = 72 x = [tex]\frac{9}{4}[/tex]
4. 7 + [tex]\sqrt{x}[/tex] = 12 5. [tex]\sqrt[4]{3x}[/tex] - 5 = -8 6. [tex]\sqrt{3x - 2}[/tex] = 5
7 - 7 + [tex]\sqrt{x}[/tex] = 12 - 7 [tex]\sqrt[4]{3x}[/tex] - 5 + 5 = -8 + 5 ([tex]\sqrt{3x - 2}[/tex])² = 5²
[tex]\sqrt{x}[/tex] = 5 [tex]\sqrt[4]{3x}[/tex] = -3 3x - 2 = 25
([tex]\sqrt{x}[/tex])² = 5² ([tex]\sqrt[4]{3x}[/tex])⁴ = -3⁴ 3x - 2 + 2 = 25 + 2
x = 25 3x = 81 3x = 27
[tex]\frac{3x}{3}[/tex] = [tex]\frac{81}{3}[/tex] [tex]\frac{3x}{3}[/tex] = [tex]\frac{27}{3}[/tex]
x = 27 x = 9
7. [tex]\sqrt[5]{x}[/tex] + 5 = 7 8. [tex]\sqrt[3]{x}[/tex] = 3[tex]\sqrt[3]{7}[/tex] 9. 8 - [tex]\sqrt[4]{x}[/tex] = 7
[tex]\sqrt[5]{x}[/tex] + 5 - 5 = 7 - 5 ([tex]\sqrt[3]{x}[/tex])³ = 3³([tex]\sqrt[3]{7}[/tex])³ 8 - 8 - [tex]\sqrt[4]{x}[/tex] = 7 - 8
[tex]\sqrt[5]{x}[/tex] = 2 x = 27(7) -[tex]\sqrt[4]{x}[/tex] = -1
([tex]\sqrt[5]{x}[/tex])⁵ = 2⁵ x = 189 (-[tex]\sqrt[4]{x}[/tex])⁴ = -1⁴
x = 32 x = 1
10. 3[tex]\sqrt[3]{x}[/tex] = 12 11. [tex]\sqrt{2x + 2}[/tex] = [tex]\sqrt{x + 10}[/tex] 12. [tex]\sqrt{x^2 - 144}[/tex] = 0
3³([tex]\sqrt[3]{x}[/tex])³ = 12³ ([tex]\sqrt{2x + 2}[/tex])² = ([tex]\sqrt{x + 10}[/tex] )² ( [tex]\sqrt{x^2 - 144}[/tex])² = 0²
27x = 1728 2x + 2 = x + 10 x² - 144 = 0
[tex]\frac{27x}{27}[/tex] = [tex]\frac{1728}{27}[/tex] 2x - x = 10 - 2 x² = 144
x = 64 x = 8 x = [tex]\sqrt{144}[/tex]
x = ±12
13. x + [tex]\sqrt{x^2 + 3}[/tex] = 3x 14. 2[tex]\sqrt{3x + 1}[/tex] = 4x
x - 3x + [tex]\sqrt{x^2 + 3}[/tex] = 3x - 3x 2² ([tex]\sqrt{3x + 1}[/tex])² = (4x)²
-2x + ([tex]\sqrt{x^2 + 3}[/tex])² = 0 4(3x + 1) = 16x²
-2x + x² + 3 = 0 12x + 4 = 16x²
x² - 2x + 3 = 0 -16x² + 12x + 4 = 0
(x - 3)(x + 1) = 0 -1[-16x² + 12x + 4 = 0]
x = 3, x = -1 16x² - 12x - 4 = 0
(8x + 2)(2x - 2) = 0
x = -1/4, x = 1
15. 2 + [tex]\sqrt{x + 1}[/tex] - [tex]\sqrt{x - 5}[/tex] = 0
2² + ([tex]\sqrt{x + 1}[/tex])² - ([tex]\sqrt{x - 5}[/tex])² = 0²
4 + x + 1 - x + 5 = 0
0x = -10
x = 0
How to find for x in a radical equations: https://brainly.ph/question/11045142
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