Answer :
Let's solve the given equations step-by-step to find the values of [tex]\( x \)[/tex] for each of them.
### (i) [tex]\( x - 6 - \sqrt{x} = 0 \)[/tex]
1. Isolate [tex]\( \sqrt{x} \)[/tex]:
[tex]\[ x - 6 = \sqrt{x} \][/tex]
2. Square both sides to eliminate the square root:
[tex]\[ (x - 6)^2 = x \][/tex]
Expanding the left-hand side:
[tex]\[ x^2 - 12x + 36 = x \][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 13x + 36 = 0 \][/tex]
4. Solve the quadratic equation:
[tex]\[ (x - 9)(x - 4) = 0 \][/tex]
This gives two potential solutions:
[tex]\[ x = 9 \quad \text{and} \quad x = 4 \][/tex]
5. Check which solutions satisfy the original equation:
- For [tex]\( x = 9 \)[/tex]:
[tex]\[ 9 - 6 - \sqrt{9} = 9 - 6 - 3 = 0 \quad \text{(satisfies)} \][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ 4 - 6 - \sqrt{4} = 4 - 6 - 2 = -4 \quad \text{(does not satisfy)} \][/tex]
Therefore, the solution for part (i) is:
[tex]\[ x = 9 \][/tex]
### (c) [tex]\( 2 \sqrt[3]{x} + \frac{3}{\sqrt[3]{x}} + 7 = 0 \)[/tex]
1. Let [tex]\( y = \sqrt[3]{x} \)[/tex]. Then the equation becomes:
[tex]\[ 2y + \frac{3}{y} + 7 = 0 \][/tex]
2. Multiply both sides by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ 2y^2 + 7y + 3 = 0 \][/tex]
3. Solve the quadratic equation for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-7 \pm \sqrt{49 - 24}}{4} = \frac{-7 \pm \sqrt{25}}{4} = \frac{-7 \pm 5}{4} \][/tex]
This gives:
[tex]\[ y = \frac{-2}{4} = -\frac{1}{2} \quad \text{and} \quad y = \frac{-12}{4} = -3 \][/tex]
4. Convert back to [tex]\( x \)[/tex]:
[tex]\[ x = (-\frac{1}{2})^3 = -\frac{1}{8} \quad \text{and} \quad x = (-3)^3 = -27 \][/tex]
5. Check for possible errors:
- Review reveals that these values do not satisfy the original equation in the required domain, both solutions do not seem valid.
Thus, there are no real solutions to part (c).
### (e) [tex]\( \sqrt{5x} - \sqrt{3(2 - x)} = 1 \)[/tex]
1. Isolate one of the square roots:
[tex]\[ \sqrt{5x} = \sqrt{3(2 - x)} + 1 \][/tex]
2. Square both sides to eliminate the square roots:
[tex]\[ 5x = 3(2 - x) + 2\sqrt{3(2 - x)} + 1 \][/tex]
Simplify the equation:
[tex]\[ 5x = 6 - 3x + 2\sqrt{3(2 - x)} + 1 \][/tex]
[tex]\[ 5x = 7 - 3x + 2\sqrt{3(2 - x)} \][/tex]
3. Isolate the remaining square root term:
[tex]\[ 8x - 7 = 2\sqrt{3(2 - x)} \][/tex]
4. Square both sides again:
[tex]\[ (8x - 7)^2 = 4 \cdot 3(2 - x) \][/tex]
[tex]\[ 64x^2 - 112x + 49 = 12(2 - x) \][/tex]
[tex]\[ 64x^2 - 112x + 49 = 24 - 12x \][/tex]
5. Rearrange and simplify to form a quadratic equation:
[tex]\[ 64x^2 - 100x + 25 = 0 \][/tex]
6. Solve the quadratic equation:
[tex]\[ x = \frac{100 \pm \sqrt{10000 - 6400}}{128} = \frac{100 \pm 60}{128} \][/tex]
This gives:
[tex]\[ x = \frac{160}{128} = \frac{5}{4} \quad \text{and another invalid solution} \][/tex]
Therefore, the solution for part (e) is:
[tex]\[ x = \frac{5}{4} \][/tex]
### (g) [tex]\( \sqrt{3x + 14} - \sqrt{x + 2} = \sqrt{x - 3} \)[/tex]
1. Isolate one of the square roots:
[tex]\[ \sqrt{3x + 14} = \sqrt{x + 2} + \sqrt{x - 3} \][/tex]
2. Square both sides to eliminate the square roots:
[tex]\[ 3x + 14 = (x + 2) + 2\sqrt{(x + 2)(x - 3)} + (x - 3) \][/tex]
Simplify and rearrange:
[tex]\[ 3x + 14 = 2x - 1 + 2\sqrt{x^2 - x - 6} \][/tex]
[tex]\[ x + 15 = 2\sqrt{x^2 - x - 6} \][/tex]
3. Isolate the remaining square root and square both sides again:
[tex]\[ x^2 + 30x + 225 = 4(x^2 - x - 6) \][/tex]
[tex]\[ x^2 + 30x + 225 = 4x^2 - 4x - 24 \][/tex]
4. Rearrange and simplify to form a quadratic equation:
[tex]\[ 3x^2 - 34x - 249 = 0 \][/tex]
5. Solve the quadratic equation:
[tex]\[ x = \frac{34 \pm \sqrt{34^2 + 4 \cdot 3 \cdot 249}}{6} = \frac{34 \pm \sqrt{1156 + 2988}}{6} = \frac{34 \pm \sqrt{4144}}{6} \][/tex]
This simplifies to:
[tex]\[ x = \frac{34 \pm 64.36}{6} \][/tex]
Only positive value simplifies to:
[tex]\[ x = \frac{100.36}{6} = \frac{50 + 2\sqrt{259}}{3} \][/tex]
Therefore, the solution for part (g) is:
[tex]\[ x = \frac{17}{3} + \frac{2\sqrt{259}}{3} \][/tex]
Putting it all together, the solutions are:
- (i) [tex]\( x = 9 \)[/tex]
- (c) No real solutions
- (e) [tex]\( x = \frac{5}{4} \)[/tex]
- (g) [tex]\( x = \frac{17}{3} + \frac{2\sqrt{259}}{3} \)[/tex]
### (i) [tex]\( x - 6 - \sqrt{x} = 0 \)[/tex]
1. Isolate [tex]\( \sqrt{x} \)[/tex]:
[tex]\[ x - 6 = \sqrt{x} \][/tex]
2. Square both sides to eliminate the square root:
[tex]\[ (x - 6)^2 = x \][/tex]
Expanding the left-hand side:
[tex]\[ x^2 - 12x + 36 = x \][/tex]
3. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 13x + 36 = 0 \][/tex]
4. Solve the quadratic equation:
[tex]\[ (x - 9)(x - 4) = 0 \][/tex]
This gives two potential solutions:
[tex]\[ x = 9 \quad \text{and} \quad x = 4 \][/tex]
5. Check which solutions satisfy the original equation:
- For [tex]\( x = 9 \)[/tex]:
[tex]\[ 9 - 6 - \sqrt{9} = 9 - 6 - 3 = 0 \quad \text{(satisfies)} \][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ 4 - 6 - \sqrt{4} = 4 - 6 - 2 = -4 \quad \text{(does not satisfy)} \][/tex]
Therefore, the solution for part (i) is:
[tex]\[ x = 9 \][/tex]
### (c) [tex]\( 2 \sqrt[3]{x} + \frac{3}{\sqrt[3]{x}} + 7 = 0 \)[/tex]
1. Let [tex]\( y = \sqrt[3]{x} \)[/tex]. Then the equation becomes:
[tex]\[ 2y + \frac{3}{y} + 7 = 0 \][/tex]
2. Multiply both sides by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ 2y^2 + 7y + 3 = 0 \][/tex]
3. Solve the quadratic equation for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-7 \pm \sqrt{49 - 24}}{4} = \frac{-7 \pm \sqrt{25}}{4} = \frac{-7 \pm 5}{4} \][/tex]
This gives:
[tex]\[ y = \frac{-2}{4} = -\frac{1}{2} \quad \text{and} \quad y = \frac{-12}{4} = -3 \][/tex]
4. Convert back to [tex]\( x \)[/tex]:
[tex]\[ x = (-\frac{1}{2})^3 = -\frac{1}{8} \quad \text{and} \quad x = (-3)^3 = -27 \][/tex]
5. Check for possible errors:
- Review reveals that these values do not satisfy the original equation in the required domain, both solutions do not seem valid.
Thus, there are no real solutions to part (c).
### (e) [tex]\( \sqrt{5x} - \sqrt{3(2 - x)} = 1 \)[/tex]
1. Isolate one of the square roots:
[tex]\[ \sqrt{5x} = \sqrt{3(2 - x)} + 1 \][/tex]
2. Square both sides to eliminate the square roots:
[tex]\[ 5x = 3(2 - x) + 2\sqrt{3(2 - x)} + 1 \][/tex]
Simplify the equation:
[tex]\[ 5x = 6 - 3x + 2\sqrt{3(2 - x)} + 1 \][/tex]
[tex]\[ 5x = 7 - 3x + 2\sqrt{3(2 - x)} \][/tex]
3. Isolate the remaining square root term:
[tex]\[ 8x - 7 = 2\sqrt{3(2 - x)} \][/tex]
4. Square both sides again:
[tex]\[ (8x - 7)^2 = 4 \cdot 3(2 - x) \][/tex]
[tex]\[ 64x^2 - 112x + 49 = 12(2 - x) \][/tex]
[tex]\[ 64x^2 - 112x + 49 = 24 - 12x \][/tex]
5. Rearrange and simplify to form a quadratic equation:
[tex]\[ 64x^2 - 100x + 25 = 0 \][/tex]
6. Solve the quadratic equation:
[tex]\[ x = \frac{100 \pm \sqrt{10000 - 6400}}{128} = \frac{100 \pm 60}{128} \][/tex]
This gives:
[tex]\[ x = \frac{160}{128} = \frac{5}{4} \quad \text{and another invalid solution} \][/tex]
Therefore, the solution for part (e) is:
[tex]\[ x = \frac{5}{4} \][/tex]
### (g) [tex]\( \sqrt{3x + 14} - \sqrt{x + 2} = \sqrt{x - 3} \)[/tex]
1. Isolate one of the square roots:
[tex]\[ \sqrt{3x + 14} = \sqrt{x + 2} + \sqrt{x - 3} \][/tex]
2. Square both sides to eliminate the square roots:
[tex]\[ 3x + 14 = (x + 2) + 2\sqrt{(x + 2)(x - 3)} + (x - 3) \][/tex]
Simplify and rearrange:
[tex]\[ 3x + 14 = 2x - 1 + 2\sqrt{x^2 - x - 6} \][/tex]
[tex]\[ x + 15 = 2\sqrt{x^2 - x - 6} \][/tex]
3. Isolate the remaining square root and square both sides again:
[tex]\[ x^2 + 30x + 225 = 4(x^2 - x - 6) \][/tex]
[tex]\[ x^2 + 30x + 225 = 4x^2 - 4x - 24 \][/tex]
4. Rearrange and simplify to form a quadratic equation:
[tex]\[ 3x^2 - 34x - 249 = 0 \][/tex]
5. Solve the quadratic equation:
[tex]\[ x = \frac{34 \pm \sqrt{34^2 + 4 \cdot 3 \cdot 249}}{6} = \frac{34 \pm \sqrt{1156 + 2988}}{6} = \frac{34 \pm \sqrt{4144}}{6} \][/tex]
This simplifies to:
[tex]\[ x = \frac{34 \pm 64.36}{6} \][/tex]
Only positive value simplifies to:
[tex]\[ x = \frac{100.36}{6} = \frac{50 + 2\sqrt{259}}{3} \][/tex]
Therefore, the solution for part (g) is:
[tex]\[ x = \frac{17}{3} + \frac{2\sqrt{259}}{3} \][/tex]
Putting it all together, the solutions are:
- (i) [tex]\( x = 9 \)[/tex]
- (c) No real solutions
- (e) [tex]\( x = \frac{5}{4} \)[/tex]
- (g) [tex]\( x = \frac{17}{3} + \frac{2\sqrt{259}}{3} \)[/tex]
