Answer :
Let's solve the equation \( x + \sqrt{5x + 10} = 8 \) step-by-step to find the correct value of \( x \).
1. Isolate the square root term:
[tex]\[ \sqrt{5x + 10} = 8 - x \][/tex]
2. Square both sides of the equation to eliminate the square root:
[tex]\[ (\sqrt{5x + 10})^2 = (8 - x)^2 \][/tex]
[tex]\[ 5x + 10 = (8 - x)^2 \][/tex]
3. Expand the squared term on the right side:
[tex]\[ 5x + 10 = 64 - 16x + x^2 \][/tex]
4. Rearrange the terms to form a standard quadratic equation:
[tex]\[ x^2 - 16x + 64 - 5x - 10 = 0 \][/tex]
[tex]\[ x^2 - 21x + 54 = 0 \][/tex]
5. Solve the quadratic equation:
To solve the quadratic equation \( x^2 - 21x + 54 = 0 \), we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \( a = 1 \), \( b = -21 \), and \( c = 54 \).
Substituting these values into the formula, we get:
[tex]\[ x = \frac{-(-21) \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 54}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{21 \pm \sqrt{441 - 216}}{2} \][/tex]
[tex]\[ x = \frac{21 \pm \sqrt{225}}{2} \][/tex]
[tex]\[ x = \frac{21 \pm 15}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{21 + 15}{2} = \frac{36}{2} = 18 \][/tex]
[tex]\[ x = \frac{21 - 15}{2} = \frac{6}{2} = 3 \][/tex]
6. Verify the solutions to ensure they satisfy the original equation:
- For \( x = 18 \):
[tex]\[ 18 + \sqrt{5(18) + 10} = 18 + \sqrt{90 + 10} = 18 + \sqrt{100} = 18 + 10 = 28 \neq 8 \][/tex]
This solution does not satisfy the original equation.
- For \( x = 3 \):
[tex]\[ 3 + \sqrt{5(3) + 10} = 3 + \sqrt{15 + 10} = 3 + \sqrt{25} = 3 + 5 = 8 \][/tex]
This solution satisfies the original equation.
Therefore, \( x = 3 \) is the correct solution.
The correct answer is [tex]\( \boxed{3} \)[/tex].
1. Isolate the square root term:
[tex]\[ \sqrt{5x + 10} = 8 - x \][/tex]
2. Square both sides of the equation to eliminate the square root:
[tex]\[ (\sqrt{5x + 10})^2 = (8 - x)^2 \][/tex]
[tex]\[ 5x + 10 = (8 - x)^2 \][/tex]
3. Expand the squared term on the right side:
[tex]\[ 5x + 10 = 64 - 16x + x^2 \][/tex]
4. Rearrange the terms to form a standard quadratic equation:
[tex]\[ x^2 - 16x + 64 - 5x - 10 = 0 \][/tex]
[tex]\[ x^2 - 21x + 54 = 0 \][/tex]
5. Solve the quadratic equation:
To solve the quadratic equation \( x^2 - 21x + 54 = 0 \), we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where \( a = 1 \), \( b = -21 \), and \( c = 54 \).
Substituting these values into the formula, we get:
[tex]\[ x = \frac{-(-21) \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 54}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{21 \pm \sqrt{441 - 216}}{2} \][/tex]
[tex]\[ x = \frac{21 \pm \sqrt{225}}{2} \][/tex]
[tex]\[ x = \frac{21 \pm 15}{2} \][/tex]
This gives two potential solutions:
[tex]\[ x = \frac{21 + 15}{2} = \frac{36}{2} = 18 \][/tex]
[tex]\[ x = \frac{21 - 15}{2} = \frac{6}{2} = 3 \][/tex]
6. Verify the solutions to ensure they satisfy the original equation:
- For \( x = 18 \):
[tex]\[ 18 + \sqrt{5(18) + 10} = 18 + \sqrt{90 + 10} = 18 + \sqrt{100} = 18 + 10 = 28 \neq 8 \][/tex]
This solution does not satisfy the original equation.
- For \( x = 3 \):
[tex]\[ 3 + \sqrt{5(3) + 10} = 3 + \sqrt{15 + 10} = 3 + \sqrt{25} = 3 + 5 = 8 \][/tex]
This solution satisfies the original equation.
Therefore, \( x = 3 \) is the correct solution.
The correct answer is [tex]\( \boxed{3} \)[/tex].
