Answer :
To solve the equation [tex]\(\sqrt{a} = a - 6\)[/tex], follow these steps:
1. Square both sides of the equation to eliminate the square root:
[tex]\[ (\sqrt{a})^2 = (a - 6)^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ a = (a - 6)^2 \][/tex]
2. Expand the right side:
[tex]\[ a = a^2 - 12a + 36 \][/tex]
3. Rearrange the equation to the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ 0 = a^2 - 12a + 36 - a \][/tex]
Simplifying, we get:
[tex]\[ 0 = a^2 - 13a + 36 \][/tex]
4. Solve the quadratic equation [tex]\(a^2 - 13a + 36 = 0\)[/tex]. Factoring, we have:
[tex]\[ a^2 - 13a + 36 = (a - 4)(a - 9) = 0 \][/tex]
5. Set each factor equal to zero and solve for [tex]\(a\)[/tex]:
[tex]\[ a - 4 = 0 \quad \Rightarrow \quad a = 4 \][/tex]
[tex]\[ a - 9 = 0 \quad \Rightarrow \quad a = 9 \][/tex]
6. Check both solutions in the original equation [tex]\(\sqrt{a} = a - 6\)[/tex]:
- For [tex]\(a = 4\)[/tex]:
[tex]\[ \sqrt{4} = 4 - 6 \quad \Rightarrow \quad 2 = -2 \quad (\text{false}) \][/tex]
Thus, [tex]\(a = 4\)[/tex] is an extraneous solution.
- For [tex]\(a = 9\)[/tex]:
[tex]\[ \sqrt{9} = 9 - 6 \quad \Rightarrow \quad 3 = 3 \quad (\text{true}) \][/tex]
Thus, [tex]\(a = 9\)[/tex] is a valid solution.
Therefore, the extraneous solution is [tex]\(\boxed{4}\)[/tex].
1. Square both sides of the equation to eliminate the square root:
[tex]\[ (\sqrt{a})^2 = (a - 6)^2 \][/tex]
Simplifying both sides, we get:
[tex]\[ a = (a - 6)^2 \][/tex]
2. Expand the right side:
[tex]\[ a = a^2 - 12a + 36 \][/tex]
3. Rearrange the equation to the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ 0 = a^2 - 12a + 36 - a \][/tex]
Simplifying, we get:
[tex]\[ 0 = a^2 - 13a + 36 \][/tex]
4. Solve the quadratic equation [tex]\(a^2 - 13a + 36 = 0\)[/tex]. Factoring, we have:
[tex]\[ a^2 - 13a + 36 = (a - 4)(a - 9) = 0 \][/tex]
5. Set each factor equal to zero and solve for [tex]\(a\)[/tex]:
[tex]\[ a - 4 = 0 \quad \Rightarrow \quad a = 4 \][/tex]
[tex]\[ a - 9 = 0 \quad \Rightarrow \quad a = 9 \][/tex]
6. Check both solutions in the original equation [tex]\(\sqrt{a} = a - 6\)[/tex]:
- For [tex]\(a = 4\)[/tex]:
[tex]\[ \sqrt{4} = 4 - 6 \quad \Rightarrow \quad 2 = -2 \quad (\text{false}) \][/tex]
Thus, [tex]\(a = 4\)[/tex] is an extraneous solution.
- For [tex]\(a = 9\)[/tex]:
[tex]\[ \sqrt{9} = 9 - 6 \quad \Rightarrow \quad 3 = 3 \quad (\text{true}) \][/tex]
Thus, [tex]\(a = 9\)[/tex] is a valid solution.
Therefore, the extraneous solution is [tex]\(\boxed{4}\)[/tex].
