Answer :
To determine which equation has exactly 1 solution, let's analyze each option:
a) [tex]\( x^2 + 116 = 116 \)[/tex]
- Simplify by subtracting 116 from both sides:
[tex]\[
x^2 = 0
\][/tex]
- The only solution to this equation is [tex]\( x = 0 \)[/tex]. So, this equation has exactly 1 solution.
b) [tex]\( x^2 + 35 = 166 \)[/tex]
- Simplify by subtracting 35 from both sides:
[tex]\[
x^2 = 131
\][/tex]
- To solve for [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[
x = \pm \sqrt{131}
\][/tex]
- This equation has 2 solutions: [tex]\( x = \sqrt{131} \)[/tex] and [tex]\( x = -\sqrt{131} \)[/tex].
c) [tex]\( -x^2 + 35 = 116 \)[/tex]
- Simplify by subtracting 35 from both sides:
[tex]\[
-x^2 = 81
\][/tex]
- Multiply both sides by -1 to solve for [tex]\( x^2 \)[/tex]:
[tex]\[
x^2 = -81
\][/tex]
- Since [tex]\( x^2 = -81 \)[/tex] is an invalid equation in the real number system (as squaring a real number cannot produce a negative result), this equation has no real solutions.
Given these results:
- Option (a) has exactly 1 solution.
- Option (b) has 2 solutions.
- Option (c) has no real solutions.
Hence, the equation that has exactly 1 solution is option a) [tex]\( x^2 + 116 = 116 \)[/tex].
a) [tex]\( x^2 + 116 = 116 \)[/tex]
- Simplify by subtracting 116 from both sides:
[tex]\[
x^2 = 0
\][/tex]
- The only solution to this equation is [tex]\( x = 0 \)[/tex]. So, this equation has exactly 1 solution.
b) [tex]\( x^2 + 35 = 166 \)[/tex]
- Simplify by subtracting 35 from both sides:
[tex]\[
x^2 = 131
\][/tex]
- To solve for [tex]\( x \)[/tex], take the square root of both sides:
[tex]\[
x = \pm \sqrt{131}
\][/tex]
- This equation has 2 solutions: [tex]\( x = \sqrt{131} \)[/tex] and [tex]\( x = -\sqrt{131} \)[/tex].
c) [tex]\( -x^2 + 35 = 116 \)[/tex]
- Simplify by subtracting 35 from both sides:
[tex]\[
-x^2 = 81
\][/tex]
- Multiply both sides by -1 to solve for [tex]\( x^2 \)[/tex]:
[tex]\[
x^2 = -81
\][/tex]
- Since [tex]\( x^2 = -81 \)[/tex] is an invalid equation in the real number system (as squaring a real number cannot produce a negative result), this equation has no real solutions.
Given these results:
- Option (a) has exactly 1 solution.
- Option (b) has 2 solutions.
- Option (c) has no real solutions.
Hence, the equation that has exactly 1 solution is option a) [tex]\( x^2 + 116 = 116 \)[/tex].
