Question
Determine the real number of solutions of the equatua \( -x^{2}+4 x-4=0 \)
Ask by Martin Bradley. in the United States
Feb 27,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The equation has one real solution: \( x=2 \).
Solution
Solve the equation \( -x^{2}+4x-4=0 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(-x^{2}+4x-4=0\)
- step1: Factor the expression:
\(-\left(x-2\right)^{2}=0\)
- step2: Divide the terms:
\(\left(x-2\right)^{2}=0\)
- step3: Simplify the expression:
\(x-2=0\)
- step4: Move the constant to the right side:
\(x=0+2\)
- step5: Remove 0:
\(x=2\)
The equation \(-x^{2}+4x-4=0\) has one real solution, which is \(x=2\).
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Simplify this solution Mind Expander
To find the real number of solutions for the equation \(-x^{2}+4x-4=0\), we can start by rearranging it into standard form: \[ x^2 - 4x + 4 = 0 \] This simplifies to: \[ (x-2)^2 = 0 \] The equation \((x-2)^2 = 0\) has exactly one solution, which is \(x = 2\). Thus, the real number of solutions for the equation is one. Taking a closer look, the graph of the quadratic function opens downwards (since the coefficient of \(x^2\) is negative), and this means that it touches the x-axis at the vertex, confirming that there is only one point of intersection.
