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Question 18 Find all real solutions of the equation by completing the square \( x^{2}-12 x+30=30 \) \( x_{1}=\square \) and \( x_{2}=\square \) with \( x_{1}
Ask by Reese Coles. in the United States
Feb 27,2025

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Tutor-Verified Answer

Answer

\( x_{1}=0 \) and \( x_{2}=12 \)

Solution

Solve the equation \( x^{2}-12x+30=30 \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-12x+30=30\) - step1: Move the expression to the left side: \(x^{2}-12x=0\) - step2: Factor the expression: \(x\left(x-12\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-12=0\\&x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=12\\&x=0\end{align}\) - step5: Rewrite: \(x_{1}=0,x_{2}=12\) The real solutions of the equation \(x^{2}-12x+30=30\) are \(x_{1}=0\) and \(x_{2}=12\), with \(x_{1}

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Beyond the Answer

To find the solutions by completing the square, we start with the equation \( x^{2} - 12x + 30 = 30 \). First, we simplify it to \( x^{2} - 12x = 0 \). Now, we take the coefficient of \( x \), which is -12, halve it to get -6, and square it to get 36. Then, we rewrite the equation as \( x^{2} - 12x + 36 = 36 \). This can be factored as \( (x - 6)^{2} = 36 \). Taking the square root of both sides gives us \( x - 6 = 6 \) or \( x - 6 = -6 \), leading to the solutions \( x_{1} = 0 \) and \( x_{2} = 12 \). So, \( x_{1} = 0 \) and \( x_{2} = 12 \).

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