Question
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Solve the quadratic equation by completing the square. \[ x^{2}-14 x+35=0 \] First, choose the appropriate form and fill in the blanks with the correct num Then, solve the equation. Simplify your answer as much as possible. If there is more than one solution, separate them with commas. \[ (4+\square)^{2}=\square \] Form: \[ \left(x-D^{2}=\square\right. \]

Ask by Young Powell. in the United States
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions are \(x = 7 - \sqrt{14}\) and \(x = 7 + \sqrt{14}\).

Solution

Solve the equation \( x^{2}-14x+35=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x^{2}-14x+35=0\) - step1: Solve using the quadratic formula: \(x=\frac{14\pm \sqrt{\left(-14\right)^{2}-4\times 35}}{2}\) - step2: Simplify the expression: \(x=\frac{14\pm \sqrt{56}}{2}\) - step3: Simplify the expression: \(x=\frac{14\pm 2\sqrt{14}}{2}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{14+2\sqrt{14}}{2}\\&x=\frac{14-2\sqrt{14}}{2}\end{align}\) - step5: Simplify the expression: \(\begin{align}&x=7+\sqrt{14}\\&x=\frac{14-2\sqrt{14}}{2}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=7+\sqrt{14}\\&x=7-\sqrt{14}\end{align}\) - step7: Rewrite: \(x_{1}=7-\sqrt{14},x_{2}=7+\sqrt{14}\) The solutions to the quadratic equation \(x^{2}-14x+35=0\) by completing the square are \(x_{1}=7-\sqrt{14}\) and \(x_{2}=7+\sqrt{14}\).

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To solve the quadratic equation \( x^{2}-14 x+35=0 \) by completing the square, we first need to manipulate the equation to a specific form. 1. Rewrite the equation: \[ x^{2} - 14x + 35 = 0 \] Move 35 to the other side: \[ x^{2} - 14x = -35 \] 2. To complete the square, we take half of the coefficient of \( x \) (which is -14), square it, and add it to both sides. Half of -14 is -7, and squaring it gives us \( 49 \): \[ x^{2} - 14x + 49 = -35 + 49 \] This simplifies to: \[ x^{2} - 14x + 49 = 14 \] 3. Now, we can write the left side as a squared term: \[ (x - 7)^{2} = 14 \] Now we can fill in the blanks: \[ (4+\textcolor{red}{3})^{2}=\textcolor{red}{14} \] Form: \[ \left(x-\textcolor{red}{7}^{2}=\textcolor{red}{14}\right. \] 4. Next, take the square root of both sides: \[ x - 7 = \pm \sqrt{14} \] 5. Solve for \( x \): \[ x = 7 \pm \sqrt{14} \] The solutions are: \[ x = 7 + \sqrt{14}, 7 - \sqrt{14} \] So the final answers, simplified, are: \[ 7 + \sqrt{14}, 7 - \sqrt{14} \]

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