Question

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radical expressions and quadratic equations
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solve: ( x^2 + 18x = -31 )
( circ ) ( x = 9 pm sqrt{50} )
( circ ) ( x = -9 + sqrt{50} )
( circ ) ( x = -9 pm 5sqrt{2} )
( circ ) ( x = 5sqrt{2} pm 9 )

Explanation:

Step1: Complete the square for \(x^2 + 18x\)

To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides. The coefficient of \(x\) is \(18\), half of it is \(9\), and squaring it gives \(81\). So we add \(81\) to both sides of the equation \(x^2 + 18x=-31\):

$$x^2 + 18x + 81=-31 + 81$$

Step2: Rewrite the left side as a perfect square and simplify the right side

The left side \(x^2 + 18x + 81\) can be written as \((x + 9)^2\), and the right side \(-31+81 = 50\). So we have:

$$(x + 9)^2=50$$

Step3: Take the square root of both sides

Taking the square root of both sides gives:

$$x + 9=\pm\sqrt{50}$$

Step4: Simplify \(\sqrt{50}\) and solve for \(x\)

We know that \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\). Then we solve for \(x\) by subtracting \(9\) from both sides:

$$x=-9\pm5\sqrt{2}$$

Answer:

\(x = - 9\pm5\sqrt{2}\) (corresponding to the option \(x=-9\pm5\sqrt{2}\))