Question

solving quadratics with square roots (lev question solve the following quadratic equation for all values of ( x ) in simplest form. ( 3(x + 7)^2 - 32 = -29 ) answer attempt 1 out of 2 + additional solution - no solution ( x = ) submit answer ( pm ) ( sqrt{} ))

Explanation:

Step1: Isolate the squared term

Start with the equation \( 3(x + 7)^2 - 32 = -29 \). First, add 32 to both sides to isolate the term with the square:
\( 3(x + 7)^2 = -29 + 32 \)
\( 3(x + 7)^2 = 3 \)

Step2: Solve for \((x + 7)^2\)

Divide both sides by 3:
\( (x + 7)^2 = \frac{3}{3} \)
\( (x + 7)^2 = 1 \)

Step3: Take the square root of both sides

Take the square root of both sides, remembering to consider both the positive and negative roots:
\( x + 7 = \pm\sqrt{1} \)
Since \( \sqrt{1} = 1 \), this simplifies to:
\( x + 7 = \pm 1 \)

Step4: Solve for \(x\)

Now, solve for \(x\) by subtracting 7 from both sides for each case:

• Case 1: \( x + 7 = 1 \)

Subtract 7 from both sides: \( x = 1 - 7 \)
\( x = -6 \)

• Case 2: \( x + 7 = -1 \)

Subtract 7 from both sides: \( x = -1 - 7 \)
\( x = -8 \)

Answer:

The solutions are \( x = -6 \) and \( x = -8 \). In boxed form (if we were to present both solutions together, though the problem might expect them separated or in a combined form, but since it's a quadratic with square roots, the solutions are \( \boldsymbol{x = -6} \) and \( \boldsymbol{x = -8} \). If we write them using the \(\pm\) notation from the steps, we can also express the solution process result as \( x = -7 \pm 1 \), which gives the two solutions as above. But directly, the solutions are \( x = -6 \) and \( x = -8 \).