What are the solutions to

x2+32x+2​48 = 0?



x=−16+22and
x=−16−22

x=​16+22and
x=16−2​2

x=−32+42and
x=−32−42

x=32+42and
x=32−42

3 answers

To solve the quadratic equation \( x^2 + 32x + 248 = 0 \), we can use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation, the coefficients are:

  • \( a = 1 \)
  • \( b = 32 \)
  • \( c = 248 \)

Let's calculate the discriminant (\( b^2 - 4ac \)) first:

\[ b^2 = 32^2 = 1024 \] \[ 4ac = 4 \times 1 \times 248 = 992 \] \[ b^2 - 4ac = 1024 - 992 = 32 \]

Now, substituting the values into the quadratic formula:

\[ x = \frac{-32 \pm \sqrt{32}}{2 \times 1} \]

Calculating \( \sqrt{32} \):

\[ \sqrt{32} = 4\sqrt{2} \]

Now substitute back into the formula:

\[ x = \frac{-32 \pm 4\sqrt{2}}{2} \]

Simplifying further:

\[ x = \frac{-32}{2} \pm \frac{4\sqrt{2}}{2} \] \[ x = -16 \pm 2\sqrt{2} \]

Thus, the solutions to the equation \( x^2 + 32x + 248 = 0 \) are:

\[ x = -16 + 2\sqrt{2} \quad \text{and} \quad x = -16 - 2\sqrt{2} \]

So none of the options presented seem to be correct in this form. The solutions can also be approximately calculated as:

\[ x \approx -16 + 2(1.414) \quad \text{and} \quad x \approx -16 - 2(1.414) \] This gives us approximately:

  1. \( x \approx -13.172 \)
  2. \( x \approx -18.828 \)

Please confirm if you need further clarification!

choose one

The correct solutions to the equation \(x^2 + 32x + 248 = 0\) can be expressed as:

\[ x = -16 + 2\sqrt{2} \quad \text{and} \quad x = -16 - 2\sqrt{2} \]

Given the options you presented earlier, the closest match is:

x = −16 + 2√2 and x = −16 − 2√2.

So, the correct choice is: \[ x = -16 + 2\sqrt{2} \quad \text{and} \quad x = -16 - 2\sqrt{2} \]