2x^2 - 8x - 7 = 0 One solution to the given equation can be written as (8-sqrt(k))/4, where k...

1. TRANSLATE the problem information

• Given information:
  • Quadratic equation: \(2x^2 - 8x - 7 = 0\)
  • One solution has the form: \(\frac{8-\sqrt{k}}{4}\)
  • Need to find: value of k

• This tells us we need to solve the quadratic and match one solution to the given form

2. INFER the approach

• Since we have a quadratic equation, we'll use the quadratic formula
• The quadratic formula will give us two solutions with ±
• We need to identify which solution matches the form \(\frac{8-\sqrt{k}}{4}\)

3. TRANSLATE equation into standard form coefficients

• Standard form: \(ax^2 + bx + c = 0\)
• From \(2x^2 - 8x - 7 = 0\): \(a = 2, b = -8, c = -7\)

4. SIMPLIFY using the quadratic formula

• Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Substitute values: \(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(-7)}}{2(2)}\)
• \(x = \frac{8 \pm \sqrt{64 + 56}}{4}\)
• \(x = \frac{8 \pm \sqrt{120}}{4}\)

5. INFER which solution matches the given form

• The two solutions are:
  • \(x = \frac{8 + \sqrt{120}}{4}\)
  • \(x = \frac{8 - \sqrt{120}}{4}\)
• The given form is \(\frac{8-\sqrt{k}}{4}\)
• This matches \(x = \frac{8 - \sqrt{120}}{4}\)

6. TRANSLATE the final comparison

• \(\frac{8-\sqrt{k}}{4} = \frac{8-\sqrt{120}}{4}\)
• Therefore: \(k = 120\)

Answer: 120

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students incorrectly identify the coefficients, especially the sign of b.

They might read \(2x^2 - 8x - 7 = 0\) and think \(b = 8\) instead of \(b = -8\). This leads to:

\(x = \frac{-8 \pm \sqrt{64 + 56}}{4} = \frac{-8 \pm \sqrt{120}}{4}\)

They then try to match \(\frac{-8-\sqrt{120}}{4}\) with \(\frac{8-\sqrt{k}}{4}\), which creates confusion since the constants don't match. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when calculating the discriminant.

They might calculate \(b^2 - 4ac = 64 - 56 = 8\) instead of \(64 + 56 = 120\) (forgetting that \(-4(2)(-7) = +56\)). This gives them \(x = \frac{8 \pm \sqrt{8}}{4}\), leading to \(k = 8\) instead of \(k = 120\).

The Bottom Line:

This problem tests whether students can systematically apply the quadratic formula while carefully tracking signs and matching algebraic forms. The key insight is recognizing that the "minus" in the given form \(\frac{8-\sqrt{k}}{4}\) corresponds to the "minus" option from the ± in the quadratic formula.