Question 7: Solving Using the Quadratic Formula

The given equation is:

$2x^2 - 3x - 4 = 0$

The quadratic formula is:

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

Here:

• $a = 2$,
• $b = -3$,
• $c = -4$.

Substituting these values into the formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-4)}}{2(2)}$

Simplify:

1. $-(-3) = 3$,
2. $(-3)^2 = 9$,
3. $4(2)(-4) = -32$, so $9 - (-32) = 9 + 32 = 41$.

Thus:

$x = \frac{3 \pm \sqrt{41}}{4}$

The correct answer is:

$\boxed{\frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-4)}}{2(2)}}$

This corresponds to Option 2 in the question.

Question 8: Identifying Imaginary Roots

The discriminant ($\Delta$) determines the type of roots: $\Delta = b^2 - 4ac$

• $\Delta > 0$: Real and distinct roots.
• $\Delta = 0$: Real and equal roots.
• $\Delta < 0$: Imaginary roots.

For each equation:

1. $x^2 - 1 = 0$:

   • $a = 1, b = 0, c = -1$, $\Delta = 0^2 - 4(1)(-1) = 4 > 0$ (Real roots).

2. $x^2 - 2 = 0$:

   • $a = 1, b = 0, c = -2$, $\Delta = 0^2 - 4(1)(-2) = 8 > 0$ (Real roots).

3. $x^2 + x + 1 = 0$:

   • $a = 1, b = 1, c = 1$, $\Delta = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0$ (Imaginary roots).

4. $x^2 - x - 1 = 0$:

   • $a = 1, b = -1, c = -1$, $\Delta = (-1)^2 - 4(1)(-1) = 1 + 4 = 5 > 0$ (Real roots).

The equation with imaginary roots is:

$\boxed{x^2 + x + 1 = 0}$

Do you want a detailed explanation or have additional questions?
Here are 5 related questions for further exploration:

1. How do you simplify the discriminant step-by-step?
2. What are the conditions for a quadratic equation to have one real root?
3. Can a quadratic equation with $a \neq 1$ have purely imaginary roots?
4. How can you graphically determine if a quadratic equation has imaginary roots?
5. What does the quadratic formula's $\pm$ symbol signify mathematically?

Tip: Always double-check discriminant calculations to ensure root types are determined accurately!