The first question in the image asks us to solve the quadratic equation $5x^2 - 4x - 5 = 0$ using the quadratic formula. Here's the solution:

The quadratic formula is:

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

For the equation $5x^2 - 4x - 5 = 0$, the coefficients are:

• $a = 5$
• $b = -4$
• $c = -5$

Step 1: Calculate the discriminant ($\Delta$):

$\Delta = b^2 - 4ac$ Substituting the values: $\Delta = (-4)^2 - 4(5)(-5) = 16 + 100 = 116$

Step 2: Substitute into the quadratic formula:

$x = \frac{-(-4) \pm \sqrt{116}}{2(5)}$ $x = \frac{4 \pm \sqrt{116}}{10}$

Step 3: Simplify $\sqrt{116}$:

$\sqrt{116} = \sqrt{4 \cdot 29} = 2\sqrt{29}$ So: $x = \frac{4 \pm 2\sqrt{29}}{10}$

Step 4: Simplify further:

$x = \frac{4}{10} \pm \frac{2\sqrt{29}}{10}$ $x = \frac{2}{5} \pm \frac{\sqrt{29}}{5}$

Final Answer:

$x = \frac{2 + \sqrt{29}}{5} \quad \text{or} \quad x = \frac{2 - \sqrt{29}}{5}$

Would you like further clarification or steps explained in detail?

Related Questions:

1. How do you calculate the discriminant for a quadratic equation?
2. What does the value of the discriminant tell us about the roots?
3. Can a quadratic equation have complex roots? If so, how do we compute them?
4. How can you verify the solution of a quadratic equation?
5. What are some real-world applications of quadratic equations?

Tip:

Always simplify square roots and fractions as much as possible to make the solution clearer!