Let's solve each problem one by one.

Problem 1: Quadratic Equation

The equation given is: $-7x^2 - 11x + 6 = 0$

To solve this quadratic equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -7$, $b = -11$, and $c = 6$.

1. Calculate the discriminant: $b^2 - 4ac = (-11)^2 - 4(-7)(6) = 121 + 168 = 289$

2. Substitute into the formula: $x = \frac{-(-11) \pm \sqrt{289}}{2(-7)}$ $x = \frac{11 \pm 17}{-14}$

3. Solve for the two possible values of $x$:

   • When $11 + 17 = 28$: $x = \frac{28}{-14} = -2$
   • When $11 - 17 = -6$: $x = \frac{-6}{-14} = \frac{3}{7}$

So, the solutions are: $x = -2 \quad \text{or} \quad x = \frac{3}{7}$

Problem 2: Right-Angled Triangle

In this right-angled triangle, the sides are:

• Hypotenuse: $15 \, \text{cm}$
• Opposite side (relative to the angle): $2x + 1$
• Adjacent side: $3x$

By the Pythagorean theorem: $\text{(Hypotenuse)}^2 = \text{(Opposite side)}^2 + \text{(Adjacent side)}^2$ Substitute the given values: $15^2 = (2x + 1)^2 + (3x)^2$

Expanding and solving for $x$:

1. $225 = (2x + 1)^2 + (3x)^2$
2. $225 = (4x^2 + 4x + 1) + 9x^2$
3. Combine terms: $225 = 13x^2 + 4x + 1$
4. Rearrange to form a quadratic equation: $13x^2 + 4x + 1 - 225 = 0$
5. Simplify: $13x^2 + 4x - 224 = 0$

Now we have a quadratic equation in the form: $13x^2 + 4x - 224 = 0$

Using the quadratic formula to solve for $x$:

1. Calculate the discriminant: $b^2 - 4ac = (4)^2 - 4 \cdot 13 \cdot (-224) = 16 + 11648 = 11664$

2. Substitute into the formula: $x = \frac{-4 \pm \sqrt{11664}}{2 \cdot 13}$ $x = \frac{-4 \pm 108}{26}$

3. Solve for the two possible values of $x$:

   • When $-4 + 108 = 104$: $x = \frac{104}{26} = 4$
   • When $-4 - 108 = -112$: $x = \frac{-112}{26} = -\frac{56}{13}$ (which would be invalid as length cannot be negative)

Thus, the value of $x$ is: $x = 4$

Would you like further explanations on any of these solutions, or do you have any questions?

Related Questions

1. What is the purpose of the discriminant in solving quadratic equations?
2. How does the Pythagorean theorem help in finding unknown sides in a triangle?
3. Could this right-angled triangle problem be solved using trigonometric ratios?
4. How would the solutions change if the quadratic formula resulted in a complex solution?
5. Why can we disregard the negative solution for $x$ in the triangle problem?

Tip

Always verify your quadratic solutions by plugging them back into the original equation to ensure accuracy.