Exercises
What are the solutions for each equation? Use the Quadratic Formula.

Ask by Weston Weber. in the United States

Jan 24,2025

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Exercises What are the solutions for each equation? Use the Quadratic Formula. $$ \begin{array}{ll}	ext { 1. }-x^{2}+7 x-3=0 & 	ext { 2. } x^{2}+6 x=10\end{array} $$

UpStudy AI · Accepted Answer

To solve each quadratic equation using the **Quadratic Formula**, we'll follow these steps:

The **Quadratic Formula** is given by:

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

where $$ a $$, $$ b $$, and $$ c $$ are coefficients from the quadratic equation $$ ax^2 + bx + c = 0 $$.

---

### **1. Solving $$-x^2 + 7x - 3 = 0$$**

First, identify the coefficients:

$$
a = -1,\quad b = 7,\quad c = -3
$$

**Step 1: Calculate the Discriminant ($$ \Delta $$)**

$$
\Delta = b^2 - 4ac = 7^2 - 4(-1)(-3) = 49 - 12 = 37
$$

**Step 2: Apply the Quadratic Formula**

$$
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-7 \pm \sqrt{37}}{2(-1)} = \frac{-7 \pm \sqrt{37}}{-2}
$$

**Step 3: Simplify the Expression**

Multiply numerator and denominator by $$-1$$ to make the denominator positive:

$$
x = \frac{7 \mp \sqrt{37}}{2}
$$

**Solutions:**

$$
x = \frac{7 + \sqrt{37}}{2} \quad \text{and} \quad x = \frac{7 - \sqrt{37}}{2}
$$

---

### **2. Solving $$x^2 + 6x = 10$$**

First, rewrite the equation in standard form:

$$
x^2 + 6x - 10 = 0
$$

Identify the coefficients:

$$
a = 1,\quad b = 6,\quad c = -10
$$

**Step 1: Calculate the Discriminant ($$ \Delta $$)**

$$
\Delta = b^2 - 4ac = 6^2 - 4(1)(-10) = 36 + 40 = 76
$$

**Step 2: Apply the Quadratic Formula**

$$
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-6 \pm \sqrt{76}}{2(1)} = \frac{-6 \pm \sqrt{76}}{2}
$$

**Step 3: Simplify the Square Root**

$$
\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}
$$

**Step 4: Substitute Back into the Formula**

$$
x = \frac{-6 \pm 2\sqrt{19}}{2}
$$

**Step 5: Simplify the Fraction**

$$
x = \frac{-6}{2} \pm \frac{2\sqrt{19}}{2} = -3 \pm \sqrt{19}
$$

**Solutions:**

$$
x = -3 + \sqrt{19} \quad \text{and} \quad x = -3 - \sqrt{19}
$$

---

### **Summary of Solutions**

1. **For $$-x^2 + 7x - 3 = 0$$:**
   $$
   x = \frac{7 + \sqrt{37}}{2} \quad \text{and} \quad x = \frac{7 - \sqrt{37}}{2}
   $$

2. **For $$x^2 + 6x = 10$$:**
   $$
   x = -3 + \sqrt{19} \quad \text{and} \quad x = -3 - \sqrt{19}
   $$

These are the exact solutions. If you need decimal approximations:

- $$\sqrt{37} \approx 6.08276$$, so:
  $$
  x \approx \frac{7 + 6.08276}{2} \approx 6.54138 \quad \text{and} \quad x \approx \frac{7 - 6.08276}{2} \approx 0.45862
  $$

- $$\sqrt{19} \approx 4.3589$$, so:
  $$
  x \approx -3 + 4.3589 \approx 1.3589 \quad \text{and} \quad x \approx -3 - 4.3589 \approx -7.3589
  $$
The solutions are: 1. For $$-x^2 + 7x - 3 = 0$$: $$ x = \frac{7 + \sqrt{37}}{2} \quad \text{and} \quad x = \frac{7 - \sqrt{37}}{2} $$ 2. For $$x^2 + 6x = 10$$: $$ x = -3 + \sqrt{19} \quad \text{and} \quad x = -3 - \sqrt{19} $$

Exercises
What are the solutions for each equation? Use the Quadratic Formula.

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