Question 3 Solve the equation: $$ 2 x^{2}+7 x+3=0 $$

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UpStudy AI · Accepted Answer

To solve the quadratic equation:

$$
2x^{2} + 7x + 3 = 0
$$

we can use the **quadratic formula**, which is a standard method for solving any quadratic equation of the form $$ ax^{2} + bx + c = 0 $$. The quadratic formula is:

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

**Step 1: Identify the coefficients**

For the given equation:

$$
a = 2,\quad b = 7,\quad c = 3
$$

**Step 2: Calculate the discriminant ($$ \Delta $$)**

The discriminant determines the nature of the roots and is given by:

$$
\Delta = b^{2} - 4ac
$$

Plugging in the values:

$$
\Delta = 7^{2} - 4 \times 2 \times 3 = 49 - 24 = 25
$$

Since the discriminant is positive ($$ \Delta > 0 $$), there are two distinct real roots.

**Step 3: Apply the quadratic formula**

$$
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-7 \pm \sqrt{25}}{2 \times 2} = \frac{-7 \pm 5}{4}
$$

**Step 4: Compute both solutions**

1. **First Root ($$ x_{1} $$)**

$$
x_{1} = \frac{-7 + 5}{4} = \frac{-2}{4} = -\frac{1}{2}
$$

2. **Second Root ($$ x_{2} $$)**

$$
x_{2} = \frac{-7 - 5}{4} = \frac{-12}{4} = -3
$$

**Step 5: Final Answer**

The solutions to the equation $$ 2x^{2} + 7x + 3 = 0 $$ are:

$$
x = -\frac{1}{2} \quad \text{and} \quad x = -3
$$

---

**Alternative Method: Factoring**

Sometimes, quadratic equations can be factored into the product of two binomials. Let's attempt to factor the given equation.

$$
2x^{2} + 7x + 3 = 0
$$

We look for two numbers that multiply to $$ a \times c = 2 \times 3 = 6 $$ and add up to $$ b = 7 $$. These numbers are 6 and 1.

Rewrite the middle term using these numbers:

$$
2x^{2} + 6x + x + 3 = 0
$$

Factor by grouping:

$$
2x(x + 3) + 1(x + 3) = 0
$$

$$
(2x + 1)(x + 3) = 0
$$

Set each factor equal to zero:

1. $$ 2x + 1 = 0 $$ → $$ x = -\frac{1}{2} $$
2. $$ x + 3 = 0 $$ → $$ x = -3 $$

This confirms the solutions obtained using the quadratic formula.
The solutions to the equation $$ 2x^{2} + 7x + 3 = 0 $$ are $$ x = -\frac{1}{2} $$ and $$ x = -3 $$.

Question 3
Solve the equation:

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