Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 2 x^{2}+5 x+3=0 $$

Short answer

The solutions are \( x = -1 \) and \( x = -\frac{3}{2} \).

Step-by-step solution

- Identify coefficients

In the quadratic equation, compare the given equation with the standard form of a quadratic equation, which is: \[ ax^2 + bx + c = 0 \]Here, the given equation is: \[ 2x^2 + 5x + 3 = 0 \]Identify the coefficients: \( a = 2 \), \( b = 5 \), \( c = 3 \)

- Write down the quadratic formula

The quadratic formula is used to find the roots of a quadratic equation and is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

- Substitute coefficients into the quadratic formula

Substitute the coefficients \( a = 2 \), \( b = 5 \), and \( c = 3 \) into the quadratic formula: \[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \]

- Simplify the expression inside the square root

Calculate the value inside the square root (the discriminant): \[ 5^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1 \]So the expression becomes: \[ x = \frac{-5 \pm \sqrt{1}}{4} \]

- Simplify the square root

Take the square root of 1: \[ \sqrt{1} = 1 \]So the expression now becomes: \[ x = \frac{-5 \pm 1}{4} \]

- Solve for the two possible values of x

There are two possible solutions for \( x \):1. \( x = \frac{-5 + 1}{4} = \frac{-4}{4} = -1 \)2. \( x = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} \)

Key concepts

quadratic formula

To solve a quadratic equation, like \[ 2x^2 + 5x + 3 = 0 \], we use the quadratic formula. This formula is a life-saver when it comes to finding the roots (solutions) of quadratic equations. The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula helps us solve any quadratic equation by plugging in the coefficients from the equation. If you remember the quadratic formula, solving quadratic equations becomes much easier!

discriminant

The discriminant is a key part of the quadratic formula. It is the part under the square root, \[b^2 - 4ac\].
The value of the discriminant tells us the nature of the solutions of the quadratic equation.

• If the discriminant is positive (like 1 in our problem), the equation has two distinct real solutions.
• If the discriminant is zero, the equation has exactly one real solution.
• If the discriminant is negative, the equation has no real solutions, only complex ones.
  In our example, the discriminant was calculated as 1, so we know there are two real solutions.

real solutions

Real solutions are the solutions of the equation which are real numbers, unlike complex numbers. When we solved the quadratic equation \[ 2x^2 + 5x + 3 = 0 \], we found the discriminant to be 1, which means there are two real solutions.
By simplifying the quadratic formula with our discriminant, we found the solutions to be:

• \[ x = -1 \]
• \[ x = -\frac{3}{2} \]
  These solutions show where the parabola represented by the equation intersects the x-axis.

coefficients

In a quadratic equation like \[ 2x^2 + 5x + 3 = 0 \], the coefficients are the numbers in front of the variables. They are crucial for solving the equation using the quadratic formula.

• \[\text{a} = 2 \] is the coefficient of \[ x^2 \],
• \[\text{b} = 5 \] is the coefficient of \[ x \],
• \[\text{c} = 3 \] is the constant term.
  Working out these coefficients correctly lets us substitute them into the quadratic formula and eventually find the solution to the equation. Always double-check your coefficients before plugging them into the formula!