Problem 29 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(2 x^{2}+x-3=0\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Quadratic Equation

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = 1 \), and \( c = -3 \).

Step 2: Use the Quadratic Formula

The quadratic formula to find the roots of the equation \( ax^2 + bx + c = 0 \) is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 3: Calculate the Discriminant

The discriminant \( \Delta \) is calculated using the formula \( b^2 - 4ac \). Substitute the values: \[ \Delta = (1)^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \]Since \( \Delta > 0 \), the equation has two distinct real solutions.

Step 4: Find the Roots

Calculate the roots by substituting the values into the quadratic formula:\[ x = \frac{-1 \pm \sqrt{25}}{2 \cdot 2} = \frac{-1 \pm 5}{4} \]This gives two solutions: \[ x = \frac{-1 + 5}{4} = 1 \]\[ x = \frac{-1 - 5}{4} = -\frac{3}{2} \]

Step 5: Verify the Solutions

Substitute \( x = 1 \) into the original equation:\[ 2(1)^2 + 1 - 3 = 0 \Rightarrow 2 + 1 - 3 = 0 \]Substitute \( x = -\frac{3}{2} \):\[ 2\left(-\frac{3}{2}\right)^2 + \left(-\frac{3}{2}\right) - 3 = 0 \Rightarrow \frac{9}{2} - \frac{3}{2} - 3 = 0 \]Both solutions satisfy the equation, confirming they are correct.

Key Concepts

Quadratic Formula

One of the building blocks for solving quadratic equations is the quadratic formula. This formula is a hero of sorts when it comes to finding the roots of these equations. You can use it when the equation takes the standard form of \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are numbers, with \( a \) not equal to zero.The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this formula:

• "\( -b \pm \sqrt{b^2 - 4ac} \)" is the part that helps us find the roots.
• The "\( \pm \)" indicates that there are usually two solutions: one with a plus sign and one with a minus sign.

With just these simple steps, you can effortlessly find the roots of any quadratic equation! You only need to know the values for \( a \), \( b \), and \( c \) from your equation to get started.

Real Solutions

When we talk about real solutions of a quadratic equation, we're essentially looking for numbers that satisfy the equation in the real number system. A quadratic equation can have either zero, one, or two real solutions.How do we determine the number of real solutions? This is where the discriminant (the part inside the square root in the quadratic formula) comes in handy. If the discriminant is:

• Greater than 0, the equation has two distinct real solutions.
• Equal to 0, there is exactly one real solution.
• Less than 0, there are no real solutions (but two complex solutions).

In our example, the discriminant was 25, which is greater than zero. This tells us that our equation, \( 2x^2 + x - 3 = 0 \), has two real solutions. These solutions are actual points on a graph where the parabola intersects the x-axis.

Discriminant

The discriminant is a key player in determining the nature of the solutions of a quadratic equation, and it's represented by \( b^2 - 4ac \). Its job is to tell us whether the roots of the equation are real or complex and how many there are.Let's break it down:

• If \( b^2 - 4ac \) (the discriminant) is positive, the roots are real and distinct. In our example, it was 25, indicating two separate real numbers where the graph of the equation crosses the x-axis.
• If the discriminant is zero, the roots are real and exactly the same. This means the parabola just touches the x-axis at one point, usually called a repeated or double root.
• If negative, the equation has no real roots, which means the parabola does not intersect the x-axis at all.

Calculating the discriminant can quickly reveal much about the equation's solutions without fully solving it.