Problem 52 - Beginning and Intermediate Algebra

Question

Find the discriminant. Use it to determine whether the solutions for each equation are A. two rational numbers B. one rational number C. two irrational numbers D. two nonreal complex numbers. Tell whether the equation can be solved using the zero-factor property, or if the quadratic formula should be used instead. Do not actually solve. $$ 4 x^{2}=4 x+3 $$

Step-by-Step Solution

Answer

Two rational numbers. Use zero-factor property.

Step 1: Convert to Standard Form

Start by converting the given equation to standard form of a quadratic equation, which is ax^{2} + bx + c = 0. Move all terms to one side of the equation: 4x^{2} - 4x - 3 = 0.

Step 2: Identify coefficients a, b, and c

Identify the coefficients from the standard form equation 4x^{2} - 4x - 3 = 0. Here, a = 4, b = -4, and c = -3.

Step 3: Calculate the Discriminant

Use the formula for the discriminant of a quadratic equation, which is D = b^{2} - 4ac. Substitute the values of a, b, and c: D = (-4)^{2} - 4(4)(-3) = 16 + 48 = 64.

Step 4: Determine the nature of the roots

Use the value of the discriminant to determine the nature of the roots: Since D > 0 and D is a perfect square (64), the solutions are two rational numbers (Case A).

Step 5: Decide how to solve the equation

Since the quadratic can be factored into rational roots, it can be solved using the zero-factor property.

Key Concepts

discriminant calculation

Starting with the discriminant calculation, you need to convert your quadratic equation into standard form. This standard form is written as \(ax^{2} + bx + c = 0\). Once you've transformed your equation into this form, identifying the coefficients \(a\), \(b\), and \(c\) becomes simpler. In the given equation \(4x^{2} = 4x + 3\), you manipulate it to obtain \(4x^{2} - 4x - 3 = 0\). Here, \(a\) is 4, \(b\) is -4, and \(c\) is -3.

The formula to find the discriminant (D) is crucial: \(D = b^{2} - 4ac\). Substitute the coefficients: \[D = (-4)^2 - 4(4)(-3) = 16 + 48 = 64\]. This number helps in determining the nature of the roots.

nature of roots

The discriminant isn't just a number; it reveals much about the roots of the quadratic equation. Let's break it down:

• If \(D > 0\) and it is a perfect square, the equation has two distinct rational roots.
• If \(D > 0\) but isn't a perfect square, the equation has two distinct irrational roots.
• If \(D = 0\), there is exactly one rational root (also called a repeated or double root).
• If \(D < 0\), the roots are complex or nonreal numbers. These come in conjugate pairs.

Given our discriminant \(D = 64\), which is greater than zero and a perfect square, this means the root nature is two rational numbers. This is derived by observing that \(D = 64\) fits the case of \(D > 0\) and being a perfect square.

quadratic formula

The quadratic formula is a powerful tool to find the roots of any quadratic equation. It is represented as \[x = \frac{{-b \pm \sqrt{{D}}}}{{2a}}\]. This formula incorporates the discriminant value you've calculated. Here’s how you use it:

Given our equation with coefficients \(a = 4\), \(b = -4\), and \(c = -3\), and knowing that \(D = 64\), we can substitute these into the formula:

\[x = \frac{{-(-4) \pm \sqrt{{64}}}}{{2(4)}} = \frac{{4 \pm 8}}{{8}}\].

Solving this would give the specific roots, but since the problem doesn’t call for solving, understanding up to this step is essential.

zero-factor property

The zero-factor property states that if a product of two factors is zero, then at least one of the factors must be zero. This property is useful when the quadratic equation can be factored into simpler binomials.

Since our equation \(4x^{2} - 4x - 3 = 0\) results in rational roots, it is factorable. You can apply the zero-factor property by expressing the equation as a product: \((4x + 3)(x - 1)=0\). Solving each factor for zero provides the roots: \(4x+3=0\) and \(x-1=0\).

This method is straightforward for equations with easily identifiable rational roots. For more complex equations, you revert to methods like the quadratic formula.