Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ \frac{3}{4} x^{2}-\frac{1}{4} x=\frac{1}{2} $$

Short answer

The solutions are x = 1 and x = -2/3.

Step-by-step solution

Simplify the Equation

Convert the given equation to standard quadratic form: divide each term by 1/4 to get $$3x^2 - x = 2.$$ Subtract 2 from both sides to set the equation to 0: $$3x^2 - x - 2 = 0.$$

Identify Coefficients

Identify the coefficients for the quadratic formula $$ax^2 + bx + c = 0.$$ Here, $$a = 3, $$b = -1, $$c = -2.$$

Apply the Quadratic Formula

The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ Substitute the identified values of a, b, and c: $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-2)}}{2(3)}.$$

Simplify Under the Square Root

Calculate the discriminant: $$b^2 - 4ac = (-1)^2 - 4(3)(-2).$$ This simplifies to $$1 + 24 = 25.$$

Calculate the Solutions

Simplify the expression in the quadratic formula: $$x = \frac{1 \pm \sqrt{25}}{6}.$$ This gives us two solutions: $$x = \frac{1 + 5}{6} = 1$$ and $$x = \frac{1 - 5}{6} = -\frac{2}{3}.$$

Key concepts

quadratic equations

Quadratic equations are polynomial equations of the second degree. This means the highest exponent of the variable (usually x) is 2. The general form of a quadratic equation is \[ax^2 + bx + c = 0\]. Here:

• \(a\) represents the coefficient of \(x^2\)
,
• \(b\) is the coefficient of \(x\)
, and
• \(c\) is the constant term.

Quadratic equations are essential in algebra as they model various real-world phenomena, such as projectile motion and area problems. Solving quadratic equations can be done through factoring, completing the square, or using the quadratic formula. In this exercise, we used the quadratic formula to find the solutions.

discriminant

The discriminant is a specific part of the quadratic formula that determines the nature of the solutions of the quadratic equation. It is found under the square root sign in the quadratic formula and is given by: \(b^2 - 4ac\).

• If the discriminant is positive (\(b^2 - 4ac > 0\), the quadratic equation has two different real solutions.
• If the discriminant is zero (\(b^2 - 4ac = 0\), the equation has exactly one real solution (also called a repeated root).
• If the discriminant is negative (\(b^2 - 4ac < 0\), the equation has no real solutions (the solutions are complex or imaginary).

In our example, we calculated the discriminant as \(25\), which is positive, indicating that there are two real solutions to the quadratic equation.

real solutions

Real solutions are the x-values that satisfy the quadratic equation and are actual numbers on the real number line, not imaginary or complex numbers. They occur when the discriminant is zero or positive. In the quadratic formula context: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], the presence of a square root of a positive number or zero in the numerator will yield real numbers for \(x\). In our solved exercise:

• We found that the discriminant was \(25\), indicating two real solutions.
• The solved values were: \(x = 1\) and \(x = -\frac{2}{3}\).

These are the real solutions, meaning they are legitimate and satisfy the original quadratic equation.