LESSON: The Quadratic Formula

DURATION: 40 minutes

LEVEL: CSEC General Mathematics

LEARNING OBJECTIVES

By the end of this lesson, students should be able to:

• solve quadratic equations using the quadratic formula.

---

INTRODUCTION

For a quadratic equation of the form \(ax^2+bx+c=0\), the quadratic formula states that

\(x={-b±\sqrt{b^2-4ac}\over2a}\)

The formula therefore gives the value(s) of the root(s) of the corresponding quadratic equation. This formula is predominantly used when the method of factorisation cannot be applied. When a question requires the roots of a quadratic equation to be given to a specified number of decimal places or significant figures, this is a clear indication that the quadratic formula should be used.

ACTIVITY 1

EXAMPLE 1

Solve the equation \(3x^2-7x+1=0\), correct to 2 decimal places.

SOLUTION

STEP 1: State the values of \(a, b\) and \(c\).

\(3x^2-7x+1=0\)

\(a=3, b=-7, c=1\)

STEP 2: Substitute these values into the quadratic formula

\(x={-b±\sqrt{b^2-4ac}\over2a}\)

\(x={-(-7)±\sqrt{(-7)^2-4(3)(1)}\over2(3)}\)

STEP 3: Simplify the calculations

\(x={7±\sqrt{49-12}\over6}\)

\(x={7±\sqrt{37}\over6}\)

STEP 4: Separate the calculation into 2 parts. One part covers the plus sign and the other covers the subtraction sign. These values will be the roots of the quadratic equation.

\(x={7+\sqrt{37}\over6}=2.18\)

\(x={7-\sqrt37\over6}=0.15\)

HINT: If your calculator indicates an error, that is most likely an indication that the calculation under the square root sign INCORRECTLY resulted in a negative value.

---

EXAMPLE 2

Solve the equation \(5x^2+4x=7+x\), correct to 2 decimal places.

SOLUTION

STEP 1: Rewrite the equation in the form \(ax^2+bx+c=0\).

\(5x^2+4x=7+x\)

\(5x^2+4x-x-7=0\)

\(5x^2+3x-7=0\)

STEP 2: State the values of \(a, b\) and \(c\).

\(a=5, b=3, c=-7\)

STEP 3: Substitute these values into the quadratic formula

\(x={-b±\sqrt{b^2-4ac}\over2a}\)

\(x={-3±\sqrt{3^2-4(5)(-7)}\over2(5)}\)

STEP 4: Simplify the calculations

\(x={-3±\sqrt{9+140}\over10}\)

\(x={-3±\sqrt{149}\over6}\)

STEP 5: Separate the calculation into 2 parts. One part covers the plus sign and the other covers the subtraction sign. These values will be the roots of the quadratic equation.

\(x={-3+\sqrt{149}\over6}=0.92\)

\(x={-3-\sqrt{149}\over6}=-1.52\)

TRY THIS

If there is an error, please reload the page.