Lesson 37, Quadratic equations: Section 2
IN LESSON 18 we saw a technique called completing the square. We will now see how to apply it to solving a quadratic equation.
Completing the square
If we try to solve this quadratic equation by factoring,
The technique is valid only when the coefficient of x² is 1.
1) Transpose the constant term to the right
x² + 6x = −2.
x² + 6x + 9 = −2 + 9.
The left-hand side is now the perfect square of (x + 3).
(x + 3)² = 7.
3 is half of the coefficient 6.
That equation has the form
That is, the solutions to
x² + 6x + 2 = 0
are the conjugate pair,
−3 + , −3 − .
For a method of checking these roots, see the theorem of the sum and product of the roots: Lesson 10 of Topics in Precalculus,
Problem 6. Solve each quadratic equation by completing the square.
To see the answer, pass your cursor from left to right
Problem 7. Find two numbers whose sum is 10 and whose product is 20.
The quadratic formula
ax² + bx + c = 0,
If we call those two roots r1 and r2 , then the quadratic can be factored as
(x − r1)(x − r2).
We will prove the quadratic formula below.
Example 4. Use the quadratic formula to solve this quadratic equation:
3x² + 5x − 8 = 0
Solution. We have: a = 3, b = 5, c = −8.
Therefore, according to the formula:
Those are the two roots. And they are rational. When the roots are rational, we could have solved the equation by factoring, which is always the simplest method.
Problem 8. Use the quadratic formula to find the roots of each quadratic.
a) x² − 5x + 5
a = 1, b = −5, c = 5.
b) 2x² − 8x + 5
a = 2, b = −8, c = 5.
c) 5x² − 2x + 2
a = 5, b = −2, c = 2.
The radicand b² − 4ac is called the discriminant. If the discriminant is
Proof of the quadratic formula
To prove the quadratic formula, we complete the square. But to do that, the coefficient of x² must be 1. Therefore, we will divide both sides of the original equation by a:
on multiplying both c and a by 4a, thus making the denominators the same (Lesson 23),
This is the quadratic formula.
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