A quadratic equation is an equation of the general form ax2 + bx + c = 0 where a ≠ 0. There are several techniques that can be used to find its roots. One of them is by factorization, which will be discussed in another article. The cons of the factorization is that it is hard to apply if the roots of the equation are irrational numbers. In this post, we will be concerned with solving quadratic equations by the quadratic formula.

Let the quadratic equation under consideration be ax2 + bx + c = 0, where a ≠ 0. Subtract c from each side of the equation. This gives

ax2 + bx = – c

Divide both sides by a. (It is allowed because a ≠ 0.)

It follows that

Note that (p + q)2 = p2 + 2pq + q2. Substituting p = x and q = \frac{b}{2a} into the formula, the left-hand side of the last equation above is equal to (x + \frac{b}{2a})^2. Consequently, we have the following.

Taking the square roots of both sides of the equation, we get:

The formula above is usually called the quadratic formula.

By defining D = b2 – 4ac the quadratic formula can be rewritten as:

D is called the discriminant. There are three possibilities for D, namely D > 0, D = 0, or D < 0.

Case I: D > 0

If D > 0 then the quadratic equation has two distinct roots, that is:

Case II: D = 0

If D = 0 then x1 = x2. In this case, the quadratic equation has exactly one root, i.e. x = - \frac{b}{2a}.

Case III: D < 0

If D < 0 then the quadratic equation has no real roots, meaning that the roots are not real numbers. The roots are complex numbers.

 

Example 1 (D > 0)

Find the roots of the equation 2x2 – 5x – 3 = 0.

Answer

In this example, a = 2, b = -5, and c = -3. Substituting these values ​​into the quadratic formula, we get:

and

In conclusion, the equation has two distinct roots, namely x1 = 3 and x2 = – ½.

 

Example 2 (D = 0)

Find the solution to the equation -x2 + 6x – 9 = 0.

Answer

In this example, a = -1, b = 6, and c = -9. Substituting these values ​​into the quadratic formula, we get:

x = 3

 

Example 3 (D < 0)

Find the roots of the quadratic equation 5x2 – 3x + 10 = 0.

Answer

In this example, a = 5, b = -3, and c = 10. Note that D = (-3)2 – 4.5.10 = – 191 < 0. Since D < 0, the quadratic equation has no real roots.

 

Example 4

Find the roots of \frac{x^2 + 1}{x^2 - x + 1} = \frac{5}{3}.

Answer

\frac{x^2 + 1}{x^2 - x + 1} = \frac{5}{3}

Multiplying both sides by 3(x2 – x + 1), we have:

3(x2 + 1) = 5(x2 – x + 1)

Expanding the expressions on both sides of the equation just-obtained and rearranging the terms in the equation, we get:

2x2 – 5x + 2 = 0

Thus, we now have a quadratic equation in the general form ax2 + bx + c = 0 where a ≠ 0. In this example, a = 2, b = -5, and c = 2. Substituting these values ​​into the quadratic formula, we get:

x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm 3}{4}

x_1 = \frac{5+3}{4} = 2

x_1 = \frac{5-3}{4} = \frac{1}{2}

Thus, the equation has two distinct roots, namely x1 = 2 and x2 = ½.

 

Example 5

Given the quadratic equation 2x2 – 8x + k = 0, determine the value of k so that the equation has exactly one root. Also, find the root of the equation.

Answer

The discriminant of the equation is D = (-8)2 – 4⋅2⋅k = 64 – 8k.

In order that the equation has exactly one root, it must hold that D = 0. Consequently, we have:

64 – 8k = 0

k = 8

Substituting k = 8 into the equation gives  2x2 – 8x + 8 = 0, which is equivalent to x2 – 4x + 4 = 0.

The only root of the equation is x = - \frac{b}{2a} = - \frac{-4}{2 \cdot 1} = 2.

 

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