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Solving Equations: (lesson 3 of 4)

Solving Quadratic Equation

Definition:

A quadratic equation in the variable x is an equation that can be written in the form:

ax2+bx+c=0a{x^2} + bx + c = 0, a≠0a \ne 0

where a, b and c represent real number coefficients.

This form is sometimes called the standard form. The term quadratic is used for any equation where the highest power of the variable x is 2. The coefficient a cannot be zero, since otherwise it would be a linear equation.

Formula for solving quadratic equations (known as the quadratic formula):

x1,2=−b±b2−4ac2a{x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, a≠0a \ne 0

To solve quadratic equations, substitute the coefficients a, b and c into the quadratic formula.

The expression b2 - 4ac shown under the square root sign is called the discriminant, because it can "discriminate" between the all possible types of answer:

type 1: If b2 - 4ac ≥ 0 ⇒ equation has two real roots;

type 2: If b2 - 4ac = 0 ⇒ equation has two real roots but they are both the same.

type 3: If b2 - 4ac ≤ 0 ⇒ equation has two complex roots;

Example 1:

Solve the following quadratic equation using the quadratic formula.

2x2 + 7x - 15 = 0

Solution:

In this case a = 2 b = 7 c= -15

The value of the discriminant is b2 - 4ac = 72 - 4(2)(-15) = 169 (Type 1)

x1,2=−b±b2−4ac2a=−7±1694=−7±134→x1=−5{x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{ - 7 \pm \sqrt {169} }}{4} = \frac{{ - 7 \pm 13}}{4} \to {x_1} = - 5 and x2=32{x_2} = \frac{3}{2}

Exercise 1: Solve quadratic equations

Level 1

x2−4x+3=0 \color{blue}{{x^2} - 4x + 3 = 0} x1=3,x2=−1 {x_1} = 3, {x_2} = - 1
x1=−3,x2=1 {x_1} = - 3, {x_2} = 1
x1=3,x2=1 {x_1} = 3, {x_2} = 1
x1=−3,x2=−1 {x_1} = - 3, {x_2} = - 1

Level 2

3x2−4x−4=0 \color{blue}{3{x^2} - 4x - 4 = 0} x1=2,x2=−23 {x_1} = 2,{x_2} = - \frac{2}{3}
x1=−2,x2=−23 {x_1} = - 2,{x_2} = - \frac{2}{3}
x1=−2,x2=23 {x_1} = - 2,{x_2} = \frac{2}{3}
x1=2,x2=23 {x_1} = 2,{x_2} = \frac{2}{3}

Example 2:

Solve the following equation using the quadratic formula.

4x2 - 20x + 25 = 0

Solution:

In this case a = 4 b = - 20 c = 25

The value of the discriminant is b2 - 4ac = 202 - 4(4)(25) = 0

x1,2=−b±b2−4ac2a=20±08=208=52→x=−52{x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{20 \pm \sqrt 0 }}{8} = \frac{{20}}{8} = \frac{5}{2} \to x = - \frac{5}{2}

That is, in this case since the value of the discriminant is zero, the two roots of the equation have the same of 2.5.

Exercise 2: Solve quadratic equations

Level 1

x2−2x+1=0 \color{blue}{{x^2} - 2x + 1 = 0} x=2 x = 2
x=−2 x = - 2
x=−1 x = - 1
x=1 x = 1

Level 2

9x2−6x+1=0 \color{blue}{9{x^2} - 6x + 1 = 0} x=13 x = \frac{1}{3}
x=−13 x = - \frac{1}{3}
x=23 x = \frac{2}{3}
x=−23 x = - \frac{2}{3}

Example 3:

Solve the following quadratic equation using the quadratic formula.

5x2 + 2x + 3 = 0

Solution:

In this case a = 5 b = 2 c = 3

The value of the discriminant is b2 - 4ac = 42 - 4(5)(3) = - 44

Since the value of the discriminant is negative, this equation has no roots that are real numbers.

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