Quadratic Equations Flashcards
Example: Which of the following are quadratic equations?
a) x2 – 2x + 1 = 0
b) x2 = 9
c) x + 2 = -2
Solution: The equations in letters a and b are quadratic equations since the highest exponent of their x (or variable) is 2. On the other hand, c is not a quadratic equation since the highest exponent of its x (or variable) is 1, making it a linear equation.
What is the Standard Form of a Quadratic Equation
The standard The standard form of a quadratic equation is ax2 + bx + c = 0
Example: Determine the values of a, b, and c (the real number parts) in 2x2 + 4x – 1 = 0
Solution: Since the 2x2 + 4x – 1 = 0 is already in standard form, then the values of a, b, and c are easy to determine:
a = 2 (the numerical coefficient of 2x2)
b = 4 (the numerical coefficient of 4x)
c = -1 (the constant term is -1)
The a, b, and c of a quadratic equation can be determined only once we have expressed it in standard form ax2 + bx + c = 0. If a quadratic equation is not yet in the standard form, we cannot immediately tell the values of a, b, and c.
What are the Different Forms of Quadratic Equation
ax2 = c or ax2 + c = 0 Form.
(x + a)(x + b) Form or the Factored Form.
Example: Which of the following are quadratic equations?
a) (x + 2)(x – 1) = 0
b) x2 + x3 = -9
c) 2x2 + 3x = -1
d) x2 = 1
Solution:
Equation a is a quadratic equation in factored form.
Equation b is NOT a quadratic equation since the highest exponent of its variable is 3.
Equation c is a quadratic equation but not yet in standard form. We can transpose -1 to the left side so that it will be in standard form.
Equation d is a quadratic equation in ax2 = c form.
Thus, equations a, c, and d are all quadratic equations.
Here are the steps to solve quadratic equations by extracting the square root:
Isolate the square variable (x2) from other quantities. This means that x2 must be the only quantity on the left-hand side and other quantities must be on the right-hand side.
Take the square root of both sides of the equation.
Example 1: Solve for x in x2 = 9
Solution:
Step 1: Isolate the square variable (x2) from other quantities.
x2 is the only quantity on the left-hand side of x2 = 9. This means that x2 is already isolated from other quantities. Thus, we can skip this step.
Step 2: Take the square root of both sides of the equation.
We get the square root of both sides of the equation.
quadratic equations 6
Notes:
If we get the square root of x2, the result will be x.
There are two square roots of a number: a positive root and a negative root, the reason why we put the sign ± when we take the square root of a number.
Thus, the answers are x1 = 3 and x2 = -3
The solutions of a quadratic equation are also called the roots of a quadratic equation. Thus, when we say the roots of x2 = 9, we are referring to the solution of x2 = 9.
Example 2: What are the roots of 2x2 = 8?
Solution:
Step 1: Isolate the square variable (x2) from other quantities.
To remove the numerical coefficient and make x2 the only quantity on the left-hand side of the equation, we can divide both sides of the equation by 2.
2x2⁄2 = 8⁄2
x2 = 4
Step 2: Take the square root of both sides of the equation.
x2 = 4
√x2 = √4
x = ±2
Thus, the roots of the equation are x1 = 2 and x2 = – 2
Example 3: What are the roots of x2 + 4 = 20?
Solution:
Although the given equation seems to be not in the ax2 = 0 or ax2 + c = 0 form, we can manipulate the equation so that we can solve it by extracting the square root.
Step 1: Isolate the square variable (x2) from other quantities.
To isolate x2 from other quantities, we can transpose 4 to the right-hand side of the equation:
x2 + 4 = 20
x2 = – 4 + 20
x2 = 16
Step 2: Take the square root of both sides of the equation.
x2 = 16
√x2 = √16
x = ±4
Thus, the roots of the equation are x1 = 4 and x2 = – 4
Example 4: Solve for the roots of 2x2 – 6 = 0
Solution:
Step 1: Isolate the square variable (x2) from other quantities.
To isolate x2 from other quantities, we can transpose -6 to the right-hand side of the equation:
2x2 – 6 = 0
2x2 = 6
2x2⁄2 = 6⁄2
x2 = 3
Step 2: Take the square root of both sides of the equation.
x2 = 3
√x2 = √3
x = ± √3
√3 is not a perfect square number (there’s no integer multiplied to itself will give 3). Thus, we just write it as √3.
Thus, the roots of the equation are x1 = √3 and x2 = – √3
Example 5: Solve for x in x2 – 19 = 6
Solution:
x2 – 19 = 6
x2 = 19 + 6 Transposition Method
x2 = 25
√x2 = √25 Taking the square root of both sides
x = ±5
The values of x are 5 and -5
Example 6: Solve for x in (x + 4)’2 = 9
Solution:
Note that it is much easier if we start extracting the square root of both sides first so that the resulting equation is just a linear equation:
(x + 4)’2 = 9
√(x + 4)2 = √9 Taking the square root of both sides
x + 4 = ±3
Since we have two square roots for 9, we are going to have to solve two linear equations:
Equation 1: x + 4 = 3
Equation 2: x + 4 = -3
Solving for Equation 1 first:
x + 4 = 3
x = – 4 + 3 Transposition Method
x = – 1
Thus, the first root is -1
Solving for Equation 2:
x + 4 = – 3
x = – 4 + (-3)
x = – 7
Thus, the second root is -7
Therefore, the roots of the quadratic equation are – 1 and – 7.
To solve quadratic equations by factoring, follow these steps:
Express the given equation in standard form. This means that one side of the quadratic equation must be 0.
Factor the expression. You must come up with two factors after factoring.
Equate each factor to zero and solve each resulting equation.
Example 1: Solve for the roots of x2 + 5x + 4 = 0 by factoring.
Solution:
Step 1: Express the given equation in standard form. The given equation is already in standard form since it is in ax2 + bx + c = 0 form and one of the sides of the quadratic equation is already 0. Hence, we can skip this step.
Step 2: Factor the expression. To factor x2 + 5x + 4, follow the steps on factoring quadratic trinomials.
x2 + 5x + 4 = 0
(x + 4)(x + 1) = 0 by Factoring
As shown above, we can come up with two factors which are x + 4 and x + 1.
Step 3: Equate each factor to zero and solve each resulting equation. We have obtained x + 4 and x + 1 as factors of x2 + 5x + 4. We set both of these factors to zero. This means that we are going to have two linear equations.
Equation 1: x + 4 = 0
Equation 2: x + 1 = 0
Solving the equations above
Equation 1: Equation 2:
x + 4 = 0 x + 1 = 0
x = – 4 x = – 1
Thus, the roots of the equation are x1 = – 4 and x2 = – 1.
Example 2: Solve for the values of x in x2 – 7x = – 10
Solution:
Step 1: Express the given equation in standard form. To express x2 – 7x = – 10, we have to transpose – 10 to the left-hand side so that the right-hand side will be 0:
x2 – 7x = – 10
x2 – 7x + 10 = 0 Transposition Method
Step 2: Factor the expression. Factoring x2 – 7x + 10 will give us (x – 5)(x – 2)
x2 – 7x + 10 = 0
(x – 5)(x – 2) = 0 by factoring
Step 3: Equate each factor to zero and solve each resulting equation. We have obtained x – 5 and x – 2 as the factors of x2 – 7x + 10. We set both of these factors to zero. This means that we are going to have two linear equations.
Equation 1: x – 5 = 0
Equation 2: x – 2 = 0
Solving the equations above
Equation 1: Equation 2:
x – 5 = 0 x – 2 = 0
x = 5 x = 2
Thus, the roots of the equation are x1 = 5 and x2 = 2.
Example 3: Solve for the roots of 2x2 + 3x – 2 = 0 by factoring.
Solution:
Step 1: Express the given equation in standard form. Since 2x2 + 3x – 2 = 0 is already in ax2 + bx + c = 0 form and one of its side is already 0, we can skip this step.
Step 2: Factor the expression. You must come up with two factors after factoring. Factoring 2x2 + 3x – 2 will give us (2x – 1)(x + 2):
2x2 + 3x – 2 = 0
(2x – 1)(x + 2) = 0 by factoring
Step 3: Equate each factor to zero and solve each resulting equation. We have obtained 2x – 1 and x + 2 as the factors of 2x2 + 3x – 2. We set both of these factors to zero. This means that we are going to have two linear equations.
Equation 1: 2x – 1 = 0
Equation 2: x + 2 = 0
Solving the equations above
Equation 1: Equation 2:
2x – 1 = 0 x + 2 = 0
2x = 1 x = – 2
To cancel the numerical coefficient in Equation 1, we divide both of its sides by 2:
2x⁄2 = ½
x = ½
Thus, the roots of the equation are x1 = ½ and x2 = -2.