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Algebra
Examples of Solving Quadratic Equations
Use factoring method to solve quadratic equation
Question 1:
Use the factoring method to solve the equation : x
^{2}
- 24 = 5x
Solution
Step 1. Write the quadratic equation in standrad form : x
^{2}
- 5x - 24 = 0
Step 2. Factor the expression on the left side of the equation : ( x - 8 ) ( x + 3 ) = 0
Step 3. Set each factor equal to zero and solve the resulting first-degree equation
the solutions of the given quadratic equation are: (x - 8) = 0 or (x + 3) = 0
when (x - 8) = 0, x = 8
when (x + 3) = 0, x = - 3
Step 4. Check the solutions
when x = 8 : x
^{2}
- 5x - 24 = 8
^{2}
- 5(8) - 24 = 64 - 40 - 24 = 64 - 64 = 0
when x = -3 : x
^{2}
- 5x - 24 = (-3)
^{2}
- 5(-3) - 24 = 9 + 15 - 24 = 24 - 24 = 0
Therefore, the solutions of the given quadratic equation are 8 and -3.
Using completing the square method to solving quadratic equation
Question 2:
Use the completing the square method to solve a quadratic equation : x
^{2}
+ 4 x - 2 = 0
Solution
Step 1. Write the quadratic equation in the equivalent form : x
^{2}
+ 4 x = 2
Step 2. Adding the square of one-half the coefficient of the x term to both sides.
x
^{2}
+ 4 x + (4/2)
^{2}
= 2 + (4/2)
^{2}
x
^{2}
+ 4 x + 4= 2 + 4
Step 3. Write the left side of the equation as the square of a binomial expression and solve the resulting equation by root extraction.
(x + 2)
^{2}
= 6
solution 1: (x + 2) = + square root of 6
solution 2: (x + 2) = - square root of 6
when (x + 2) = + square root of 6, x = - 2 + square root of 6
when (x + 2) = - square root of 6, x = - 2 - square root of 6
Step 4. Check if the solutions are satisfy the given equation.
Therefore, x = -2 + square root of 6 and x = -2 - square root of 6 are two solutions of the given quadratic equation.
Using completing the square method to solving quadratic equation
Question 3:
Use the completing the square method to solve a quadratic equation : 3 x
^{2}
+ 2 x - 1 = 0
Solution
Step 1. Write the quadratic equation in the equivalent form : 3 x
^{2}
+ 2 x = 1
Step 2. Divide both sides of the equation by the coefficient 3 of the x
^{2}
term
Step 3. Adding the square of one-half the coefficient of the x term to both sides.
Step 4. Write the left side of the equation as the square of a binomial expression and solve the resulting equation.
Step 5. Check if the solutions are satisfy the given equation.
when x = 1/3,
3x
^{2}
+ 2x - 1 = 3(1/3)
^{2}
+ 2(1/3) - 1 = 3(1/9) + 2/3 - 1 = 3/9 + 2/3 - 1 = 1/3 + 2/3 - 1 = 0
so x = 1/3 satisfy the given quadratic equation. x = 1/3 is one of the solutions of the given quadratic equation.
when x = - 1,
3x
^{2}
+ 2x - 1 = 3(-1)
^{2}
+ 2(-1) -1 = 3 - 2 - 1 = 0
so x = -1 satisfy the given quadratic equation. x = -1 is one of the solutions of the given quadratic equation.
Therefore, x = 1/3 and x = -1 are the solutions of the given quadratic equation.
Using substitute a variable method to solve quadratic equation
Question 4:
Solving the equation x
^{4}
- 10x
^{2}
+ 9 = 0
Solution
Let t = x
^{2}
, then equation x
^{4}
- 10x
^{2}
+ 9 = 0 can be expressed as t
^{2}
- 10t + 9 = 0
Solve this equation by factoring t
^{2}
- 10t + 9 = 0 as:
( t - 9 ) ( t - 1 ) = 0
t - 9 = 0 and t - 1 = 0
t = 9 and t = 1
when t = 9, x
^{2}
= 9, then x
_{1}
= 3 and x
_{2}
= -3
when t = 1, x
^{2}
= 1, then x
_{3}
= 1 and x
_{4}
= -1
Therefore, 1, -1, 3 and -3 are the solutions.
Using substitute a variable method to solve quadratic equation
Question 5:
Solving the equation t
^{-2}
+ 2t
^{-1}
- 15 = 0
Solution
x = t
^{-1}
, then t
^{-2}
+ 2t
^{-1}
- 15 = 0 can be expressed as x
^{2}
+ 2x - 15 = 0
Solve this equation by factoring t
^{-2}
+ 2t
^{-1}
- 15 = 0 as
( x + 5 ) ( x - 3 ) = 0
x + 5 = 0 and x - 3 = 0
x = - 5 and x = 3
when x = - 5, t
^{-1}
= 1/t = -5, then t = -1/5
when x = 3, t
^{-1}
= 1/t = 3, then t = 1/3
Check if the solutions are correct
when t = -1/5, t
^{-1}
= 1/t = -5 , t
^{-2}
= (1/t)
^{2}
= (-5)
^{2}
= 25
then t
^{-2}
+ 2t
^{-1}
- 15 = 25 + 2(-5) - 15 = 25 - 10 - 15 = 0
when t = 1/3, t
^{-1}
= 1/t = 3 , t
^{-2}
= (1/t)
^{2}
= 3
^{2}
= 9
then t
^{-2}
+ 2t
^{-1}
- 15 = 9 + 2(3) - 15 = 9 + 6 - 15 = 0
Therefore, the solution of the equation is t = -1/5 and t = 1/3
Using substitute a variable method to solve quadratic equation
Question 6:
Solving the equation 2( x + 2 )
^{2}
+ 5( x + 2 ) - 3 = 0
Solution
Let t = x + 2, then 2( x + 2 )
^{2}
+ 5( x + 2 ) - 3 = 0 can be expressed as
2t
^{2}
+ 5t - 3 = 0
Solve this equation by factoring 2t
^{2}
+ 5t - 3 = 0 as
( 2t - 1 ) ( t + 3 ) = 0
2t - 1 = 0 and t + 3 = 0
2t = 1 or t = 1/2 and t = - 3
when t = 1/2 , x + 2 = 1/2 or x = -2 + 1/2 = - 4/2 + 1/2 = - 3/2
when t = - 3 , x + 2 = - 3 or x = - 2 - 3 = - 5
Check if x = - 3/2 and x = - 3 satisfy the given equation
when x = - 3/2, x + 2 = - 3/2 + 2 = - 3/2 + 4/2 = 1/2, (x + 2)
^{2}
= (1/2)
^{2}
= 1/4
2 ( x + 2 )
^{2}
+ 5 ( x + 2 ) - 3 = 2 ( 1/4) + 5 ( 1/2) - 3 = 1/2 + 5/2 - 3 = 6/2 - 3 = 0
Thus, x = - 3/2 is a solution of given equation.
when x = - 5, x + 2 = - 5 + 2 = - 3, (x + 2)
^{2}
= ( - 3 )
^{2}
= 9
2 ( x + 2 )
^{2}
+ 5 ( x + 2 ) - 3 = 2 ( 9 ) + 5 ( - 3 ) - 3 = 18 - 15 - 3 = 0
Thus, x = - 3 is also a solution of the given equation.
Therefore, solutions of the given equation are x = - 3/2 and x = - 5
Using quadratic formula method to solve quadratic equation
Question 7:
Use the quadratic formula to solve a quadratic equation : 3 x
^{2}
+ 2 x - 1 = 0
Solution
Therefore, x = 1/3 and x = -1 are the solutions of the equation.