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# Examples of Solving Quadratic Equations

### Use factoring method to solve quadratic equation

### Using completing the square method to solving quadratic equation

### Using completing the square method to solving quadratic equation

### Using substitute a variable method to solve quadratic equation

### Using substitute a variable method to solve quadratic equation

### Using substitute a variable method to solve quadratic equation

### Using quadratic formula 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.

- 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) = + Sqrt (6)
- solution 2: (x + 2) = - Sqrt (6)
- when (x + 2) = + Sqrt (6), x = - 2 + Sqrt (6)
- when (x + 2) = - Sqrt (6), x = - 2 - Sqrt (6)
- Step 4. Check if the solutions are satisfy the given equation.
- Therefore, x = -2 + Sqrt (6) and x = -2 - Sqrt (6) are two solutions of the given 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.

- 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.

- 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

- 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

- 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.

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