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Examples of Solving Quadratic Equations

Solution: Step 1. Write the quadratic equation in standrad form : x² - 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 : x² - 5x - 24 = 8² - 5(8) - 24 = 64 - 40 - 24 = 64 - 64 = 0; when x = -3 : x² - 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 : x² + 4 x - 2 = 0

Solution
Step 1. Write the quadratic equation in the equivalent form : x² + 4 x = 2
Step 2. Adding the square of one-half the coefficient of the x term to both sides.: x² + 4 x + (4/2)²= 2 + (4/2)²; x² + 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 x² + 2 x - 1 = 0

Solution: Step 1. Write the quadratic equation in the equivalent form : 3 x² + 2 x = 1; Step 2. Divide both sides of the equation by the coefficient 3 of the x² 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,; 3x² + 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,; 3x² + 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.

Solution: Let t = x² , then equation x⁴ - 10x² + 9 = 0 can be expressed as t² - 10t + 9 = 0; Solve this equation by factoring t² - 10t + 9 = 0 as:; ( t - 9 ) ( t - 1 ) = 0; t - 9 = 0 and t - 1 = 0; t = 9 and t = 1; when t = 9, x² = 9, then x₁ = 3 and x₂ = -3; when t = 1, x² = 1, then x₃ = 1 and x₄ = -1; Therefore, 1, -1, 3 and -3 are the solutions.

Solution: x = t^-1, then t^-2 + 2t^-1 - 15 = 0 can be expressed as x² + 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 = 3² = 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

Solution: Let t = x + 2, then 2( x + 2 )² + 5( x + 2 ) - 3 = 0 can be expressed as; 2t² + 5t - 3 = 0; Solve this equation by factoring 2t² + 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 x² + 2 x - 1 = 0