Solve each quadratic function using the quadratic formula. Remember to get the equation equal to zero if it is not already and to leave the answer in simplest form.
a. [tex]$-8 x^2-11 x-2=0$[/tex]
b. [tex]$3 x^2+4 x-11=-12$[/tex]
c. [tex]$5 x^2+10 x-16=-12$[/tex]
d. [tex]$5 x^2-12 x+9=0$[/tex]
Answer (2)
Apply the quadratic formula x = 2 a − b ± b 2 − 4 a c to each equation.
For equation a: − 8 x 2 − 11 x − 2 = 0 , the solutions are x = 16 − 11 ± 57 .
For equation b: 3 x 2 + 4 x + 1 = 0 , the solutions are x = − 3 1 and x = − 1 .
For equation c: 5 x 2 + 10 x − 4 = 0 , the solutions are x = 5 − 5 ± 3 5 .
For equation d: 5 x 2 − 12 x + 9 = 0 , the solutions are x = 5 6 ± 3 i .
Explanation
Understanding the Quadratic Formula We are given four quadratic equations to solve using the quadratic formula. The quadratic formula is given by: x = 2 a − b ± b 2 − 4 a c where a , b , and c are the coefficients of the quadratic equation a x 2 + b x + c = 0 . We need to ensure each equation is in the standard form before applying the formula.
Solving Equation a a. The equation is − 8 x 2 − 11 x − 2 = 0 . Here, a = − 8 , b = − 11 , and c = − 2 . Substituting these values into the quadratic formula, we get: x = 2 ( − 8 ) − ( − 11 ) ± ( − 11 ) 2 − 4 ( − 8 ) ( − 2 ) x = − 16 11 ± 121 − 64 x = − 16 11 ± 57 x = 16 − 11 ± 57 So the solutions are x = 16 − 11 + 57 and x = 16 − 11 − 57 .
Solving Equation b b. The equation is 3 x 2 + 4 x − 11 = − 12 . First, we rewrite it in the standard form: 3 x 2 + 4 x + 1 = 0 . Here, a = 3 , b = 4 , and c = 1 . Substituting these values into the quadratic formula, we get: x = 2 ( 3 ) − 4 ± 4 2 − 4 ( 3 ) ( 1 ) x = 6 − 4 ± 16 − 12 x = 6 − 4 ± 4 x = 6 − 4 ± 2 So the solutions are x = 6 − 4 + 2 = 6 − 2 = − 3 1 and x = 6 − 4 − 2 = 6 − 6 = − 1 .
Solving Equation c c. The equation is 5 x 2 + 10 x − 16 = − 12 . First, we rewrite it in the standard form: 5 x 2 + 10 x − 4 = 0 . Here, a = 5 , b = 10 , and c = − 4 . Substituting these values into the quadratic formula, we get: x = 2 ( 5 ) − 10 ± 1 0 2 − 4 ( 5 ) ( − 4 ) x = 10 − 10 ± 100 + 80 x = 10 − 10 ± 180 x = 10 − 10 ± 36 × 5 x = 10 − 10 ± 6 5 x = 5 − 5 ± 3 5 So the solutions are x = 5 − 5 + 3 5 and x = 5 − 5 − 3 5 .
Solving Equation d d. The equation is 5 x 2 − 12 x + 9 = 0 . Here, a = 5 , b = − 12 , and c = 9 . Substituting these values into the quadratic formula, we get: x = 2 ( 5 ) − ( − 12 ) ± ( − 12 ) 2 − 4 ( 5 ) ( 9 ) x = 10 12 ± 144 − 180 x = 10 12 ± − 36 x = 10 12 ± 6 i x = 5 6 ± 3 i So the solutions are x = 5 6 + 3 i and x = 5 6 − 3 i .
Final Answer Therefore, the solutions to the given quadratic equations are: a. x = 16 − 11 ± 57 b. x = − 3 1 and x = − 1 c. x = 5 − 5 ± 3 5 d. x = 5 6 ± 3 i
Examples
The quadratic formula is a fundamental tool used in various fields, such as physics and engineering. For instance, when calculating the trajectory of a projectile, the quadratic formula helps determine the time it takes for the projectile to reach a certain height or distance. Similarly, in electrical engineering, it can be used to analyze circuits and determine the values of components needed to achieve specific performance characteristics. Understanding and applying the quadratic formula allows engineers and scientists to solve real-world problems involving parabolic relationships.
To solve the given quadratic equations, we used the quadratic formula and found the solutions for each equation. The results include both real and complex solutions, depending on the discriminant of each quadratic. Specifically, we found two solutions for some equations and complex solutions for others.
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