Match each quadratic equation with its solution set.

[tex]2 x^2-10 x-3=0[/tex]
[tex]2 x^2-9 x+6=0[/tex]
[tex]2 x^2-8 x+5=0[/tex]
[tex]2 x^2-9 x-1=0[/tex]
[tex]2 x^2-8 x-3=0[/tex]

[tex]\frac{9 \pm \sqrt{33}}{4}[/tex]
[tex]\frac{4 \pm \sqrt{6}}{2}[/tex]
[tex]\frac{9 \pm \sqrt{89}}{4}[/tex]
[tex]\frac{4 \pm \sqrt{22}}{2}[/tex]

Apply the quadratic formula x = 2 a − b ± b 2 − 4 a c ​ ​ to each quadratic equation.
Simplify the solutions obtained from the quadratic formula.
Match the simplified solution set with the given solution sets.
The matched pairs are: 2 x 2 − 9 x + 6 = 0 → 4 9 ± 33 ​ ​ , 2 x 2 − 8 x + 5 = 0 → 2 4 ± 6 ​ ​ , 2 x 2 − 9 x − 1 = 0 → 4 9 ± 89 ​ ​ , 2 x 2 − 8 x − 3 = 0 → 2 4 ± 22 ​ ​ .

Explanation

Problem Analysis We are given five quadratic equations and five solution sets. Our goal is to match each equation with its correct solution set. We will use the quadratic formula to find the solutions for each equation and then match them to the given sets. The quadratic formula is given by: x = 2 a − b ± b 2 − 4 a c ​ ​

Solving Equation 1 For the equation 2 x 2 − 10 x − 3 = 0 , we have a = 2 , b = − 10 , and c = − 3 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − ( − 10 ) ± ( − 10 ) 2 − 4 ( 2 ) ( − 3 ) ​ ​ = 4 10 ± 100 + 24 ​ ​ = 4 10 ± 124 ​ ​ = 4 10 ± 2 31 ​ ​ = 2 5 ± 31 ​ ​ However, none of the provided solutions match this form directly. Let's re-examine the solutions provided and see if we can manipulate them to match.

Solving Equation 2 For the equation 2 x 2 − 9 x + 6 = 0 , we have a = 2 , b = − 9 , and c = 6 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − ( − 9 ) ± ( − 9 ) 2 − 4 ( 2 ) ( 6 ) ​ ​ = 4 9 ± 81 − 48 ​ ​ = 4 9 ± 33 ​ ​ This matches the solution 4 9 ± 33 ​ ​ .

Solving Equation 3 For the equation 2 x 2 − 8 x + 5 = 0 , we have a = 2 , b = − 8 , and c = 5 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − ( − 8 ) ± ( − 8 ) 2 − 4 ( 2 ) ( 5 ) ​ ​ = 4 8 ± 64 − 40 ​ ​ = 4 8 ± 24 ​ ​ = 4 8 ± 2 6 ​ ​ = 2 4 ± 6 ​ ​ This matches the solution 2 4 ± 6 ​ ​ .

Solving Equation 4 For the equation 2 x 2 − 9 x − 1 = 0 , we have a = 2 , b = − 9 , and c = − 1 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − ( − 9 ) ± ( − 9 ) 2 − 4 ( 2 ) ( − 1 ) ​ ​ = 4 9 ± 81 + 8 ​ ​ = 4 9 ± 89 ​ ​ This matches the solution 4 9 ± 89 ​ ​ .

Solving Equation 5 For the equation 2 x 2 − 8 x − 3 = 0 , we have a = 2 , b = − 8 , and c = − 3 . Plugging these values into the quadratic formula, we get: x = 2 ( 2 ) − ( − 8 ) ± ( − 8 ) 2 − 4 ( 2 ) ( − 3 ) ​ ​ = 4 8 ± 64 + 24 ​ ​ = 4 8 ± 88 ​ ​ = 4 8 ± 2 22 ​ ​ = 2 4 ± 22 ​ ​ This matches the solution 2 4 ± 22 ​ ​ .

Final Answer Therefore, the correct matches are:


2 x 2 − 10 x − 3 = 0 No match in the given options. 2 x 2 − 9 x + 6 = 0 → 4 9 ± 33 ​ ​ 2 x 2 − 8 x + 5 = 0 → 2 4 ± 6 ​ ​ 2 x 2 − 9 x − 1 = 0 → 4 9 ± 89 ​ ​ 2 x 2 − 8 x − 3 = 0 → 2 4 ± 22 ​ ​
Examples
Quadratic equations are incredibly useful in various real-world scenarios. For example, engineers use them to design bridges and calculate the forces acting upon them. Economists use them to model supply and demand curves, helping to predict market behavior. Even in sports, quadratic equations can help athletes optimize their trajectories, such as calculating the best angle to throw a ball for maximum distance. Understanding quadratic equations allows us to solve problems related to optimization, prediction, and design across many disciplines.

Answered by GinnyAnswer | 2025-07-03

The quadratic equations are matched with their solution sets using the quadratic formula. The matches are: 2 x 2 − 9 x + 6 = 0 → 4 9 ± 33 ​ ​ , 2 x 2 − 8 x + 5 = 0 → 2 4 ± 6 ​ ​ , 2 x 2 − 9 x − 1 = 0 → 4 9 ± 89 ​ ​ , and 2 x 2 − 8 x − 3 = 0 → 2 4 ± 22 ​ ​ .
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Answered by Anonymous | 2025-07-04