Solve [tex]x^2+5 x-3=0[/tex] using the quadratic formula.
A) [tex]x=-\frac{5}{2}+\sqrt{13}, x=-\frac{5}{2}-\sqrt{13}[/tex]
B) [tex]x=-\frac{5}{2}+\frac{\sqrt{37}}{2}, x=-\frac{5}{2}-\frac{\sqrt{37}}{2}[/tex]
C) [tex]x=-\frac{5}{2}+\frac{\sqrt{13}}{2}, x=-\frac{5}{2}-\frac{\sqrt{13}}{2}[/tex]
D) [tex]x=-\frac{5}{2}+\sqrt{37}, x=-\frac{5}{2}-\sqrt{37}[/tex]

Identify the coefficients: a = 1 , b = 5 , c = − 3 .
Apply the quadratic formula: x = 2 a − b ± b 2 − 4 a c ​ ​ .
Calculate the discriminant: b 2 − 4 a c = 37 .
Find the solutions: x = − 2 5 ​ + 2 37 ​ ​ and x = − 2 5 ​ − 2 37 ​ ​ .

The answer is x = − 2 5 ​ + 2 37 ​ ​ , x = − 2 5 ​ − 2 37 ​ ​ ​
Explanation

Understanding the Problem We are given the quadratic equation x 2 + 5 x − 3 = 0 . Our goal is to solve for x using the quadratic formula. The quadratic formula is a general method for finding the roots of any quadratic equation of the form a x 2 + b x + c = 0 .

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 .

Identifying Coefficients In our equation, x 2 + 5 x − 3 = 0 , we can identify the coefficients as follows:

a = 1 b = 5 c = − 3

Substituting into the Formula Now, we substitute these values into the quadratic formula:

x = 2 ( 1 ) − 5 ± 5 2 − 4 ( 1 ) ( − 3 ) ​ ​
Let's simplify this expression.

Calculating the Discriminant First, we calculate the discriminant, which is the term inside the square root:

b 2 − 4 a c = 5 2 − 4 ( 1 ) ( − 3 ) = 25 + 12 = 37

Finding the Solutions Now, we substitute the discriminant back into the quadratic formula:

x = 2 − 5 ± 37 ​ ​
This gives us two possible solutions for x :

The Final Solutions The two solutions are:

x 1 ​ = 2 − 5 + 37 ​ ​ = − 2 5 ​ + 2 37 ​ ​
x 2 ​ = 2 − 5 − 37 ​ ​ = − 2 5 ​ − 2 37 ​ ​
So, the solutions are x = − 2 5 ​ + 2 37 ​ ​ and x = − 2 5 ​ − 2 37 ​ ​ .

Selecting the Correct Option Comparing our solutions to the given options, we see that option B matches our result:

B) x = − 2 5 ​ + 2 37 ​ ​ , x = − 2 5 ​ − 2 37 ​ ​
Examples
The quadratic formula is incredibly useful in many real-world scenarios. For example, imagine you are designing a bridge and need to calculate the exact points where the supporting cables should be anchored to ensure stability. The curve of the cable can often be modeled by a quadratic equation, and using the quadratic formula, you can find the precise anchor points. Similarly, in physics, when calculating the trajectory of a projectile, such as a ball thrown in the air, the quadratic formula helps determine the range and maximum height of the projectile. These calculations are crucial for ensuring safety and accuracy in engineering and physics applications.

Answered by GinnyAnswer | 2025-07-04