Problem

Find the zeros of the function. Give exact answers and approximate solutions rounded to three decimal places when possible.
\[
x^{2}+7 x-3=0
\]
The exact solutions are $x=$
(Simplify your answer Use integers or fractions for any numbers in the expression. Type an exact answer, using radicals and $i$ as needed. Use a comma to separate answers as needed )
The approximate solutions to three decimal places are $x \approx$
(Use a comma to separate answers as needed.)

Answer

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Answer

\(\boxed{\text{Approximate solutions: } x \approx 0.405, x \approx -7.405}\)

Steps

Step 1 :Given the quadratic equation \(x^2 + 7x - 3 = 0\), we can use the quadratic formula to find the exact solutions:

Step 2 :\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 3 :Here, \(a = 1\), \(b = 7\), and \(c = -3\).

Step 4 :Calculate the discriminant: \(\Delta = b^2 - 4ac = 7^2 - 4(1)(-3) = 61\)

Step 5 :Plug the values into the quadratic formula: \(x = \frac{-7 \pm \sqrt{61}}{2}\)

Step 6 :The exact solutions are \(x = \frac{-7 + \sqrt{61}}{2}\) and \(x = \frac{-7 - \sqrt{61}}{2}\)

Step 7 :Approximate the solutions to three decimal places: \(x \approx 0.405\) and \(x \approx -7.405\)

Step 8 :\(\boxed{\text{Exact solutions: } x = \frac{-7 + \sqrt{61}}{2}, x = \frac{-7 - \sqrt{61}}{2}}\)

Step 9 :\(\boxed{\text{Approximate solutions: } x \approx 0.405, x \approx -7.405}\)

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