Question

question
solve the equation for all real solutions in simplest form.
$b^2 + 10b + 1 = -6$
answer attempt 1 out of 4

Explanation:

Step1: Rewrite the equation in standard form

First, we need to rewrite the given equation \(b^{2}+10b + 1=-6\) in the standard quadratic form \(ax^{2}+bx + c = 0\). We add 6 to both sides of the equation:
\(b^{2}+10b+1 + 6=0\)
\(b^{2}+10b + 7=0\)

Step2: Use the quadratic formula

For a quadratic equation of the form \(ax^{2}+bx + c = 0\), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). In our equation \(a = 1\), \(b=10\), and \(c = 7\).
First, we calculate the discriminant \(\Delta=b^{2}-4ac\). Substituting the values of \(a\), \(b\), and \(c\):
\(\Delta=(10)^{2}-4\times1\times7\)
\(=100 - 28\)
\(=72\)

Then we find the square root of the discriminant: \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\)

Now we substitute into the quadratic formula:
\(b=\frac{-10\pm6\sqrt{2}}{2\times1}=\frac{-10\pm6\sqrt{2}}{2}\)
We can simplify this by dividing each term in the numerator by 2:
\(b=- 5\pm3\sqrt{2}\)

Answer:

\(b=-5 + 3\sqrt{2}\) or \(b=-5-3\sqrt{2}\)