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#1**+1 **

I'm assuming that x_{1} and x_{2 } are the roots

If the roots are 2 and -2, the we have that, using either one of them.....

(2)^2 + b - 10 = 0

4 + b - 10 = 0

-6 + b = 0

b = 6

Then our function becomes

x^2 + 6 - 10 =0

x^2 - 4 = 0

Check to see that x = 2 and x = - 2 make this = 0

CPhill Aug 16, 2017

#3**+1 **

Here's another method that works, too. Let's use the zero-product theorem. If haven't heard of the theorem, then you have probably used it before, unbeknown to you. With this theorem, we can make the following equation with the roots of \(\pm2\)

\((x+2)(x-2)=0\)

The zero-product theorem states that at least one (or both) factors must be equal to zero for the left-hand side to equal 0. I have made an equation that contains both of the desired roots. Since we know that this equation has the desired roots and \(x^2+b-10=0\) has the same desired roots, we can conclude that \((x+2)(x-2)=x^2+b-10\). Now, solve for *b*:

\((x+2)(x-2)=x^2+b-10\) | Expand the multiplication of both binomials by realizing that \((a+b)(a-b)=a^2-b^2\) |

\(x^2-4=x^2+b-10\) | Subtract x^2 from both sides of the equation. |

\(-4=b-10\) | Add 10 to both sides. |

\(6=b\) | |

This is simply another method of solving.

TheXSquaredFactor Aug 16, 2017