Answer :

Explanation

We must solve the following equation for x:

[tex]\sqrt{x^2-12x+44}=3[/tex]

We can square both sides of the equation:

[tex]\begin{gathered} (\sqrt{x^2-12x+44})^2=3^2 \\ x^2-12x+44=9 \end{gathered}[/tex]

Then we substract 9 from both sides:

[tex]\begin{gathered} x^2-12x+44-9=9-9 \\ x^2-12x+35=0 \end{gathered}[/tex]

Now we have a quadratic expression equalized to 0. The solutions to this equation are given by the quadratic solving formula. For an equation ax²+bx+c=0 the quadratic formula states that its solutions are:

[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

For the equation we found before we have a=1, b=-12 and c=35. Then its solutions are:

[tex]\begin{gathered} x=\frac{-(-12)\pm\sqrt{(-12)^2-4*1*35}}{2*1}=\frac{12\pm\sqrt{144-140}}{2}=\frac{12\pm\sqrt{4}}{2} \\ x=\frac{12\pm2}{2} \end{gathered}[/tex]

So there are two solutions:

[tex]\begin{gathered} x=\frac{12+2}{2}=\frac{14}{2}=7 \\ x=\frac{12-2}{2}=\frac{10}{2}=5 \end{gathered}[/tex]Answer

Then the answers are x=5 and x=7.

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