So first we have the expression "six times a number". Let's use x to designate the number, then this expression can be written as 6x. We are being told that this term is decresed by 2 which means that now we have 6x-2. Then we have a certain condition on the principal square root, this would be:
[tex]\sqrt[]{6x-2}[/tex]We are being told that this square root is equal to "3 less than the number". This last term can be written as x-3. Knowing that x-3 has to be equal to the square root I mentioned before we can build an equation for x:
[tex]\sqrt[]{6x-2}=x-3[/tex]Know let's find x. First I'm going to get rid of the square root by squaring both sides of the equation:
[tex]\begin{gathered} (\sqrt[]{6x-2})^2=6x-2=(x-3)^2 \\ 6x-2=(x-3)^2=x^2-6x+9 \\ 6x-2=x^2-6x+9 \end{gathered}[/tex]Know let's move all terms to the right side:
[tex]\begin{gathered} 6x-2=x^2-6x+9 \\ 0=x^2-6x+9-6x+2 \\ 0=x^2-12x+11 \end{gathered}[/tex]So we have a cuadratic function equalizing 0. This means that we can use the cuadratic formula:
Where a, b and c are the coefficients of the cuadratic equation.In this case a=1, b=-12 and c=11 so we have:
[tex]\begin{gathered} x=\frac{-(-12)\pm\sqrt[]{(-12)^2-4\cdot1\cdot11}}{2\cdot1}=\frac{12\pm\sqrt[]{144-44}}{2}=\frac{12\pm\sqrt[]{100}}{2} \\ x=\frac{12\pm10}{2}=6\pm5 \end{gathered}[/tex]So we have two possible values for x, 1 and 11. Then the solution set is {1,11}
