Question

If 6 times a number is decreased by 2, the principal square root of this difference is 3 less than the number. Find the number(s).

Answer 1

To solve the problem, we can set up an equation using the given information and then solve for the variable 'x' using solving techniques such as factoring, completing the square, or the quadratic formula. The possible values for 'x' are 11 or 1.

Step-by-step explanation:

To solve this problem, we first need to translate the given information into an equation. Let's assume the number is represented by 'x'.

We are told that '6 times the number is decreased by 2'. This can be written as 6x - 2.

The principal square root of this difference is 3 less than the number, which can be represented as √(6x - 2) = x - 3.

To find the value(s) of 'x', we need to solve this equation for 'x'.

Squaring both sides of the equation gives us 6x - 2 = (x - 3)^2.

Expanding the square on the right side gives us 6x - 2 = x^2 - 6x + 9.

Combining like terms and rearranging the equation gives us x^2 - 12x + 11 = 0.

We can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula.

Using the quadratic formula, we can find the roots (values of 'x').

x = (-(-12) ± √((-12)^2 - 4(1)(11))) / (2(1))

Simplifying this expression gives us x = (12 ± √(144 - 44)) / 2.

x = (12 ± √(100)) / 2.

x = (12 ± 10) / 2.

Therefore, the possible values for 'x' are x = 11 or x = 1.

Answer 2

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:

`\sqrt[]{6x-2}`

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:

`\sqrt[]{6x-2}=x-3`

Know let's find x. First I'm going to get rid of the square root by squaring both sides of the equation:

`\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}`

Know let's move all terms to the right side:

`\begin{gathered} 6x-2=x^2-6x+9 \\ 0=x^2-6x+9-6x+2 \\ 0=x^2-12x+11 \end{gathered}`

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:

`\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=(12\pm10)/(2)=6\pm5 \end{gathered}`

So we have two possible values for x, 1 and 11. Then the solution set is {1,11}