Question

A whole number is squared and doubled to give a result which is four more than the product of seven and the number.

Answer 1

Let x be the number.

We know that we squared ir and doubled it, this means:

`2x^2`

This have to be equal to four more than the product of seven and the number:

`2x^2=7x+4`

this is the same as:

`2x^2-7x-4=0`

this can be solved for the general formula:

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

in our case a=2, b=-7 and c=-4; then:

`\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(2)(-4)}}{2(2)} \\ x=\frac{7\pm\sqrt[]{49+32}}{4} \\ x=\frac{7\pm\sqrt[]{81}}{4} \\ x=(7\pm9)/(4) \end{gathered}`

hence:

`\begin{gathered} x=(7+9)/(4)=(16)/(4)=4 \\ or \\ x=(7-9)/(4)=-(2)/(4)=-(1)/(2) \end{gathered}`

Since we are looking for a whole number, then the number is 4.