Question

Find a positive real number such that its square is equal to 11 times the number, increased by 42.

Answer 1

Let x be the unknown positive real number. Then, we can write our statement mathematically as

$x^2=11x+42$

then, by moving the right hand side, we have

$x^2-11x-42=0$

and we can solve this equation by means of the quadratic formula

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

where, in our case, a=1 (the coefficient of x squared), b =-11 and c=-42. By substituting these values into the last formula, we get

$\begin{gathered} x=\frac{-(-11)\pm\sqrt[]{(-11)^2-4(1)(-42)}}{2(1)} \\ x=\frac{11\pm\sqrt[]{121+168}}{2} \\ x=\frac{11\pm\sqrt[]{289}}{2} \\ x=\frac{11\pm17}{2} \end{gathered}$

the first solution is given by taking the + sign and the secon solution with the - sign. Then, we have

$\begin{gathered} \text{First solution: } \\ x=\frac{11+17}{2}=\frac{28}{2}=14 \\ \text{Second solution} \\ x=\frac{11-17}{2}=-\frac{6}{2}=-3 \end{gathered}$

However, we need the positive real number. Then, the solution must be x=14.