Step 1
Given; Stephen purchases a square-shaped canvas to use for a painting. The diagonal of the square is 10 inches longer than the length of its sides.
Step 2
By Pythagoras theorem we will have;
[tex]\begin{gathered} (x+10)^2=x^2+x^2 \\ 2x^2-(x^2+20x+100)=0 \\ x^2-20x-100=0 \end{gathered}[/tex]Hence, x will be;
[tex]\begin{gathered} x_{1,\:2}=\frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:1\cdot \left(-100\right)}}{2\cdot \:1} \\ x_{1,\:2}=\frac{-\left(-20\right)\pm \:20\sqrt{2}}{2\cdot \:1} \\ x_1=\frac{-\left(-20\right)+20\sqrt{2}}{2\cdot \:1},\:x_2=\frac{-\left(-20\right)-20\sqrt{2}}{2\cdot \:1} \\ x=10\left(1+\sqrt{2}\right),\:x=10\left(1-\sqrt{2}\right) \\ Since\text{ x must be positive, x=10\lparen1+}\sqrt{2})in \end{gathered}[/tex]Therefore, the answers will be;
The side length of the canvas is best found by using the quadratic formula
because the equation is prime.
Solving the equation produces two approximate measurements, and one must be discarded for being unreasonable.
