91Ó°ÊÓ

The area of a square is defined as $A=s^2$, where s is the side of the square.

Let x be the original side length of the square, then the area of the square with side x is $A=x^2$.

Now if the side lengths of the larger square are 4 inches longer than the side lengths of the original square, so, the side length of the new square will be $x+4$. And so, the area of the larger square is$A={(x+4)}^2$.

It is given that the larger square has an area that is 4 times the area of the original square so, the equation thus obtained will be ${(x+4)}^2=4x^2$.

Rewrite the equation ${(x+4)}^2=4x^2$in standard form.

$$\begin{gathered} {(x+4)}^2=4x^2 \\ (x+4)(x+4)=4x^2 \\ {x}^2+8x+16=4x^2 \\ −4x^2+x^2+8x+16=4x^2−\text{ â¶Ä‰}4x^2\text{ â¶Ä‰â€‰}\left[{\text{ â¶Ä‰Subtract â¶Ä‰â€‰â¶Ä‰4x}}^{\text{2}}\text{ â¶Ä‰from â¶Ä‰both â¶Ä‰sides â¶Ä‰}\right] \\ −3x^2+8x+16=0 \end{gathered}$$

Now solve for x by using the quadratic formula. Substitute $-3$for a, 8 for b, and 16 for c into the expression $x=\frac{−bÂ\pm \sqrt{b^2−4ac}}{2a}$.

$$\begin{gathered} x=\frac{−8Â\pm \sqrt{8^2−4\left(−3\right)\left(16\right)}}{2\left(−3\right)} \\ =\frac{−8Â\pm \sqrt{64+192}}{−6} \\ =\frac{−8Â\pm \sqrt{256}}{−6} \\ =\frac{−8Â\pm 16}{−6} \\ x=\frac{−8−16}{−6}\text{ â¶Ä‰â€‰â¶Ä‰and â¶Ä‰â€‰â¶Ä‰}\frac{−8+16}{−6} \\ x=4\text{ â¶Ä‰and â¶Ä‰}\frac{−8}{3} \end{gathered}$$

Since x represents the side length of the original square which must be a positive number, only $x=4$is a solution. Therefore, the side of the original square is then 4 inches.