SOLUTION: This is an application problem from a chapter on the quadratic formula.
The diagonal of a square is 2 m longer than a side. Find the length of a side.
Thank you.
Algebra.Com
Question 277181: This is an application problem from a chapter on the quadratic formula.
The diagonal of a square is 2 m longer than a side. Find the length of a side.
Thank you.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let 
represent the measure of one of the sides. Then 
is the measure of the diagonal which is the hypotenuse of an isosceles right triangle.
There are two ways to go about this one. The 'charge straight at it' way uses the Pythagorean Theorem.
^2\ =\ x^2\ +\ x^2)
Using the Quadratic Formula:
\ \pm\ \sqrt{(-4)^2\ -\ 4(1)(-4)}}{2(1)}\ =\ \frac{4\ \pm\ \sqrt{32}}{2}\ =\ \frac{4\ \pm\ 4\sqrt{2}}{2}\ =\ 2\ \pm\ 2\sqrt{2})
Exclude the negative root because we are looking for a positive measure of length and we are left with:
+++++++++++++++++++++++++++++++++++++++++++++++++
A slightly simpler method requires that you recall that an isosceles right triangle has three sides in proportion:
Hence, if is the measure of one side, then 
is the measure of the hypotenuse. But we are given that the measure of the hypotenuse is , therefore:
Solving for :
\ =\ 2)
Finally, using the conjugate of the denominator to rationalize the denominator:
\left(\frac{\sqrt{2}\ +\ 1}{\sqrt{2}\ +\ 1}\right)\ =\ 2\ +\ 2\sqrt{2})
(Verification of that last step is left as an exercise for the student)
John
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