SOLUTION: I need help solving this equation: The hypotenuse of a right isosceles triangle is one meter greater than either of it's sides. Find the length of both.

Question 612307: I need help solving this equation: The hypotenuse of a right isosceles triangle is one meter greater than either of it's sides. Find the length of both.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
A "right isosceles" triangle is a right triangle with two congruent legs.
Let x = the length of each of these legs
And if the hypotenuse is 1 meter greater than the sides, then it must be: x+1

With these expressions we can use the Pythagorean equation to solve the problem. (Remember to put the hypotenuse by itself.)

Now we simplify. (Be careful to use FOIL or the to square the right side:

Since this is a quadratic equation and since quadratic equations are a little easier to solve when the squared terms has a positive coefficient, I'm going to subtract the entire right side from both sides:

Next we factor or use the quadratic formula. This will not factor so we must use the formula:

Simplifying...

So
or
Since the second x is negative. So the seoncd x is negative. Since x is the length of a side of a triangle and since we cannot have negative lengths, we must reject that solution. So

is the only solution to the equation and, therefore, the length of the two legs. The length of the hypotenuse, x+1, would then be

Another way to solve this is to recognize and take advantage of the fact that right isosceles triangles are 45-45-90 triangles. And the sides of 45-45-90 triangle have a fixed, known relationship to each other: If x if the length of a leg, then will always be the hypotenuse. But we already have the expression of x+1 for the hypotenuse. Therefore these expression must be equal:

We can solve this for x. Subtracting x from each side we get:

Factoring out x:

Dividing by :

Rationalizing the denominator:

and
which is the solution we got earlier.
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