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Difficulty: MediumQuadratic Equations and the Quadratic Formula

A right triangle has legs of length x−0.5x - 0.5 inches and 2x2x inches, and a hypotenuse of length 2x+0.52x + 0.5 inches. What is the value of xx?

  1. A
    1
  2. B
    1.5
  3. C
    2
  4. 3Answer
  5. E
    4

Answer

3
Applying the Pythagorean theorem to the right triangle yields (x−0.5)2+(2x)2=(2x+0.5)2(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2. Expanding the terms gives x2−x+0.25+4x2=4x2+2x+0.25x^2 - x + 0.25 + 4x^2 = 4x^2 + 2x + 0.25. Subtracting 4x24x^2 and 0.250.25 from both sides simplifies the equation to x2−x=2xx^2 - x = 2x. Subtracting 2x2x from both sides gives the standard quadratic equation x2−3x=0x^2 - 3x = 0. Factoring this expression gives x(x−3)=0x(x - 3) = 0, which yields solutions x=0x = 0 and x=3x = 3. Since the side length x−0.5x - 0.5 must be positive, xx must be greater than 0.50.5. Therefore, the only valid solution is 33.

Step-by-Step Solution

1
Set up the equation using the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, with the given side lengths.
(x−0.5)2+(2x)2=(2x+0.5)2(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2
The Pythagorean theorem relates the legs and hypotenuse of any right triangle.
2
Expand each squared term algebraically.
(x2−x+0.25)+4x2=4x2+2x+0.25(x^2 - x + 0.25) + 4x^2 = 4x^2 + 2x + 0.25
Expanding the binomials allows us to combine like terms and simplify the equation.
3
Subtract 4x24x^2 and 0.250.25 from both sides of the equation.
x2−x=2xx^2 - x = 2x
Simplifying the equation makes it easier to solve.
4
Move all terms to the left side to write the quadratic equation in standard form.
x2−3x=0x^2 - 3x = 0
A quadratic equation must be set to zero to be solved by factoring.
5
Factor the quadratic expression.
x(x−3)=0x(x - 3) = 0
Factoring allows us to find the roots of the equation.
6
Solve for xx and choose the value that makes all side lengths positive.
x=3x = 3 (since x=0x = 0 is not a valid length because a side length x−0.5x - 0.5 must be greater than 00)
Only a positive value of xx greater than 0.50.5 yields physically possible side lengths for the triangle.

Key Concept

Setting up and solving quadratic equations using algebraic expansion and the Pythagorean theorem
Estimated Time:1m 30s
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Topics

Quadratic Equations and the Quadratic Formula

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