Video Transcript
Find the value of 𝑥 given that a
right triangle has a hypotenuse of length two 𝑥 and sides of lengths 𝑥 plus one
and 𝑥 plus three.
We will begin by sketching the
right triangle. We are told that the hypotenuse has
a length two 𝑥 and the other two sides have lengths 𝑥 plus one and 𝑥 plus
three. We can apply the Pythagorean
theorem to create a quadratic equation. This states that 𝑎 squared plus 𝑏
squared is equal to 𝑐 squared, where 𝑐 is the length of the hypotenuse and 𝑎 and
𝑏 are the lengths of the two shorter sides. Substituting in the given lengths,
we have 𝑥 plus one squared plus 𝑥 plus three squared is equal to two 𝑥
squared.
We can distribute the parentheses
on the left-hand side using the FOIL method. 𝑥 plus one all squared is equal to
𝑥 squared plus two 𝑥 plus one. And 𝑥 plus three all squared is
equal to 𝑥 squared plus six 𝑥 plus nine. The sum of these two expressions is
equal to four 𝑥 squared. By collecting like terms, the
left-hand side simplifies to two 𝑥 squared plus eight 𝑥 plus 10. We can then subtract all three of
these terms from both sides, leaving us with two 𝑥 squared minus eight 𝑥 minus 10
equals zero. Next, we can divide through by
two. We now have the quadratic equation
𝑥 squared minus four 𝑥 minus five equals zero, which we can solve by
factoring.
As the first term has a coefficient
of one, the first term in each of our parentheses will be 𝑥. We then need to find two integers
that have a product of negative five and a sum of negative four. Negative five multiplied by one is
negative five, and negative five plus one is equal to negative four. Our quadratic therefore factors to
𝑥 minus five multiplied by 𝑥 plus one, and this is equal to zero.
To solve this equation, we set each
factor equal to zero. This gives us two solutions: 𝑥
equals five and 𝑥 is equal to negative one. Whilst these are both valid roots
of the given quadratic equation, they are not both valid solutions to the
problem. The expression two 𝑥 represents
the length of the hypotenuse of the right triangle, and this must be strictly
positive. We can therefore disregard the
solution 𝑥 is equal to negative one, and the value of 𝑥 is five.
Whilst it is not required in this
question, we could substitute this value of 𝑥 back into the expressions for each of
the side lengths of the triangle. The right triangle has side lengths
six, eight, and 10. And since six squared plus eight
squared is equal to 10 squared, this is a right triangle.
