Question

for the pair of similar triangles, find the appropriate value(s) of x. the appropriate value(s) of x is/are \boxed{}. (simplify your answer. use a comma to separate answers as needed.)

Explanation:

Step1: Identify corresponding sides

Since the triangles are similar, their corresponding sides are proportional. The hypotenuse of the first triangle is 39, and the hypotenuse of the second triangle is \(x + 8\). The leg of the first triangle is \(2x - 6\), and the corresponding leg of the second triangle is 20. So we set up the proportion: \(\frac{2x - 6}{20}=\frac{39}{x + 8}\)

Step2: Cross - multiply

Cross - multiplying gives \((2x - 6)(x + 8)=20\times39\)
Expand the left - hand side: \(2x(x + 8)-6(x + 8)=780\)
\(2x^{2}+16x-6x - 48 = 780\)
\(2x^{2}+10x-48 - 780 = 0\)
\(2x^{2}+10x-828 = 0\)
Divide the entire equation by 2: \(x^{2}+5x - 414 = 0\)

Step3: Solve the quadratic equation

We use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 5\), and \(c=-414\)
First, calculate the discriminant \(\Delta=b^{2}-4ac=(5)^{2}-4\times1\times(-414)=25 + 1656 = 1681\)
Then, \(x=\frac{-5\pm\sqrt{1681}}{2}=\frac{-5\pm41}{2}\)
We have two solutions:
\(x_1=\frac{-5 + 41}{2}=\frac{36}{2}=18\)
\(x_2=\frac{-5-41}{2}=\frac{-46}{2}=-23\)
But we need to check if the side lengths are positive.
For \(x=-23\):
\(2x - 6=2\times(-23)-6=-46 - 6=-52\) (negative, not valid)
\(x + 8=-23 + 8=-15\) (negative, not valid)
For \(x = 18\):
\(2x-6=2\times18-6=36 - 6 = 30\) (positive)
\(x + 8=18 + 8 = 26\) (positive)

Answer:

18