Question

what is the value of x?

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(a = 5x\), \(b=4x + 5\), and \(c=8x-3\).
So, \((5x)^{2}+(4x + 5)^{2}=(8x-3)^{2}\).

Step2: Expand the equations

Expand \((5x)^{2}=25x^{2}\), \((4x + 5)^{2}=16x^{2}+40x + 25\), and \((8x-3)^{2}=64x^{2}-48x + 9\).
Then \(25x^{2}+16x^{2}+40x + 25=64x^{2}-48x + 9\).

Step3: Simplify the equation

Combine like terms: \(41x^{2}+40x + 25=64x^{2}-48x + 9\).
Move all terms to one side: \(64x^{2}-48x + 9-(41x^{2}+40x + 25)=0\).
\(64x^{2}-48x + 9 - 41x^{2}-40x - 25=0\).
\(23x^{2}-88x - 16=0\).
Factor or use quadratic formula. Another way:
\(25x^{2}+16x^{2}+40x + 25=64x^{2}-48x + 9\)
\(41x^{2}+40x + 25-64x^{2}+48x - 9 = 0\)
\(-23x^{2}+88x + 16=0\)
Multiply by \(- 1\): \(23x^{2}-88x - 16=0\)
Or check by substituting the values from the options.
Substitute \(x = 3\):
Left - hand side of \(a^{2}+b^{2}=(5\times3)^{2}+(4\times3 + 5)^{2}=15^{2}+17^{2}=225+289 = 514\).
Right - hand side of \(c^{2}=(8\times3-3)^{2}=21^{2}=441\).
Substitute \(x = 2\):
\(a^{2}+b^{2}=(5\times2)^{2}+(4\times2 + 5)^{2}=10^{2}+13^{2}=100 + 169=269\).
\(c^{2}=(8\times2-3)^{2}=13^{2}=169\).
Substitute \(x = 4\):
\(a^{2}+b^{2}=(5\times4)^{2}+(4\times4 + 5)^{2}=20^{2}+21^{2}=400+441 = 841\).
\(c^{2}=(8\times4-3)^{2}=29^{2}=841\).

Answer:

\(x = 4\)