Question

practice in \\( \triangle a b c \\) above, suppose \\( m \angle a b d = \left( x ^ { 2 } - 5 \
ight) ^ { \circ } \\) and \\( m \angle c b d = 4 x ^ { \circ } \\). solve for \\( x \\)

Explanation:

Step1: Set up the equation

Since \(\angle ABD\) and \(\angle CBD\) are complementary (assuming \(\angle ABC = 90^{\circ}\)), we have \((x^{2}-5)+4x = 90\).

Step2: Rearrange the equation

\(x^{2}+4x - 95=0\).

Step3: Solve the quadratic equation

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b = 4\), \(c=-95\).
\(x=\frac{-4\pm\sqrt{4^{2}-4\times1\times(-95)}}{2\times1}=\frac{-4\pm\sqrt{16 + 380}}{2}=\frac{-4\pm\sqrt{396}}{2}=\frac{-4\pm6\sqrt{11}}{2}=-2\pm3\sqrt{11}\).
Since \(x>0\), \(x=-2 + 3\sqrt{11}\approx - 2+3\times3.317=7.95\approx8\) (another root \(x=-2-3\sqrt{11}<0\) is rejected as angle measure cannot be negative when considering the context of the problem).

Answer:

\(x = 8\)