Question

1. $overrightarrow{at}$ is the bisector of $angle car$. $mangle cat = 2x^2 + x$. $mangle car = 5x^2 - 8$. solve for all possible values for $x$ & $mangle car$.

$x = \underline{quadquad}$
$m\angle car = \underline{quadquad}$
$2(2x^2 + x) = 5x^2 - 8$

Explanation:

Step1: Use Angle Bisector Property

Since \(\overrightarrow{AT}\) bisects \(\angle CAR\), we know that \(m\angle CAR = 2 \times m\angle CAT\). Substituting the given expressions:
\[2(2x^2 + x) = 5x^2 - 8\]

Step2: Expand and Simplify

Expand the left - hand side:
\[4x^2 + 2x = 5x^2 - 8\]
Rearrange the terms to form a quadratic equation:
\[5x^2 - 4x^2 - 2x - 8 = 0\]
\[x^2 - 2x - 8 = 0\]

Step3: Factor the Quadratic Equation

Factor \(x^2 - 2x - 8\):
We need two numbers that multiply to \(- 8\) and add up to \(-2\). The numbers are \(-4\) and \(2\). So,
\[(x - 4)(x + 2)=0\]

Step4: Solve for \(x\)

Set each factor equal to zero:

• If \(x - 4 = 0\), then \(x = 4\).
• If \(x + 2 = 0\), then \(x=-2\).

Step5: Check for Valid Angles

We need to check if the angle measures are non - negative.

• When \(x = 4\):

\(m\angle CAT=2(4)^2 + 4=2\times16 + 4 = 32 + 4=36\)
\(m\angle CAR = 5(4)^2-8=5\times16 - 8 = 80 - 8 = 72\) (valid, since \(72>0\) and \(36>0\))

• When \(x=-2\):

\(m\angle CAT=2(-2)^2+( - 2)=2\times4-2 = 8 - 2 = 6\)
\(m\angle CAR = 5(-2)^2-8=5\times4 - 8 = 20 - 8 = 12\) (valid, since \(12>0\) and \(6>0\))

Answer:

\(x = 4\) or \(x=-2\); when \(x = 4\), \(m\angle CAR = 72\); when \(x=-2\), \(m\angle CAR = 12\)