To solve this problem, we will use the fact that complementary angles sum up to 90°.

Step 1: Set up the equation

The measures of the angles are:

• $m\angle A = 2x^2 + 35$
• $m\angle B = x + 10$

Since $\angle A$ and $\angle B$ are complementary: $m\angle A + m\angle B = 90$

Substituting the expressions: $(2x^2 + 35) + (x + 10) = 90$

Step 2: Simplify the equation

Combine like terms: $2x^2 + x + 45 = 90$

Subtract 90 from both sides: $2x^2 + x - 45 = 0$

Step 3: Solve the quadratic equation

The equation is: $2x^2 + x - 45 = 0$

We solve this using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 2$,
• $b = 1$,
• $c = -45$.

Substitute into the formula: $x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-45)}}{2(2)}$ $x = \frac{-1 \pm \sqrt{1 + 360}}{4}$ $x = \frac{-1 \pm \sqrt{361}}{4}$ $x = \frac{-1 \pm 19}{4}$

Step 4: Find the two possible values for $x$

1. $x = \frac{-1 + 19}{4} = \frac{18}{4} = 4.5$
2. $x = \frac{-1 - 19}{4} = \frac{-20}{4} = -5$

Step 5: Determine valid solutions

Since $x$ represents part of an angle measure, it must be positive. Thus, $x = 4.5$.

Step 6: Find the measures of the angles

Substitute $x = 4.5$ into the angle expressions:

• $m\angle A = 2x^2 + 35 = 2(4.5)^2 + 35 = 2(20.25) + 35 = 40.5 + 35 = 75.5^\circ$
• $m\angle B = x + 10 = 4.5 + 10 = 14.5^\circ$

Final Answer:

• $x = 4.5$
• $m\angle A = 75.5^\circ$
• $m\angle B = 14.5^\circ$

Do you want further clarification or detailed steps on any part of the solution?

Here are 5 related questions to deepen your understanding:

1. What are supplementary angles, and how would this problem differ if the angles were supplementary?
2. What are the general properties of quadratic equations, and how do they apply here?
3. Can we solve this problem graphically? How would the graph of $2x^2 + x - 45 = 0$ look?
4. How does the discriminant ($b^2 - 4ac$) help determine the number of solutions in a quadratic equation?
5. What happens if one angle expression includes a higher power term, like $x^3$?

Tip: Always verify the validity of your solutions by substituting them back into the original problem!