Find the sum of the squares of the solutions of $x^2-13x+4=0$.
Let $r_1$ and $r_2$ be the roots of this polynomial. Therefore, $r_1+r_2=13$ and $r_1r_2=4$. Notice that $r_1^2+2r_1r_2+r_2^2=169$. This means that the sum of the squares of the roots can be obtained by subtracting the term containing the product of $r_1$ and $r_2$, so $r_1^2+r_2^2=169-2(4)=\boxed{161}$.
161
Level 5
Algebra
Problem: Find the sum of the squares of the solutions of $x^2-13x+4=0$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the squares of the solutions of the quadratic equation, we first need to identify the solutions. The equation is in the standard form \( ax^2 + bx + c = 0 \) where \( a = 1 \), \( b = -13 \), and \( c = 4 \). 

We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots. First, we calculate the discriminant:

1. Calculate \( b^2 - 4ac \):
   - \( b^2 = (-13)^2 = 169 \)
   - \( 4ac = 4 \cdot 1 \cdot 4 = 16 \)
   - Thus, \( b^2 - 4ac = 169 - 16 = 153 \).

2. Now, apply the quadratic formula:
   - \( x = \frac{13 \pm \sqrt{153}}{2} \).

Let the solutions be \( r_1 = \frac{13 + \sqrt{153}}{2} \) and \( r_2 = \frac{13 - \sqrt{153}}{2} \).

Next, we need to find the sum of the squares of these solutions, which can be expressed as \( r_1^2 + r_2^2 \). We can use the identity \( r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2 \).

3. Calculate \( r_1 + r_2 \) and \( r_1 r_2 \):
   - From Vieta's formulas, \( r_1 + r_2 = -\frac{b}{a} = 13 \).
   - \( r_1 r_2 = \frac{c}{a} = 4 \).

4. Substitute into the identity:
   - \( r_1^2 + r_2^2 = (13)^2 - 2 \cdot 4 \).
   - Calculate \( (13)^2 = 169 \).
   - Calculate \( 2 \cdot 4 = 8 \).
   - Thus, \( r_1^2 + r_2^2 = 169 - 8 = 161 \).

The final answer is \( 161 \).<|END_OF_LATENT|> The final answer is $161$.