4х^4-37х^2+9=0
Let's consider the equation:
\[ 4x^4 - 37x^2 + 9 = 0 \]
This is a quartic equation, but it can be reduced to a quadratic in terms of \( y = x^2 \):
\[ 4y^2 - 37y + 9 = 0 \]
Now, solve this quadratic equation for \( y \):
Use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 4 \), \( b = -37 \), and \( c = 9 \):
\[ y = \frac{37 \pm \sqrt{(-37)^2 - 4 \times 4 \times 9}}{2 \times 4} \]
Calculate discriminant:
\[ (-37)^2 = 1369 \] \[ 4 \times 4 \times 9 = 144 \] \[ \text{Discriminant} = 1369 - 144 = 1225 \]
Since \( \sqrt{1225} = 35 \):
\[ y = \frac{37 \pm 35}{8} \]
Now, find the two solutions:
- \( y = \frac{37 + 35}{8} = \frac{72}{8} = 9 \)
- \( y = \frac{37 - 35}{8} = \frac{2}{8} = \frac{1}{4} \)
Recall that \( y = x^2 \). Now, solve for \( x \):
Case 1: \( y = 9 \)
\[ x^2 = 9 \]
\[ x = \pm 3 \]
Case 2: \( y = \frac{1}{4} \)
\[ x^2 = \frac{1}{4} \]
\[ x = \pm \frac{1}{2} \]
Final solutions:
\[ x = \pm 3, \quad x = \pm \frac{1}{2} \]
If you'd like, I can help verify these solutions or clarify any step!
