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!
