Question

1. ( x^4 + 5x^2 - 36 = 0 )

Explanation:

Step1: Substitute \( y = x^2 \)

Let \( y = x^2 \), then the equation \( x^4 + 5x^2 - 36 = 0 \) becomes \( y^2 + 5y - 36 = 0 \).

Step2: Solve the quadratic equation

For the quadratic equation \( ay^2 + by + c = 0 \) (here \( a = 1 \), \( b = 5 \), \( c = -36 \)), we use the quadratic formula \( y=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) or factor it. Factoring: \( y^2 + 5y - 36=(y + 9)(y - 4)=0 \). So \( y + 9 = 0 \) or \( y - 4 = 0 \), which gives \( y=-9 \) or \( y = 4 \).

Step3: Substitute back \( y = x^2 \)

• For \( y=-9 \): \( x^2=-9 \), so \( x=\pm\sqrt{-9}=\pm3i \) (where \( i \) is the imaginary unit, \( i^2=-1 \)).
• For \( y = 4 \): \( x^2 = 4 \), so \( x=\pm\sqrt{4}=\pm2 \).

Answer:

The solutions are \( x = 2 \), \( x = -2 \), \( x = 3i \), \( x = -3i \)