Solving of biquadratic equations

An equation of the form ax⁴ + bx² + с = 0, where a, b, c are some numbers, and a ≠ 0, is called biquadratic. To solve it, you need to use a variable replacement method. Let's introduce a new variable t = x². Then x⁴ = t², and the original equation will be equivalent to a quadratic equation: at² + bt + с = 0. We already know how to find the values ​​of the roots of a quadratic equation. Using the known values ​​of t, solving the equations x² = t, we find the roots of the original biquadratic equation.
If t < 0, then the equation x² = t has no roots.
If then

Examples of solving of biquadratic equations

Example 1.
Solve the equation x⁴ - 17x² + 16 = 0.
Solution.
The original equation is biquadratic. Replacing the variable t = x² => x⁴ = t², we get an equivalent original quadratic equation:
x⁴ - 17x² + 16 = 0 <=> t² - 17t + 16 = 0
Let's calculate the discriminant of a square trinomial: a = 1, b = -17, c = 16,
D = b² - 4ac = (-17)² - 4*1*16 = 289-64 = 225 > 0, therefore, the equation has two real roots:


Using the found values of t, solving the equation x² = t, we find the roots of the original biquadratic equation:

Thus, the original equation has 4 real roots.

Answer: -4, -1, 1, 4.

Example 2.
Solve the equation 9x⁴ + 32x² - 16 = 0.
Solution.
The original equation is biquadratic. Replacing the variable t = x² => x⁴ = t², we get an equivalent original quadratic equation:
9x⁴ + 32x² - 16 = 0 <=> 9t² + 32t - 16 = 0
Let's calculate the discriminant of a square trinomial. We have a = 9, b = 32, c = -16.
Since b = 32, that is, b is divisible by 2 ( = 16), let's calculate the discriminant D₁:
Therefore, the equation has two real roots.

Using the found values of t, solving the equation x² = t, we find the roots of the original biquadratic equation:

The first equation x² = -4 has no roots, and the second, and therefore the original, has two real roots x = .
Answer: -2/3, 2/3.

Example 3.
Solve the equation x⁴ + 3x² - 10 = 0.
Solution.
The original equation is biquadratic. Replacing the variable t = x² => x⁴ = t², we get an equivalent original quadratic equation:
x⁴ + 3x² - 10 = 0 <=> t² + 3t - 10 = 0
Let's calculate the discriminant of a square trinomial: a = 1, b = 3, c = -10,
D = b² - 4ac = 3² - 4*1*(-10) = 9+40 = 49 > 0, Therefore, the equation has two real roots:

Using the found values of t, solving the equation x² = t, we find the roots of the original biquadratic equation:

The first equation x² = -5 has no roots, and the second, and therefore the original, has two real roots
Answer: -√2, √2.