Examples of solving quadratic and biquadrate equations

Example 1.
Solve the quadratic equation x² - 3x = 0.
Solution.
The equation x² - 3x = 0 is an incomplete quadratic equation, so we will solve it by factoring it:
x² - 3x = 0 <=> x(x-3)=0 <=> x-3 = 0, x = 0 <=> x = 3, x = 0.
Answer: 0, 3.

Example 2.
Solve the equation Solution.
Let's bring the left side of the equation to a common denominator:

Multiply both parts of the equation by 4:

Answer: -2, 2.

Example 3.
Solve the quadratic equation x² + 3x + 10 = 0.
Solution.
Let's calculate the discriminant of the quadratic trinomial. We have a = 1, b = 3, c = 10.
D = b² - 4ac = 3² - 4*1*10 = 9 - 40 = -31 < 0, therefore, there are no real roots.
Answer: there are no roots.

Example 4.
Solve the quadratic equation x² + 12x + 36 = 0.
Solution.
Let's calculate the discriminant of the quadratic trinomial. We have a = 1, b = 12, c = 36.
Since b = 12 is an even number, we calculate the discriminant D₁ :
D₁ = (b/2)² - ac = 6² - 1*36 = 0, therefore, the equation has a unique root x = (-b/2)/a = (-6)/1 = -6.
This equation can be solved without calculating the discriminant by converting the quadratic trinomial using the formula of reduced multiplication:
x² + 12x + 36 = 0 <=> (x+6)² = 0 <=> x = -6.
Answer: -6.

Example 5.
Solve the quadratic equation 4x² -28x + 49 = 0.
Solution.
Let's calculate the discriminant of the quadratic trinomial. We have a = 4, b = -28, c = 49.
Since b = -28 - is an even number, we calculate the discriminant D₁ :
D₁ = (b/2)² - ac = (-14)² - 4*49 = 196-196 = 0, therefore, the equation has a unique root x = (-b/2)/a = 14/4 = 7/2.
This equation can also be solved without calculating the discriminant by converting the square trinomial using the reduced multiplication formula:
4x² -28x + 49 = 0 <=> (2x-7)² = 0 <=> 2x = 7 <=> x = 7/2.
Answer: 7/2.

Example 6.
Solve the equation
Solution.
Let's bring the left side of the equation to a common denominator:
Multiplying both parts of the equation by -6, we get x² + 3x = 0. We will solve this incomplete quadratic equation by factoring it:
Answer: -3, 0.

Example 7.
Solve the equation Solution.
Let's bring the left and right sides of the equation to a common denominator:


Multiplying both parts of the equation by 15, we get:
6x² + 3x = 20x-10 <=> 6x² + 3x - 20x + 10 = 0 <=> 6x² - 17x + 10 = 0.
Let's calculate the discriminant of the quadratic trinomial: a = 6, b = -17, c = 10.
D = b² - 4ac = (-17)² - 4*6*10 = 289 - 240 = 49 > 0, therefore, the equation has two real roots:

Answer: 5/6, 2.

Example 8.
Solve the equation
Solution.
Let's calculate the discriminant of the quadratic trinomial: a = 1, b = 2√2, c = 1.
Since b = 2√2, that is, b is divisible by 2 (b/2 = √2), let's calculate the discriminant D₁:
D₁ = ()² - ac = (√2)² - 1*1 = 1 > 0. Cледовательно, уравнение имеет два действительных корня.

Answer: -√2-1, -√2+1.

Example 9.
Solve the quadratic equation
Solution.
Multiply the left and right sides of the equation by 6:


Let's calculate the discriminant of the resulting quadratic trinomial. We have a = 3, b = -6, c = 2.
Since b = -6, that is, b is divisible by 2 (b/2 = -3), let's calculate the discriminant D₁:
D₁ = (b/2)² - ac = 3² - 3*2 = 3 > 0. Therefore, the equation has two real roots.


Answer:

Example 10.
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 11.
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 12.
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.