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Quadratic equations

This page is part of the section “Task 1 USE profile level. Elementary equations”. Here are solutions to quadratic equations with neat roots (including fractional ones). Each problem includes the condition, a step-by-step solution, and the answer.

Problems with solutions

Problem 1. Reduced equation (Vieta’s theorem)

Solve the equation:

x2−5x+6=0

The equation is reduced (the coefficient of x² is 1). By Vieta’s theorem, the sum of the roots is 5 and the product is 6:

x1+x2=5,x1·x2=6

We look for two numbers whose sum is 5 and whose product is 6. These are 2 and 3.

Answer: x1=2, x2=3.

Problem 2. Complete equation (via the discriminant)

Solve the equation:

2x2−7x+3=0

Find the discriminant using the formula D = b² − 4ac:

D=b2−4ac=(−7)2−4·2·3=49−24=25 x1,2=7±252·2=7±54

Roots: x₁ = (7 + 5)/4 = 3, x₂ = (7 − 5)/4 = 1/2.

Answer: x1=3, x2=12.

Problem 3. Incomplete quadratic equation

Solve the equation:

3x2−12=0

Move the constant term to the right and divide both sides by 3:

3x2=12⇒x2=4⇒x=±2

Answer: x1=−2, x2=2.

Problem 4. Factoring out the common factor

Solve the equation:

x2+4x=0

Factor out the common factor x. The product is zero when at least one of the factors is zero:

x(x+4)=0⇒x=0илиx+4=0

Answer: x1=−4, x2=0.

Frequently asked questions

How to solve quadratic equations from Task 1 of the USE?

There are three main methods: via the discriminant (universal), via Vieta’s theorem (for reduced equations with integer roots), and via factoring out the common factor or the difference of squares (for incomplete equations). The choice depends on the form of the equation.

When is it convenient to use Vieta’s theorem?

Vieta’s theorem is convenient when the equation is reduced (a = 1) and the roots are integers. Then the sum of the roots is −b and the product is c, and the roots can be found mentally. If the roots are fractional or the equation is complete, it is easier to use the discriminant.

What to do if the discriminant is negative?

If D < 0, the equation has no real roots. In Task 1 of the USE profile level such equations do not occur — the answer is always real. If D = 0, there is one root (of multiplicity 2).

How to solve incomplete quadratic equations?

If the equation has the form ax² + c = 0 (no term with x), move c to the right and take the root: x = ±√(−c/a). If the form is ax² + bx = 0 (no constant term), factor out x: x(ax + b) = 0, hence x = 0 or x = −b/a.

See also