DTM mathematics formulas: a core guide for the OTM entrance test

The phrase "DTM mathematics" is still frequently searched for, but entrance tests are now organized by the Agency for Assessment of Knowledge and Skills. The official program and the assessment criterion may be updated from year to year; therefore this article does not estimate the number of questions, but explains the core formulas used in algebra, progression, geometry, and trigonometry.

Memorizing a formula alone is not enough. In a problem you need to recognize which formula is required, check its conditions, and confirm the answer by an independent method. Each example below shows exactly these three steps.

Short multiplication formulas

These formulas appear on the exam both directly and in a "hidden" form inside other topics:

• $(a+b)^2 = a^2 + 2ab + b^2$
• $(a-b)^2 = a^2 - 2ab + b^2$
• $a^2 - b^2 = (a-b)(a+b)$
• $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$

The difference of squares sharply speeds up the calculation: for example, $47^2 - 43^2 = (47-43)(47+43) = 4 \cdot 90 = 360$ — without a calculator, in one line.

1-example. If $x + \dfrac{1}{x} = 5$ and $x \ne 0$, find $x^2 + \dfrac{1}{x^2}$.

Solution. We square both sides of the given equality:

$\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 \cdot x \cdot \dfrac{1}{x} + \dfrac{1}{x^2} = x^2 + 2 + \dfrac{1}{x^2} = 25.$

Therefore, $x^2 + \dfrac{1}{x^2} = 25 - 2 = 23$. Note: here it is not necessary to find $x$ itself — this very idea is the key to not losing time on the test.

The quadratic equation and Vieta's theorem

For the equation $ax^2 + bx + c = 0$ $(a \ne 0)$, the discriminant is $D = b^2 - 4ac$. In the real numbers, if $D<0$, there are no roots; when $D\ge0$, the roots are $x_{1,2} = \dfrac{-b \pm \sqrt{D}}{2a}$. By Vieta's theorem:

$x_1 + x_2 = -\dfrac{b}{a}, \qquad x_1 \cdot x_2 = \dfrac{c}{a}.$

On tests, what is often asked for is not the roots themselves, but an expression in them — in that case Vieta's theorem works faster than the discriminant.

2-example. Calculate the sum of the squares of the roots of the equation $x^2 - 7x + 10 = 0$ without finding the roots.

Solution. By Vieta, $x_1 + x_2 = 7$ and $x_1 x_2 = 10$. We use the short multiplication formula:

$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = 49 - 20 = 29.$

Check: the roots of the equation are $x_1 = 2$, $x_2 = 5$ (because $2 + 5 = 7$, $2 \cdot 5 = 10$), and $2^2 + 5^2 = 4 + 25 = 29$. The answer is confirmed.

Progressions

For an arithmetic progression:

• $n$-term: $a_n = a_1 + (n-1)d$
• sum: $S_n = \dfrac{(a_1 + a_n) \cdot n}{2}$

For a geometric progression: $b_n = b_1 q^{n-1}$. If $q \ne 1$, then $S_n = \dfrac{b_1(q^n - 1)}{q - 1}$; if $q=1$, all terms are equal to $b_1$, and $S_n=nb_1$.

3-example. In an arithmetic progression, $a_1 = 5$, $d = 3$. Find $a_{20}$ and $S_{20}$.

Solution. First we find the twentieth term:

$a_{20} = a_1 + 19d = 5 + 19 \cdot 3 = 5 + 57 = 62.$

Now the sum: $S_{20} = \dfrac{(a_1 + a_{20}) \cdot 20}{2} = \dfrac{(5 + 62) \cdot 20}{2} = 67 \cdot 10 = 670$.

Be careful: the formula contains $19d$, that is, $(n-1)d$. If $nd$ is written, the progression shifts by one step.

Core formulas from geometry and trigonometry

Core formulas from geometry: the Pythagorean theorem $c^2 = a^2 + b^2$, the area of a triangle $S = \dfrac{1}{2}ah$ and $S = \dfrac{1}{2}ab\sin C$, the area of a circle $S = \pi r^2$, the circumference $C = 2\pi r$. In trigonometry, $\sin^2\alpha + \cos^2\alpha = 1$ and $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$ are used; in the tangent formula, $\cos\alpha \ne 0$ must hold.

4-example. Find the area of the triangle with sides 5, 12, and 13.

Solution. We check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. By the converse of the Pythagorean theorem, this is a right triangle with legs 5 and 12. Therefore,

$S = \dfrac{1}{2} \cdot 5 \cdot 12 = 30.$

Recognizing Pythagorean triples such as 5–12–13 (as well as 3–4–5 and 8–15–17) shortens the calculation. Still, check the equality: a triple indicates a right triangle only when $a^2+b^2=c^2$ holds.

Independent exercise

If $x+\dfrac{1}{x}=4$ and $x\ne0$, find $x^2+\dfrac{1}{x^2}$. Short answer: We square the equality and subtract $2$; the answer is $14$.

Typical mistakes: do not do these

• writing $(a+b)^2 = a^2 + b^2$ — the term $2ab$ is omitted.
• A sign error in the discriminant: if $b = -7$, then $b^2 = (-7)^2 = 49$, and it is never $-49$.
• In a progression, writing $a_n = a_1 + nd$ — the correct form is $(n-1)d$; the result shifts by one step.
• assuming $\sqrt{a^2} = a$ — in the general case $\sqrt{a^2} = |a|$, because $a$ can be negative.
• Canceling addends in a fraction: $\dfrac{a+c}{b+c}$ cannot be "simplified" to $\dfrac{a}{b}$ — cancellation works only on factors.

Conclusion and the next step

• Remember each formula together with its condition: restrictions such as $x\ne0$, $a\ne0$, $q\ne1$, and $\cos\alpha\ne0$ are part of the answer.
• If finding the roots is not required, use Vieta's theorem and identities.
• In a progression, check the index $(n-1)$, and in geometry check the condition for applying the formula.
• Practice the formulas in thematic mathematics tests and rework the mistakes in the mistake notebook.

Sources

• UzBMB: details of the test tasks in general-education subjects
• UzBMB: the number of tests and the assessment criterion for the 2026/2027 admission (PDF)