Daily math problems for September 25, 2026 — Limits, Analytic Geometry, Algebra & more

$\bullet$ Factor the numerator. Roots of $x^2+5x-14=0$ are $x=2$ and $x=-7$:

$$x^2+5x-14=(x-2)(x+7)$$

$\bullet$ Cancel and evaluate:

$$\underset{x\to 2}{\lim }\frac{(x-2)(x+7)}{x-2}=\underset{x\to 2}{\lim }(x+7)=9$$

$d=\sqrt{3^2+4^2+0^2}=\sqrt{25}=5$. (3-4-5 triple in the $xy$-plane.)

Lagrange: in a finite group, the order of any subgroup (and hence the order of any element, which is the order of the cyclic subgroup it generates) divides $\mathrm{\mid }G\mathrm{\mid }$.

$\mathrm{\mid }AB\mathrm{\mid }=3$, $\mathrm{\mid }AC\mathrm{\mid }=4$, $\mathrm{\mid }BC\mathrm{\mid }=\sqrt{9+16}=5$. Perimeter $=3+4+5=12$.

${\int }_0^1{e}^{x}dx=e^1-e^0=e-1$.

$a^2=16\Rightarrow a=4$; $b^2=9\Rightarrow b=3$.

Area $={\displaystyle \frac{1}{2}}\mathrm{\mid }1(6-10)+3(10-2)+5(2-6)\mathrm{\mid }={\displaystyle \frac{1}{2}}\mathrm{\mid }-4+24-20\mathrm{\mid }=0$. Hence collinear.

$\bullet$ Eliminate the first column using $2R_2-R_1$ and $2R_3-R_1$:

$$2R_2-R_1=(0,3,2a-1,3)$$

$$2R_3-R_1=(0,-1,1,2b+1)$$

$\bullet$ For rank $2$, these two rows must be proportional. Matching the second components gives ratio $-3$:

$$2a-1=-3\cdot 1\Rightarrow a=-1$$

$$3=-3(2b+1)\Rightarrow b=-1$$

$\bullet$ So the answer is:

$$(a,b)=(-1,-1)$$

By Pascal: $\left(\genfrac{}{}{0px}{}{7}{3}\right)+\left(\genfrac{}{}{0px}{}{7}{4}\right)=\left(\genfrac{}{}{0px}{}{8}{4}\right)={\displaystyle \frac{8!}{4!4!}}=70$.

$\sqrt{x+5}=2+\sqrt{x-3}$. Square: $x+5=4+4\sqrt{x-3}+x-3\Rightarrow 4=4\sqrt{x-3}\Rightarrow \sqrt{x-3}=1\Rightarrow x=4$. Check: $\sqrt{9}-\sqrt{1}=3-1=2$ ✓.