Daily math problems for July 13, 2026 — Quadratic equations, Trigonometry, Algebra & more

By Viète's formulas, $x_1+x_2=-(-7)=7$. (The product is $10$, so the roots are $2$ and $5$.)

$s={\displaystyle \frac{5+6+7}{2}}=9$. $A=\sqrt{9\cdot 4\cdot 3\cdot 2}=\sqrt{216}=6\sqrt{6}\approx 14.70$.

The factor $(X-2)$ appears with exponent $3$, so the multiplicity is $3$.

$\bullet$ Evaluate the function values:

$$f(3)=C_3^3+{\log }_{2}4=1+2=3$$

$$f(7)=C_7^3+{\log }_{2}8=35+3=38$$

$\bullet$ So $f(7)\mathrm{/}38=1$ and ${\log }_{\sqrt{3}}f(3)={\log }_{\sqrt{3}}3=2$.

$\bullet$ Reduce the trig / limit / log / factorial entries:

$$\tan \frac{\pi }{4}=1$$

$$2^{\lim x^{23}\mathrm{/}x!}=2^0=1$$

$$i^4=1,\lg {10}^{-48}=-48,P_4=4!=24,\ln 1=0$$

$\bullet$ Assemble the matrix:

$$A=\left(\begin{array}{ccc}3& 1& 1\\ 1& 2& 1\\ -48& 24& 0\end{array}\right)$$

$\bullet$ Expand along row 1:

$$\det A=3(2\cdot 0-1\cdot 24)-1(1\cdot 0-1\cdot (-48))+1(1\cdot 24-2\cdot (-48))$$

$$\det A=3(-24)-1(48)+1(120)$$

$$\det A=-72-48+120=0$$

$V=\pi {\int }_0^1(x^2-x^4)dx=\pi \text{\hspace{0.17em}⁣}{[{\displaystyle \frac{x^3}{3}}-{\displaystyle \frac{x^5}{5}}]}_0^1=\pi \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{5}})={\displaystyle \frac{2\pi }{15}}$.

By Fermat, $3^6\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}7)$, so ${\widehat{3}}^6=\widehat{1}$.

$\bullet$ Rewrite in power form using $\ln \sqrt[3]{x^{10}}=\frac{10}{3}\ln x$:

$$f(x)=x^{-7\mathrm{/}3}+\frac{10}{3}\ln x$$

$\bullet$ Differentiate once:

$$f^{\mathrm{\prime }}(x)=-\frac{7}{3}{x}^{-10\mathrm{/}3}+\frac{10}{3x}$$

$\bullet$ Differentiate again:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{70}{9}{x}^{-13\mathrm{/}3}-\frac{10}{3x^2}$$

$\bullet$ Plug in $x=2$ in $f^{\mathrm{\prime }}$ and $x=3$ in $f^{\mathrm{\prime }\mathrm{\prime }}$:

$$f^{\mathrm{\prime }}(2)=-\frac{7}{3\cdot 2^{10\mathrm{/}3}}+\frac{5}{3}$$

$$f^{\mathrm{\prime }\mathrm{\prime }}(3)=\frac{70}{9\cdot 3^{13\mathrm{/}3}}-\frac{10}{27}$$

$P(-1)>0$ and $P(1)<0$: $2+m>0$ and $-2+m<0$, giving $-2<m<2$.

$3\cdot (-5)=(m-1)(-(m+1))$, i.e. $-15=-(m-1)(m+1)=1-m^2$, so $m^2=16$ and $m\in \{-4,4\}$.

$\int \ln xdx=x\ln x-x+C$. Evaluate: $(e\cdot 1-e)-(1\cdot 0-1)=0-(-1)=1$.