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Results tagged with algebra-precalculus
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user 187799
For questions about algebra and precalculus topics, which include linear, exponential, logarithmic, polynomial, rational, and trigonometric functions; conic sections, binomial, surds, graphs and transformations of graphs, solving equations and systems of equations; and other symbolic manipulation topics.
1
vote
Simplifying Logarithm
Hint
$$\log ab=\log a+\log b$$
$$\log \frac{a}{b}=\log a-\log b$$
$$k\log a=\log a^k$$
- 23.4k
1
vote
Accepted
Problem with finding solutions to polynomial equation
Hint:
$$x^3(x^2+2)=-297$$
or
$$x^3(x^2+2)=(-3)^3(11)$$
$$x^3(x^2+2)=(-3)^3((-3)^2+2)$$
- 23.4k
1
vote
Accepted
Estimate this expression $\sqrt{\sin(15/0.5)\times 5000-1}?$
the series of $\sqrt{x^2-1}$
$$\sqrt{x^2-1}= x-\frac{1}{2x}+O(\frac{1}{x})^3$$
so
$$\sqrt{2500-1}=\sqrt{50^2-1}\approx 50-\frac{1}{100}=49.99$$
- 23.4k
2
votes
How can you solve $x^2+2x^{-1}-1=0$?
$$x^3-x+2=0$$
use the Newton_Raphson Method
$$y=x^3-x+2$$
$$x_{n+1}=x_n-\frac{y_n}{y'_n}$$
when you find the first root, use the long division to reduce the cubic equation to second and use the quadra …
- 23.4k
2
votes
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$
Hint:
$$(2-\sqrt{3})(2+\sqrt{3})=1$$
$$(2-\sqrt{3})(x)=1$$
so
$$\frac{1}{x}=2-\sqrt{3}$$
$$(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$$
$$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2$$
$$x^2+\frac{1}{x^2}=(2+\sqr …
- 23.4k
-1
votes
Adding similar elements basic misunderstanding
$$(10^n) + 9 (10^n)=(10^n) + (10-1) (10^n)=???$$
- 23.4k
0
votes
Accepted
$x+1$ is a factor of $2x^3-5x^2-3x+K$. Find K.
Or we can use the long division to get:
$2x^3-5x^2-3x+K=(x+1)(2x^2-7x+4)+(K-4)=0$
- 23.4k
0
votes
Find the integer $m$ such that $(\frac{1}{5})^m \cdot (\frac{1}{4})^ {24} = \frac{1}{2(10)^{...
$$\frac{1}{2(10)^{47}}=\frac{1}{2}\frac{1}{2^{47}}\frac{1}{5^{47}}=\frac{1}{2^{48}}\frac{1}{5^{47}}=\frac{1}{4^{24}}\frac{1}{5^{47}}=\left(\frac{1}{4}\right)^{24}\left(\frac{1}{5}\right)^{47}$$
- 23.4k
1
vote
Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$
Hint
$$\frac{a^2}{a^4+a^2+1}=\frac{a}{
(a^2+a+1)}\frac{a}{(a^2-a+1)}=\frac{1}{(\frac{a^2+1}{a}+1)}\frac{1}{(\frac{a^2+1}{a}-1)}=\frac{1}{((\frac{a^2+1}{a})^2-1)}$$
- 23.4k
0
votes
Engineering Problem With natural logs
I think the Newton's method is suitable for same as equation.
rearrange the equation to get
$$y=234-350e^{-870k}+116e^{-9k}$$
and use the following to find the roots
$$k_{n+1}=k_n-\frac{y(k)_n}{y(k) …
- 23.4k
3
votes
Accepted
How do I solve the expression $-(x-1)^3=0$ over $\mathbb{R}$
$$(x-1)(x-1)(x-1)=0$$
so there are three roots
$$x_1=1$$
$$x_2=1$$
$$x_3=1$$
- 23.4k
2
votes
Solving a quadratic equation with parameters
$$x^2+(2-2a)x+a^2-2a=0$$
use the quadratic formula according to the following form
$$x^2+Ax+B=0$$
$$x=-0.5A\pm\sqrt{0.25A^2-B}$$
hence
$$x=(a-1)\pm\sqrt{(a-1)^2-a^2+2a}$$
$$x=(a-1)\pm\sqrt{a^2-2a+1-a^ …
- 23.4k
1
vote
How would i go about solving this exponential equation?
$$2^{x-2}+2^{-(x+2)}=\frac{17}{16}$$
$$2^{x}+2^{-x}=\frac{17}{4}$$
$$2^{x}+\frac{1}{2^{x}}=\frac{17}{4}$$
then assume $y=2^x$ to get
$$y+\frac{1}{y}=\frac{17}{4}$$
$$y^2+1=\frac{17}{4}y$$
- 23.4k
1
vote
9
votes
How do I prove that $x^2 + y^2 = (x + y) ^2 – 2xy$ geometrically/intuitively?
.........
$$(x+y)^2=x^2+2xy+y^2$$
- 23.4k