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22

You can learn this by expanding the factored form of the cubic. If we suppose $a$, $b$ and $c$ are the roots, then: $$(x-a)(x-b)(x-c)=x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$$ It's a good thought to do this out by hand if you've not seen it before. Notice that when we multiply monomials like $(x-a)$, we just "take" one term from each monomial and multiply them ...


21

Clearly there is no negative root as all terms are positive for $x < 0$. The question remains if there are positive roots. Here is a simple way which often works. Case 1: $0 < x <1$. $$P(x) = (15-x) + (x^2-x^7) + x^8 > 0$$ as each term is positive. Case 2: $ x > 1$. Similarly $$P(x) = (x^8-x^7) + (x^2-x) + 15 > 0$$ as $x=1$ is not a ...


17

You can use the rational root theorem to guess some roots. Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient. For example, in the integer case, one can take $q=1$, and one only has to test $p = \pm 1$, $p = \pm 2$, $p= \pm 5$, and then you have it. ...


16

Let, $$y=x^2+2$$ Now, $y^2-6xy+8x^2=0$ $\implies \left(\frac{y}{x}\right)^2 -6\left(\frac{y}{x}\right)+8=0$ $\implies y = 4x$ or $y=2x$ $\implies x^2-4x+2=0$ or $x^2-2x+2=0$ $\implies x = 2 \pm \sqrt{2}, 1 \pm i$


14

We look at the equation $$ax^2-bx(x-1)+c(x-1)^2=0.\tag{1}$$ If $a\ne 0$, then $1$ is not a solution, so Equation (1) is equivalent to $$a\left(\frac{x}{x-1}\right)^2-b\frac{x}{x-1}+c=0.\tag{2}$$ If we put $y=-\frac{x}{x-1}$ then the solutions of (2) are the solutions of $ay^2+by+c=0$. It follows that $$\frac{-x}{x-1}=\alpha\quad\text{or}\quad ...


13

The polynomial can be rewritten $$ x^7(x-1) + x(x-1) + 15. $$ Unless $x$ is between $0$ and $1$, the first two terms are positive, and so the polynomial is positive. Even if $x$ is between $0$ and $1$, the first two terms are tiny in magnitude, certainly each individually greater than $-1$, so that when $15$ is added to their sum, the result is positive. ...


9

HINT: Divide by $4^x$ to get $$a^2-a-2=0$$ where $a=\left(\dfrac32\right)^x$ Can you solve for $a?$ Now for real $x,a>0$ See also : Exponent Combination Laws


7

You could work by locating the roots approximately by computing at some easy values and noting sign changes - Sturm's Theorem is a heavy duty resource, and Descartes Rule of Signs can be indicative. There are more basic observations too, which can help to narrow the search amongst rational roots - using changes in the sign of the value and the intermediate ...


7

Hint: If you considered the two roots as $p$ and $p^2$, then you can write the quadratic as $$ax^2 + bx + c = a(x-p)(x-p^2)$$ Alternatively, for a quadratic, using the formula for the sum and products of roots we have $$p + p^2 = -\frac{b}{a}$$ and $$p \cdot p^2 = p^3 = \frac{c}{a}$$ Now you just need to find a way to eliminate $p$ from the two equations. ...


6

To expand on wythagoras' excellent answer, let me point out that when some of the coefficients are reasonably large (as in this case), we can sometimes relatively quickly extract some more information by reducing the given polynomial $f$ modulo a small integer $m$. This exploits the idea that if $r$ is a root of $f$, so that $f(r) = 0$, then $f(r) \equiv 0 ...


6

I think graphing would be the best option, the graph below shows the curves $3^{|x|}$ in blue and $1 + |2-|x||$ in red, they intersect in two places, so your equation has two real solutions. It is quite easy to sketch these graphs. $3^{|x|}$ is simply an exponential function that goes to infinity fast after $x>1$ and you simply reflect the shape in ...


6

It should be clear that on the interval $[-1,1]$ you have $|x^8-x^7+x^2-x|\leq |x^8|+|x^7|+|x^2|+|x|\leq 4$ and so $x^8-x^7+x^2-x+15\geq 11$ Further you should notice that $x^8-x^7>0$ when $|x|>1$ and that $x^2-x>0$ when $|x|>1$, so $x^8-x^7+x^2-x+15\geq 11$ for all $x$


6

Yes, polynomial equations of degree four (or less) have formulas that will let you compute the (possible) roots. For more on how to do this in this case, you can take a look at this Wikipedia article about solving the quartic equation. Another way would be to guess a root. Try for example $x=0,-1, \color{red}{1}$. If you find a root $\alpha$, then you can ...


5

Hint: Factor it as $$(k+1)2^x=8$$ and divide by $k+1$ to obtain $$2^x=\frac{8}{k+1}$$ provided that $k\neq-1$. What can we say about above equation?


5

hint: $\dfrac{2c-b}{a-b+c} = \dfrac{2\dfrac{c}{a}-\dfrac{b}{a}}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2\alpha\cdot \beta+(\alpha+\beta)}{1+\alpha+\beta+\alpha\cdot \beta}$, can you continue?


5

That problem has a unique solution, up to permutations of the variables. That happens since the power sums $p_k=x^k+y^k+z^k$ for $k=1,2,3$ give you the values of the elementary symmetric functions $e_1=x+y+z,e_2=xy+xz+yz,e_3=xyz$ through Newton's identities. Then $x,y,z$ can be identified with the roots of the polynomial: $$ p(w) = w^3 - e_1 w^2 + e_2 w - ...


4

First note that $$x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)=35$$ Now substituting $x+y+z=1$, we have $1^{2}-2(xy+yz+zx)=35$, and thus $xy+yz+zx=-17$. Then, $$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)+3xyz=97$$ and substituting with $1$ and our above result once again leads us to have $xyz=15$. Now we need a product of $15$ and sum of $1$ for ...


4

$$f(x)=x-2+\frac{a-3}{x}$$ $$\color{blue}{x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$$ Distinct real solutions iff $b^2-4ac\geq0$ You should get $$\boxed{\color{red}{x_{1,2}=1\pm\sqrt{4-a}}}$$


4

Let $f(x)=\sin x-x\cos x$. You have $f'(x)=x\sin x$. Since $\sin x$ has the same sign as $x$ for $x\in[-\pi/2,\pi/2]$, we know that $f'(x)\geq0$ in this interval and $f'(x)>0$ for $x\in[-\pi/2,\pi/2]\setminus\{0\}$. Therefore $f$ is strictly increasing and can have at most one zero.


4

The more elegant way to solve this equation was given by lab bhattacharjee, but I'll simply elaborate on how to solve it using your method, which was still correct. Hint: We have that $$a^2 - ab -2b^2 = 0 \iff b = \frac{a}{2}, \text{or }b= -a.$$ You now have a system of two equations, $\log_3 a = \log_2 b$ and from this you can see that neither $a$ or ...


3

Take the equation $$ax^3+bx^2+cx+d=0$$ which has solutions $\alpha$, $\beta$, $\gamma$. Group the odd and even powers on each side and factor to obtain $$x(ax^2+c) = -(bx^2+d).$$ Now squaring both sides gives $$x^2(ax^2+c)^2 = (bx^2+d)^2.$$ This equation has solutions $\alpha$, $\beta$, $\gamma$ but only contains even powers of $x$ so we can make the ...


3

Following what wythagoras did in his answer, we arrive at $$2^x = \frac{8}{k+1}$$ Now, the function $2^x$ is strictly increasing and is always positive. (*) So, we must have $$\frac{8}{k+1} > 0$$ and $k > -1$. (*) To see what I mean, look at the graph below: the graph is the function $2^x$, as you can see the function hits every value greater than ...


3

$$(x^2+2)^2+8x^2=6x(x^2+2)\Longleftrightarrow$$ $$x^4+4x^2+4+8x^2=6x(x^2+2)\Longleftrightarrow$$ $$x^4+12x^2+4=6x(x^2+2)\Longleftrightarrow$$ $$x^4+12x^2+4=6x^3+12x\Longleftrightarrow$$ $$x^4+12x^2-6x^3-12x+4=0\Longleftrightarrow$$ $$(x^2-4x+2)(x^2-2x+2)=0\Longleftrightarrow$$ $$(x^2-4x+2)=0\vee (x^2-2x+2)=0\Longleftrightarrow$$ $$x^2-4x=-2\vee ...


3

If the two quadratic polynomials $f(x) = px^2 + qx + r$ and $g(x) = qx^2 + rx + p$ have a common root then this is also a root of the linear polynomial $$h(x) = qf(x) - pg(x) = (q^2-pr)x - (p^2-qr)$$ which means that either $q^2-pr=0$ and $p^2-qr=0$ for which $p=q=r$ or that $x = \frac{p^2-qr}{q^2-pr}$ is the common root. Inserting this into either $f(x)$ ...


3

See edit below I don't think the solution should be independent of $c$. Maybe you typed into Wolfram $$ a^x+b^x=0 $$ In that case, it's not too hard to see where the solution comes from. Move $b^x$ to one side to get: \begin{align} b^x&=-a^x\\ &=(-1)\times a^x\\ &=e^{i\pi}a^x \end{align} Then take $\ln$ of both sides: $$ x\ln(b)=i\pi ...


3

Let $\alpha = \dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta = \dfrac{-b-\sqrt{b^2-4ac}}{2a}$. Note that $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$. We want to calculate: $$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} $$ In terms of $\alpha$ and $\beta$. $$\dfrac{2c-b\pm \sqrt{(b-2c)^2 - 4(a-b+c)c}}{2(a-b+c)} = \dfrac{2c-b\pm ...


2

Consider the case of $(a_1,\ldots,a_n)=(0,\ldots,0)$ first, where it's obvious. Then observe that $$f(a_1,\ldots,a_n)=0\iff g(0,\ldots,0)=0$$ and $$f\in (x_1-a_1,\ldots,x_n-a_n)\iff g\in (x_1,\ldots,x_n)$$ where $$g(x_1,\ldots,x_n)=f(x_1+a_1,\ldots,x_n+a_n)$$


2

By symmetry, if $x$ is a root, so is $-x$. Let us assume $x>0$ and solve $$3^x-|2-x|=1,$$ i.e for $$x<2,\ 3^x-x-3=0$$ and for $$x>2,\ 3^x+x+1=0.$$ The latter relation is not possible, by positivity of the terms. As $(3^x-x-3)'=\ln(3)3^x-1$, the function is increasing and has at most one real root. There is indeed one in the interval $(1,2)$, as ...


2

The discriminant of the equation $x^2+kx+(k-3)=0$ is $k^2-4k+12=(k-2)^2+8$, which is positive, so the roots of the equation are both real, for any choice of $k$. Let the roots be $r_1$ and $r_2$. Writing $x^2+kx+(k-3)=(x-r_1)(x-r_2)$ and expanding, we see the product of the roots is $k-3$. We want this product to be negative, i.e. $k<3$.


2

The "if" part is clear, so we'll deal with the "only if". Moreover, we take the polynomials to be "true" quadratics ---that is, $p$ and $q$ are non-zero--- since otherwise the proposition is false (as @Winther mentions in a comment to the OP). First, note that quadratics with a common root could have both roots in common, in which case they are ...



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