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7

Hint First let $a=\sqrt{x^2-7x-2}.$ Note that this means $a\ge0$. Solve $a=a^2-12$. Then solve for $x$ from $a^2=x^2-7x-2.$

7

We may check each possible value of $C$, brute force, quick answer. Well, it is simpler maybe to check the relation modulo nine, i.e. find possible values for $C$ so that $$(2+C+8)(3+C+1) = (9+0+C+5+8)\text{ modulo nine.}$$ We get the simpler equation $(C+1)(C+4)=(C+4)$ modulo nine, so $C(C+4)=0$ modulo nine. If one factor is divisible by $3$, the other ...

5

Let's assume $A$ is positive definite. Multiplying on left and right by $A^{1/2}$ (the positive definite square root of $A$) the equation becomes $$(A^{1/2} X A^{1/2})^2 = A^{1/2} B A^{1/2}$$ Now $A^{1/2} B A^{1/2}$ is positive semidefinite, so has a positive semidefinite square root $C$, and we can take $X = A^{-1/2} C A^{-1/2}$ so that $C = A^{1/2} X A^{... 4 Consider in$\mathbb C[z]$the equation $$az^2+bz+c=0.$$ As in$\mathbb R[z]$, the solutions are given by $$z_{1,2}=\frac{-b\pm\sqrt{\Delta }}{2a},$$ where$\Delta =b^2-4ac.$4 Hint: It is better to expres$n$in terms of$m$and then use divisibility properties $$n = {m^2+8\over m+11} \implies m+11\mid m^2+8$$ Notice$m+11\mid m^2-121$so$m+11\mid 129=3\cdot 43$so$m+11\in\{1,3,43,129\}$so$m=32$or$m=118$... 4 Here is an easy way to finish the problem once you got the quadratic: $$499 C = 2745- 10C^2$$ Now since the RHS is a multiple of$5$so is the LHS, and hence$5|C$. Since$C$is a digit,$C=0$or$C=5$, but$0$is not a solution to your equation. All you have to do is check if$5$is a solution. P.S. The solution by dan_fulea becomes much shorter if ... 2 Another way... Let$u = x^2 - 7x -2$so you are looking at $$\sqrt{u} = u-12 \text{.}$$ This is a quadratic in the variable$\sqrt{u}. \begin{align*} u - \sqrt{u} - 12 &= 0 \\ (\sqrt{u} - 4)(\sqrt{u}+3) &= 0 \end{align*} For the product of two numbers to be zero, at least one of them is zero, so either\sqrt{u} = 4$or$\sqrt{u} = -3$(or ... 2 Let$p(x)=x^2-7x-2$. Then your equation becomes$\sqrt{p(x)}=p(x)-12$. So, start by solving the equation$\sqrt y=y-12$. Use the fact that its only solution is$16$. 2 Once you've built that quadratic, you can avoid factoring it by taking it modulo 10, since you know the answer you want is a single digit integer. $$10C^2+499C−2745\equiv5-C\equiv0\pmod {10}$$ From which C=5 quickly. 2 Complete the square:$x^2-3x=\left(x-\dfrac 32\right)^2-\dfrac 94$. Then$f(x)=\lfloor\left(x-\dfrac 32\right)^2-\dfrac 94\rfloor$. So the range is$\left[-3,\infty\right)\cap \Bbb Z$. 2 Use that your equation can be written as $$(z-(2+i)) (z-(1-3 i))$$ 2 $$m(n-m)=-11n+8\implies n =\frac {m^2+8}{m+11}\implies$$ $$n=m-11+\frac {129}{m+11}\implies$$ $$m-n=11-\frac {129}{m+11}$$ The only integral positive solutions are $$(m,n)=(32,24),(118,108)$$ Thus the desired sum is$18$1 Hint: Write your equation in the form $$n=m-11+\frac{129}{m+11}$$ 1 You could have nonreal and complex coefficient for A and B. They're just numbers! $$\frac{x-3}{x^2-7}=\frac{A}{x-\sqrt{7}}+\frac{B}{x+\sqrt{7}}$$ So by equality we have: $$\frac{x-3}{x^2-7}=\frac{A(x+\sqrt{7})+B(x-\sqrt{7})}{(x+\sqrt{7})(x-\sqrt{7})}$$ S0: $$A+B=1 \; and \; (A-B)\sqrt{(7)}=-3$$ Finally : $$A=\frac{\sqrt{7}-3}{2\sqrt{7}}$$ $$B=\frac{\... 1 Yes, you can do it, provided you don't mind using irrational real numbers. That is, if a is positive, then you have that$$x^2-a=(x-\sqrt a)(x+\sqrt a).$$1 The OP made two errors in the exercise: As the quadratic formula is \lambda= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}, the -(3-2i) should be replaced by (3-2i) for -b. The calculation of \Delta=b^2-4ac is incorrect. We have$$\Delta =b^2-4ac=(3-2i)^2-4(1)(5-5i)=(5-12i)-(20+20i)=-15+8i$$so that$$\lambda=\frac{3-2i\pm \sqrt{-15+8i}}{2}$$where we ... 1 Hint You had the correct idea but you did the square root wrong.$$\sqrt{(3-2i)^2-4(5-5i)}\ne 5$$in fact,$$\sqrt{(3-2i)^2-4(5-5i)}=\sqrt{-15+8i}=a+bi$$with, a,b \in \Bbb R, so,$$(a+bi)^2=-15+8i$$solve the above equation and get your result. 1 Let’s complete the square:$$(z^2-(3-2i)z+\left(\dfrac{3-2i}{2}\right)^2=\left(\dfrac{3-2i}{2}\right)^2-(5-5i).$$This gives us$$\left(z-\dfrac{3-2i}{2}\right)^2=\dfrac{-15+8i}{4}.$$It is not hard to realize that the RHS of the last equation is the square of$$\dfrac{1+4i}{2}.$$Finally take the square root of both sides (do not forget both+ and -) and ... 1 As long as a_1a_3-a_2^2= 0 we have$$ a_1+a_3-2a_2= 16\\ 64a_1 -16 a_2 =64 $$or$$ a_1+a_3-2a_2=16\\ 4a_1 - a_2 =4 $$solving for a_1,a_2 we have$$ a_1 = \frac 17(a_3-8)\\ a_2 = \frac 47(a_3-15) $$substituting now into a_1a_3-a_2^2= 0 we have$$ 9a_3^2-424a_3-3600=0 $$and solving gives$$ a_3 = \left\{\frac{100}{9},\ 36\right\} $$etc. 1 Subtract the first row from the last row of the system: then you get$$ \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 64a_1 -16 a_2 -64 &=0 \end{split} $$Then multiply by 8 the second row and subtract it again from the third row:$$ \begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 64a_1 -16 a_2 -64 &=0 \end{split}\... 1 The correct way to do this manually is to complete the square which will put your quadratics in turning point form (also sometimes called ‘vertex form’) at which point they can easily be graphed. Khan Academy has a good video on this. If you want to verify your answers, or just have a play around to understand what the graphs of quadratics look like, you ... 1 As the comments point out, it's not really any faster to use algebraic number theory to solve the problem. On the other hand, you can consider $$y^2=-27x^2 + 18px + p^2$$ where$p$is a prime number (your question is the special case$p=13$). Then solving this equation can be reduced to finding$a$,$b\$ such that $$a^2+ab+b^2=p^2\text{.}$$ Finding ...

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