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## New answers tagged quadratics

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The quadratic equation: $$Kx^2+Lx+M$$ has 2 roots $r1,r2$ given by: $$r1=\frac{-L+\sqrt{L^2-4KM}}{2K}$$ $$r2=\frac{-L-\sqrt{L^2-4KM}}{2K}$$ When the roots are equal you get: $$-L+\sqrt{L^2-4KM}=-L-\sqrt{L^2-4KM}$$ $$\sqrt{L^2-4KM}=-\sqrt{L^2-4KM}$$ This can happen when: $$L^2-4KM=0 ...... EQ. (1)$$ Given your equation: $$f(x)=x^2 + a(a+2b+c) = 2(a+b)x... 0 Hint: Set the discriminant [2(a+b)]^2-4a(a+2b+c)=0. 0 Using the quadratic formula we get$$x_{1,2}=a+b\pm\sqrt{b^2-ac}$$and then it must be$$a+b+\sqrt{b^2-ac}=a+b+\sqrt{b^2-ac}$$0 You just have to compute the distance of the origin from the line y=3x+1 and from the parabola y=x^2+x-2. The first distance is \frac{1}{\sqrt{10}}, the second one can be found by identifying the minimum of$$ d^2(x) = x^2+(x^2+x-2)^2 $$which occurs at a root of a cubic equation. In a simpler way, you may just check that the circle centered at the ... 0 The circle is tangential to either the line or the parabola. Given that the origin (0,0), the center of the circle, is closer to the line than to the parabola, the upper limit of the circle is just the distance from the origin to the line, which is given by$$r \le \frac{m}{\sqrt{1+k^2}}=\frac{1}{\sqrt{10}}$$where k=3 is the slope of the line and m = ... 1 You could have nonreal and complex coefficient for A and B. They're just numbers!$$ \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{\...

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Yes, you can do it, provided you don't mind using irrational real numbers. That is, if $a$ is positive, then you have that $$x^2-a=(x-\sqrt a)(x+\sqrt a).$$

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Hint: $f(x)$ is continuous and $\lim\limits_{x\to\pm\infty} f(x)=\infty$ and $f(x)$ has a minimum when $f'(x)=2x-3=0$.

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Take the derivative of $f(x)$. Then we have $$\dfrac{d}{dx}f(x)=2x-3.$$ Solve for $x$ when this derivative equals $0$ and we have $x=\dfrac{3}{2}.$ Now plug this back into your equation $f(x)=x^{2}-3x$ and we have $f\big(\dfrac{3}{2}\big)={\big(\dfrac{3}{2}\big)^2}-3\big(\dfrac{3}{2}\big)=-\dfrac{9}{4}.$ Therefore your range is $y\geq-\dfrac{9}{4}.$

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If we assume that $x$ can have any real value, then $y = x^2 - 3x = (x-\frac{3}{2})^2 - \frac{9}{4} \geq - \frac{9}{4}$

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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 $\left[-3,\infty\right)\cap \Bbb Z$.

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I'm going to use $x=a_1, \ y=a_2, \ z=a_3$ $$xz - y^2=0 \tag A$$ $$x + z - 2y - 16 = 0 \tag B$$ $$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$ Substitute $xz = y^2$ from (A) into (C) \begin{align} xz + 64x - y^2 - 16y - 64 &= 0 \\ y^2 + 64x - y^2 - 16y - 64 &= 0 \\ 64x - 16y - 64 &= 0 \\ y &= 4x - 4 \tag D \end{align} Substitute $y =... 1 The OP made two errors in the exercise: As the quadratic formula is$\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 ... 2 Use that your equation can be written as $$(z-(2+i)) (z-(1-3 i))$$ 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.$1 As long as$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= 0we 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: $$\begin{split} a_1a_3-a_2^2 &=0\\ a_1+a_3-2a_2-16&=0\\ 64a_1 -16 a_2 -64 &=0 \end{split}\... 0 By the second equation we get$$a_3=2a_2+16-a_1$$so we get in (1)$$2a_1a_2+16a_1-a_1^2-a_2^2=0$$(I) and in (3)$$2a_1a_2+16a_1-a_1^2+64a_1-a_2^2-16a_2-64=0$$(II) with$$2a_1a_2=a_1^2-16a_1+a_2^2$$we get in (II): We get$$4a_1-a_2=4$$Now e can eliminate a_2$$2a_1(4a_1-4)+16a_1-a_1^2-(4a_1-4)^2=0$$Now you can solve this equation for a_1. 1 As the comments point out, it's not really any faster to use algebraic number theory to solve the problem. On the other hand, you can consider$$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{.}$$Finding ... 1 The correct way to do this manually is to complete the square which will put your quadratics in turning point form (also sometimes called ‘vertex form’) at which point they can easily be graphed. Khan Academy has a good video on this. If you want to verify your answers, or just have a play around to understand what the graphs of quadratics look like, you ... 5 Let's assume A is positive definite. Multiplying on left and right by A^{1/2} (the positive definite square root of A) the equation becomes$$ (A^{1/2} X A^{1/2})^2 = A^{1/2} B A^{1/2}$$Now A^{1/2} B A^{1/2} is positive semidefinite, so has a positive semidefinite square root C, and we can take X = A^{-1/2} C A^{-1/2} so that C = A^{1/2} X A^{... 2$$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}$$4 Hint: It is better to expres n in terms of m and then use divisibility properties$$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$$Now since the RHS is a multiple of 5 so is the LHS, and hence 5|C. Since C is a digit, C=0 or C=5, but 0 is not a solution to your equation. All you have to do is check if 5 is a solution. P.S. The solution by dan_fulea becomes much shorter if ... 2 Once you've built that quadratic, you can avoid factoring it by taking it modulo 10, since you know the answer you want is a single digit integer.$$10C^2+499C−2745\equiv5-C\equiv0\pmod {10}$$From which C=5 quickly. 7 We may check each possible value of C, brute force, quick answer. Well, it is simpler maybe to check the relation modulo nine, i.e. find possible values for C so that$$ (2+C+8)(3+C+1) = (9+0+C+5+8)\text{ modulo nine.} $$We get the simpler equation (C+1)(C+4)=(C+4) modulo nine, so C(C+4)=0 modulo nine. If one factor is divisible by 3, the other ... 0 We need to solve$$x^2-7x-2-\sqrt{x^2-7x-2}-12=0$$or$$\left(\sqrt{x^2-7x-2}+3\right)\left(\sqrt{x^2-7x-2}-4\right)=0$$or$$\sqrt{x^2-7x-2}=4,$$which gives$$x^2-7x-2=16$$and since for these values of x we have$$x^2-7x-14=4>0,$$we'll get all roots from the equation:$$x^2-7x-18=0,$$which gives your answer:$$\{-2,9\}$$2 Another way... Let u = x^2 - 7x -2 so you are looking at$$ \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 ... 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. 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. 0 Those called homogeneous problems can be easily solved Calling d^2 = x^2+y^2 and making y = \lambda x we have d^2 = x^2(1+\lambda^2) $$but from 3x^2+4 x y +6 y^2 = 140 we obtain$$ x^2(3+4\lambda+6\lambda^2) = 140 $$then$$ d^2 = \frac{140(1+\lambda^2)}{3+4\lambda+6\lambda^2} $$now deriving$$ \frac{d}{d\lambda}d^2 = \frac{200(2\lambda^2-... 0 It appears to be a straightforward problem using Lagrange multiplier. $$F(x,y)= 3x^2+4 xy +6y -140 ; \quad G(x,y)=x^2+y^2; \tag1$$ Partially differentiate the Lagrangian $$\frac{F_x}{F_y}=\frac{G_x}{G_y}=\lambda \tag2$$ $$\frac{6x+ 4y}{4 x}=\frac{3x+2y}{2 x}=\frac{x}{y}=\lambda \tag3$$ Cross-multiply , let\dfrac{y}{x}=t,$and simplify $$3 xy - 2 ... 0 Okay, I figured out my problem. My brain somehow thought of subtracting (a - b) instead of dividing (a - b). After dividing, the resulting inequality is 1 > (a + b) Careless mistake! Edit: Since (a - b) is not negative here, the inequality sign should be kept the same. However, if (a - b) was negative, it would be necessary to reverse the inequality sign. 1 Let's go with your idea about factoring the difference of squares, and thus see that:$$a-b > a^2 - b^2 \implies a-b > (a-b)(a+b)$$Since a>b is given, then a-b \ne 0 (and in particular, a-b > 0) and we can divide both sides of the right-hand inequality by a-b. Doing so, we see a cancellation and obtain the following:$$a-b > (a-b)(... 0 Let$f(x)=ax^2+bx+6$and$f(0)=6>0$So function$f(x)\geq 0$for all real$x$So$f(3)=9a+3b+6\geq0\Rightarrow 3a+b\geq -2$0 Your equation can be re-written as $$f(x)=(x-4)(x-10)+k(x-7)(x-13)$$ So it can be seen that$f(4)=27kf(7)=-9f(10)=-9kf(13)=27$Since$f(7)$and$f(13)$are of opposite signs,clearly one root must be lying on the interval$(7,13)$. Therefore we can say that another real root must also exist since in a quadratic equation with real coefficients ... 0 Look at it in an another way. The equation can be re-written as $$x-1+\frac{k}{3}\left(x^2+x+1\right)=0$$ which can be said to be a family of curves passing through points of intersection of the curves$y=x-1$and$y=x^2+x+1$. But then the two curves neither touch nor intersect in real points. So no value of$k$can create the condition you are ... 2 The condition gives: $$2a^2-2a(b+1)-b^2+b=0,$$ which gives $$a=\frac{b+1+\sqrt{3b^2+1}}{2}$$ or $$a=\frac{b+1-\sqrt{3b^2+1}}{2}.$$ In the first case we need to prove that $$\frac{(1+\sqrt3)b}{2}<\frac{b+1+\sqrt{3b^2+1}}{2}<1+\frac{(1+\sqrt3)b}{2}$$ or $$\sqrt3b<1+\sqrt{1+3b^2}<2+\sqrt3b,$$ which is obvious. In the second case we need $$\frac{b+1-... 0 Note: I will convert to LaTex/MathJax at a later time when I am able to... Note I had to alter the value 2*\sqrt{5}+4 to be -0.47124 instead of -0.47123 and the value 2*sqrt{5}+4 to 8.47214 instead of the value 8.47213 for the answer to work. I can't tell why the original values don't work! 1 If construct a Cartesian plane in which you label one axis k and the other axis y, then the graph of the equation y = k^2−8k−4 is a parabola with vertex (-4,-20), which is what you find when you apply the vertex form y = a(k-h)^2 + L to the equation y = k^2−8k−4. But I think the parabola whose vertex you are supposed to find is the graph of$$y =... 0 Hint: Write your function in the form $$f(x)=\left(x-\frac{k}{2}\right)^2-\left(\frac{k^2}{4}-2k-1\right)$$ 1 Step 7 is incorrect, because (as you showed) it leads to$k > 4 \pm2\sqrt{5}$, which leads to $$k > 4- 2\sqrt{5} \quad\text{or}\quad k > 4 +2\sqrt{5},$$ from which you would infer that the range is$\{k: k \color{red}{>} 4- 2\sqrt{5}\} \cup \{k: k > 4 +2\sqrt{5}\},$which is not the correct range; the first$>$should be$<$instead. ... 0 Step 7 is wrong as we cannot write that way rather it should be again like parabola and written in factor form as $${\ ( k-4-2\sqrt{5})(k-4+2\sqrt{5}) }> 0$$ Hence either $${\ k < 4-2 {\sqrt{5}} }$$ or $${\ k > 4 + 2{\sqrt {5}} }$$ 3 Take care with signs - you need$-32$not$+32$. There are various ways of doing this, but one is simply to write: $$(x^2+ax-32)(x^2+bx+62)=x^4+(a+b)x^3+(62-32+ab)x^2+(62a-32b)x-1984$$ and equate coefficients.$a+b=-3762a-32b=-808$Solve for$a$and$b$and thus find the coefficient of$x^2$. 2 Let WLOG $$x_1x_2=62$$ then, $$x_3x_4=32$$ Step$1$$$808=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4$$ $$=62(x_3+x_4)+32(x_1+x_2)$$ We also know that,$x_1+x_2+x_3+x_4=37$Thus, $$x_1+x_2=\frac{62\times37-808}{30}$$ Likewise, $$x_3+x_4=\frac{808-32\times37}{30}$$ Step$2$$$k=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$$ $$=62+32+(x_1+x_2)(x_3+x_4)$$ ... 3 Here is a simpler solution, obtained by recognizing a certain symmetry. Let us give names to the LHS and RHS of the given equation : $$\underbrace{x(x+1)(x+a)(x+a+1)}_{f_a(x)}=\underbrace{a^2}_{g_a(x)} \tag{1}$$ Here are, for two cases ($a=-3$and$a=-4.5$), the joint graphical representations of the curves of$f_a$(blue) and$g_a$(red). The midpoint$...

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Derivative is also one of tool, $\frac{dy}{dx}=2x-2 =0$ At $x=1$: $\frac{d^2y}{dx^2} = 2 >0$ We get min. At $x=1$

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