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AlgebraQuestion and Answers: Page 242

Question Number 105536 Answers: 1 Comments: 0

Question Number 105534 Answers: 1 Comments: 2

Question Number 105456 Answers: 2 Comments: 0

Prove that the curve y=x^4 +3x^2 +2x does not meet the straight line y=2x−1 and find the distace between their nearest points.(Answer (1/(√5)))

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{2x} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{y}=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distace}\:\mathrm{between} \\ $$$$\mathrm{their}\:\mathrm{nearest}\:\mathrm{points}.\left(\mathrm{Answer}\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right) \\ $$

Question Number 105448 Answers: 3 Comments: 2

Given { ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z)),((((4z^2 )/(4z^2 +1)) = x)) :} . find x+y+z ?

$$\mathcal{G}{iven}\:\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases}\:\:\:.\:{find} \\ $$$${x}+{y}+{z}\:?\: \\ $$

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Question Number 105340 Answers: 1 Comments: 0

1\Show that the polynomial, P(x)=x^n +ax+b, n≥2 (a,b)∈R^2 has at most 3 real distinct roots. 2\Show that the equation x^4 +4ax+b=0 has no more than 2 real distinct roots.

$$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polynomial},\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} +\mathrm{ax}+\mathrm{b},\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{at}\:\mathrm{most}\:\mathrm{3}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{4ax}+\mathrm{b}=\mathrm{0}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{2}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$

Question Number 105322 Answers: 0 Comments: 1

Question Number 105320 Answers: 0 Comments: 0

find C kx^2 +ky^2 +z^2 ≥C x,y,z,k∈R such that xy+yz+zx=1

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{C}} \\ $$$${kx}^{\mathrm{2}} +{ky}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{C} \\ $$$${x},{y},{z},{k}\in\mathbb{R}\:\:{such}\:{that} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$

Question Number 105234 Answers: 0 Comments: 0

f:R→R x,y∈R f(f(x)f(y))+f(x+y)=f(xy) f(x)=?

$${f}:\mathbb{R}\rightarrow\mathbb{R}\:\:{x},{y}\in\mathbb{R} \\ $$$${f}\left({f}\left({x}\right){f}\left({y}\right)\right)+{f}\left({x}+{y}\right)={f}\left({xy}\right) \\ $$$${f}\left({x}\right)=? \\ $$

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Question Number 105158 Answers: 4 Comments: 0

{ ((xy+x+y = 20)),((x^2 +y^2 = 40)) :}find x and y

$$\begin{cases}{{xy}+{x}+{y}\:=\:\mathrm{20}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{40}}\end{cases}{find}\:{x}\:{and}\:{y} \\ $$

Question Number 105124 Answers: 1 Comments: 0

x^4 +ax^3 +bx^2 +cx+d=0 Find x.

$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${Find}\:{x}. \\ $$

Question Number 105120 Answers: 1 Comments: 0

f(2f(x)+f(y))=2x+y f:R→R f(x)=? x,y∈R

$${f}\left(\mathrm{2}{f}\left({x}\right)+{f}\left({y}\right)\right)=\mathrm{2}{x}+{y}\:\:\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=?\:\:\:\:{x},{y}\in\mathbb{R} \\ $$$$ \\ $$

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Question Number 105023 Answers: 1 Comments: 0

(((x^2 −y^2 )^2 )/((1−x^2 )(y^2 −1)))+(((1−x^2 )^2 )/((x^2 −y^2 )(y^2 −1)))+(((y^2 −1)^2 )/((1−x^2 )(x^2 −y^2 )))=

$$\frac{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}= \\ $$$$ \\ $$

Question Number 104973 Answers: 1 Comments: 1

2(√3) + i is a cubic root for 18(√3) + 35i find the other 2 cubic roots

$$\mathrm{2}\sqrt{\mathrm{3}}\:+\:\boldsymbol{{i}}\:\:\:{is}\:{a}\:{cubic}\:{root}\:{for} \\ $$$$\mathrm{18}\sqrt{\mathrm{3}}\:+\:\mathrm{35}\boldsymbol{{i}}\: \\ $$$$\boldsymbol{{find}}\:\:\boldsymbol{{the}}\:\boldsymbol{{other}}\:\mathrm{2}\:\boldsymbol{{cubic}}\:\boldsymbol{{roots}} \\ $$

Question Number 104859 Answers: 2 Comments: 0

Question Number 104832 Answers: 1 Comments: 0

{ ((x+y+(x^2 /y^2 ) = 7)),(((((x−y)x^2 )/y^2 ) = 12 )) :}

$$\begin{cases}{{x}+{y}+\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{7}}\\{\frac{\left({x}−{y}\right){x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:=\:\mathrm{12}\:}\end{cases} \\ $$

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Question Number 104811 Answers: 2 Comments: 0

Given { ((a+b(√3)−2c = 1)),((3b^2 +c^2 = 2a^2 )),((a^2 +4ac = 5c^2 )) :} find b

$${Given}\:\begin{cases}{{a}+{b}\sqrt{\mathrm{3}}−\mathrm{2}{c}\:=\:\mathrm{1}}\\{\mathrm{3}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{2}{a}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\mathrm{4}{ac}\:=\:\mathrm{5}{c}^{\mathrm{2}} }\end{cases} \\ $$$${find}\:{b} \\ $$

Question Number 104717 Answers: 0 Comments: 0

In the a sport camp, 65% children know playing the football,70%−in voleyball,75%−in basketball.What is least number of children who know playing all above three sport games? (Answer 10%)

$$\mathrm{In}\:\mathrm{the}\:\mathrm{a}\:\:\mathrm{sport}\:\mathrm{camp},\:\mathrm{65\%}\:\mathrm{children}\:\mathrm{know} \\ $$$$\mathrm{playing}\:\mathrm{the}\:\mathrm{football},\mathrm{70\%}−\mathrm{in}\:\mathrm{voleyball},\mathrm{75\%}−\mathrm{in} \\ $$$$\mathrm{basketball}.\mathrm{What}\:\mathrm{is}\:\mathrm{least}\:\mathrm{number}\:\mathrm{of}\:\mathrm{children}\:\mathrm{who} \\ $$$$\mathrm{know}\:\mathrm{playing}\:\mathrm{all}\:\mathrm{above}\:\mathrm{three}\:\mathrm{sport}\:\mathrm{games}? \\ $$$$\left(\mathrm{Answer}\:\mathrm{10\%}\right) \\ $$

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