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AlgebraQuestion and Answers: Page 223

Question Number 128329 Answers: 2 Comments: 0

(2)Solution set : x ∣2x−6 ∣ < 3x is _ (2) If lim_(x→2) ((2−(√(a+bx^3 )))/(x−2)) = H , then lim_(x→2) ((x^2 −4)/( (√(a+bx^3 ))−1)) = _

$$\left(\mathrm{2}\right)\mathrm{Solution}\:\mathrm{set}\::\:{x}\:\mid\mathrm{2}{x}−\mathrm{6}\:\mid\:<\:\mathrm{3}{x}\: \\ $$$$\mathrm{is}\:\_ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{If}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}−\sqrt{{a}+{bx}^{\mathrm{3}} }}{{x}−\mathrm{2}}\:=\:{H}\:,\:{then} \\ $$$$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\:\sqrt{{a}+{bx}^{\mathrm{3}} }−\mathrm{1}}\:=\:\_ \\ $$

Question Number 128290 Answers: 0 Comments: 4

find 10−(1/(10−(1/(10−(1/(10−(1/(10−......))))))))=? (A) 5−2(√6) (B) 5+2(√6) (C) 5±2(√6) (D) none of above please give your answer and explain why!

$${find}\:\mathrm{10}−\frac{\mathrm{1}}{\mathrm{10}−\frac{\mathrm{1}}{\mathrm{10}−\frac{\mathrm{1}}{\mathrm{10}−\frac{\mathrm{1}}{\mathrm{10}−......}}}}=? \\ $$$$\left({A}\right)\:\:\:\:\:\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left({B}\right)\:\:\:\:\:\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left({C}\right)\:\:\:\:\:\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left({D}\right)\:\:\:\:\:{none}\:{of}\:{above} \\ $$$$ \\ $$$${please}\:{give}\:{your}\:{answer}\:{and}\:{explain} \\ $$$${why}! \\ $$

Question Number 128264 Answers: 1 Comments: 3

((2207−(1/(2207−(1/(2207−(1/(2207−(1/(2207−...))))))))))^(1/(8 ))

$$\sqrt[{\mathrm{8}\:\:\:\:}]{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−\frac{\mathrm{1}}{\mathrm{2207}−...}}}}} \\ $$

Question Number 128232 Answers: 0 Comments: 0

1.For n∈N,define s_n =1+2^2^n +2^2^(n+1) .Which of the following is false for some values of n∈N? A.3∣s_n B.7∣s_n C.s_n ∣s_(n+1) D.s_n ^( 2) >s_(n+1) E.12s_1 s_2 ...s_n <s_(n+1)

Question Number 128204 Answers: 0 Comments: 0

Let x, y, z be pairwise distinct real numbers, if (x + y − 7)[z(x + y) + 24] = (y + z − 7)[x(y + z) + 24] = (z + x − 7)[y(z + x) + 24] Find x^2 + y^2 + z^2

$$\mathrm{Let}\:\:\:\mathrm{x},\:\:\mathrm{y},\:\:\mathrm{z}\:\:\:\mathrm{be}\:\mathrm{pairwise}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{if} \\ $$$$\:\:\left(\mathrm{x}\:\:+\:\:\mathrm{y}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{z}\left(\mathrm{x}\:\:+\:\:\mathrm{y}\right)\:\:+\:\:\mathrm{24}\right]\:\:\:=\:\:\left(\mathrm{y}\:\:+\:\:\mathrm{z}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{x}\left(\mathrm{y}\:\:+\:\:\mathrm{z}\right)\:\:+\:\:\mathrm{24}\right]\:\:\:=\:\:\:\left(\mathrm{z}\:\:+\:\:\mathrm{x}\:\:−\:\:\mathrm{7}\right)\left[\mathrm{y}\left(\mathrm{z}\:\:+\:\:\mathrm{x}\right)\:\:+\:\:\mathrm{24}\right] \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{y}^{\mathrm{2}} \:\:+\:\:\mathrm{z}^{\mathrm{2}} \\ $$

Question Number 128171 Answers: 0 Comments: 0

Question Number 128146 Answers: 1 Comments: 0

4<x<5 a=(√(x−3+2(√(x−4)) )) b=(√(x−4))−1 x−a×b=?

$$\mathrm{4}<\mathrm{x}<\mathrm{5}\:\mathrm{a}=\sqrt{\mathrm{x}−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{x}−\mathrm{4}}\:\:}\:\:\:\mathrm{b}=\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{x}−\mathrm{a}×\mathrm{b}=? \\ $$

Question Number 128132 Answers: 1 Comments: 0

If f(x + (1/x)) = x^4 + (1/x^4 ) , find f(5)

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{x}^{\mathrm{4}} \:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:,\:\:\:\:\:\mathrm{find}\:\:\mathrm{f}\left(\mathrm{5}\right) \\ $$

Question Number 128166 Answers: 1 Comments: 0

(√(x−2)) + (√(x+3)) +(√(4x+1)) = 10 for x∈R

$$\:\sqrt{\mathrm{x}−\mathrm{2}}\:+\:\sqrt{\mathrm{x}+\mathrm{3}}\:+\sqrt{\mathrm{4x}+\mathrm{1}}\:=\:\mathrm{10} \\ $$$$\:\mathrm{for}\:\mathrm{x}\in\mathbb{R}\: \\ $$

Question Number 128090 Answers: 0 Comments: 2

someone help cheking this: x^3 =x+c let x=(p−2q)+(q−2p) = −(p+q) p^3 −8q^3 −6pq(p−2q)+ q^3 −8p^3 −6pq(q−2p)+ 3(p−2q)(q−2p)[(p−2q)+(q−2p)] = (p−2q)+(q−2p)+c ⇒ −7(p^3 +q^3 )+6pq(p+q) −3(p+q)[5pq−2(p^2 +q^2 )] +(p+q) = c let p^3 +q^3 =−(c/7) And since x=−(p+q) so with −x^3 =−x−c we have p^3 +q^3 +3pq(p+q)=p+q−c ⇒ (3pq−1)(p+q)=−((6c)/7) 1−9pq+6(p^2 +q^2 )=0 (p^2 +q^2 )(p+q)−pq(p+q)=−(c/7) say p^2 +q^2 =t , then pq=((1+6t)/9) ; p+q=(((−((6c)/7)))/(((1+6t)/3)−1)) p+q=((9c)/(7(1−3t))) p^2 +q^2 =(p+q)^2 −2pq t=[((9c)/(7(1−3t)))]^2 −((2(1+6t))/9) [((9c)/(7(1−3t)))]^2 =((21t+2)/9) ⇒ (21t+2)(3t−1)^2 =(((27c)/7))^2 .....

$${someone}\:{help}\:{cheking}\:{this}: \\ $$$$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c}\:\: \\ $$$${let}\:\:{x}=\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:−\left({p}+{q}\right) \\ $$$${p}^{\mathrm{3}} −\mathrm{8}{q}^{\mathrm{3}} −\mathrm{6}{pq}\left({p}−\mathrm{2}{q}\right)+ \\ $$$${q}^{\mathrm{3}} −\mathrm{8}{p}^{\mathrm{3}} −\mathrm{6}{pq}\left({q}−\mathrm{2}{p}\right)+ \\ $$$$\mathrm{3}\left({p}−\mathrm{2}{q}\right)\left({q}−\mathrm{2}{p}\right)\left[\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:=\:\left({p}−\mathrm{2}{q}\right)+\left({q}−\mathrm{2}{p}\right)+{c} \\ $$$$\Rightarrow \\ $$$$−\mathrm{7}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)+\mathrm{6}{pq}\left({p}+{q}\right) \\ $$$$−\mathrm{3}\left({p}+{q}\right)\left[\mathrm{5}{pq}−\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\right] \\ $$$$+\left({p}+{q}\right)\:=\:{c} \\ $$$${let}\:\:{p}^{\mathrm{3}} +{q}^{\mathrm{3}} =−\frac{{c}}{\mathrm{7}} \\ $$$${And}\:{since}\:\:\:{x}=−\left({p}+{q}\right) \\ $$$${so}\:{with}\:\:−{x}^{\mathrm{3}} =−{x}−{c}\:\:\:{we}\:{have} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right)={p}+{q}−{c} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{3}{pq}−\mathrm{1}\right)\left({p}+{q}\right)=−\frac{\mathrm{6}{c}}{\mathrm{7}} \\ $$$$\:\:\mathrm{1}−\mathrm{9}{pq}+\mathrm{6}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}+{q}\right)−{pq}\left({p}+{q}\right)=−\frac{{c}}{\mathrm{7}} \\ $$$${say}\:\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={t}\:\:,\:{then} \\ $$$${pq}=\frac{\mathrm{1}+\mathrm{6}{t}}{\mathrm{9}}\:\:;\:\:{p}+{q}=\frac{\left(−\frac{\mathrm{6}{c}}{\mathrm{7}}\right)}{\frac{\mathrm{1}+\mathrm{6}{t}}{\mathrm{3}}−\mathrm{1}} \\ $$$${p}+{q}=\frac{\mathrm{9}{c}}{\mathrm{7}\left(\mathrm{1}−\mathrm{3}{t}\right)} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq} \\ $$$${t}=\left[\frac{\mathrm{9}{c}}{\mathrm{7}\left(\mathrm{1}−\mathrm{3}{t}\right)}\right]^{\mathrm{2}} −\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{6}{t}\right)}{\mathrm{9}} \\ $$$$\left[\frac{\mathrm{9}{c}}{\mathrm{7}\left(\mathrm{1}−\mathrm{3}{t}\right)}\right]^{\mathrm{2}} =\frac{\mathrm{21}{t}+\mathrm{2}}{\mathrm{9}} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{21}{t}+\mathrm{2}\right)\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} =\left(\frac{\mathrm{27}{c}}{\mathrm{7}}\right)^{\mathrm{2}} \\ $$$$..... \\ $$

Question Number 128065 Answers: 0 Comments: 1

Question Number 128048 Answers: 3 Comments: 0

Σ_(n=0) ^∞ (((−1)^n )/(8n+3)) =?

$$\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{8n}+\mathrm{3}}\:=? \\ $$

Question Number 128007 Answers: 0 Comments: 2

Can anyone find any error in my attempt to improve upon the Cardano′s cubic formula.. or as an alternate to the trigonometric solution... here it goes: X^3 −pX−q=0 let (X/( (√p))) = x ; and with (q/(p(√p))) = c , x^3 −x−c=0 let x=(k+1)(p+q) =(p+kq) + (kp+q) p^3 +k^3 q^3 +3kpq(p+kq)+ k^3 p^3 +q^3 +3kpq(kp+q) −(p+kq)−(kp+q)=c ⇒ (1+k^3 )(p^3 +q^3 )+ (p+kq)(3kpq−1)+ (kp+q)(3kpq−1)= c let 3kpq=1 ⇒ p^3 +q^3 =(c/(1+k^3 )) & p^3 q^3 =(1/(27k^3 )) ; hence p^3 , q^3 = (c/(2(1+k^3 )))±(√((c^2 /(4(1+k^3 )^2 ))−(1/(27k^3 )))) lets choose upon a value of k such that D=0 ⇒ 4(1+k^3 )^2 =27c^2 k^3 .....(I) (just a quadratic..) first for c^2 >(8/(27)) we always can get two real k values, and even p=q then. x=(k+1)(p+q) = 2(k+1)p x=2(k+1)[(c/(2(1+k^3 )))]^(1/3) but simply pq=(1/(3k)) hence x=2(k+1)p = ((2(k+1))/( (√(3k)))) . ________________________ even for c=1 i dint get a correct answer, please help error-freeing it. (Thanks!)

$${Can}\:{anyone}\:{find}\:{any}\:{error}\:{in} \\ $$$${my}\:{attempt}\:{to}\:{improve}\:{upon} \\ $$$${the}\:{Cardano}'{s}\:{cubic}\:{formula}.. \\ $$$${or}\:{as}\:{an}\:{alternate}\:{to}\:{the} \\ $$$${trigonometric}\:{solution}... \\ $$$${here}\:{it}\:{goes}: \\ $$$$\:\:\:\:{X}^{\mathrm{3}} −{pX}−{q}=\mathrm{0} \\ $$$${let}\:\:\frac{{X}}{\:\sqrt{{p}}}\:=\:{x}\:;\:\:{and}\:{with}\:\frac{{q}}{{p}\sqrt{{p}}}\:=\:{c}\:, \\ $$$$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:\boldsymbol{{x}}=\left(\boldsymbol{{k}}+\mathrm{1}\right)\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\boldsymbol{{p}}+\boldsymbol{{kq}}\right)\:+\:\left(\boldsymbol{{kp}}+\boldsymbol{{q}}\right) \\ $$$$\:\:\:\:{p}^{\mathrm{3}} +{k}^{\mathrm{3}} {q}^{\mathrm{3}} +\mathrm{3}{kpq}\left({p}+{kq}\right)+ \\ $$$$\:\:\:\:{k}^{\mathrm{3}} {p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{kpq}\left({kp}+{q}\right) \\ $$$$\:\:\:\:\:−\left({p}+{kq}\right)−\left({kp}+{q}\right)={c} \\ $$$$\Rightarrow \\ $$$$\:\:\:\left(\mathrm{1}+{k}^{\mathrm{3}} \right)\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)+ \\ $$$$\:\:\:\left({p}+{kq}\right)\left(\mathrm{3}{kpq}−\mathrm{1}\right)+ \\ $$$$\:\:\:\left({kp}+{q}\right)\left(\mathrm{3}{kpq}−\mathrm{1}\right)=\:{c} \\ $$$${let}\:\:\:\mathrm{3}\boldsymbol{{kpq}}=\mathrm{1}\:\:\:\Rightarrow \\ $$$$\:\:\:{p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\frac{{c}}{\mathrm{1}+{k}^{\mathrm{3}} } \\ $$$$\&\:\:\:{p}^{\mathrm{3}} {q}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}{k}^{\mathrm{3}} }\:\:\:;\:\:{hence} \\ $$$$\:\:\:{p}^{\mathrm{3}} ,\:{q}^{\mathrm{3}} \:\:=\: \\ $$$$\:\:\frac{{c}}{\mathrm{2}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{k}^{\mathrm{3}} }} \\ $$$${lets}\:{choose}\:{upon}\:{a}\:{value}\:{of}\:\boldsymbol{{k}} \\ $$$${such}\:{that}\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{4}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{27}{c}^{\mathrm{2}} {k}^{\mathrm{3}} \:\:\:.....\left({I}\right) \\ $$$$\:\:\left({just}\:{a}\:{quadratic}..\right) \\ $$$${first}\:{for}\:\:\:{c}^{\mathrm{2}} >\frac{\mathrm{8}}{\mathrm{27}}\:\:{we}\:\:{always}\:{can} \\ $$$${get}\:{two}\:{real}\:{k}\:{values},\:{and}\:{even} \\ $$$${p}={q}\:{then}. \\ $$$$\:\:{x}=\left({k}+\mathrm{1}\right)\left({p}+{q}\right) \\ $$$$\:\:\:\:=\:\mathrm{2}\left({k}+\mathrm{1}\right){p}\: \\ $$$$\:{x}=\mathrm{2}\left({k}+\mathrm{1}\right)\left[\frac{{c}}{\mathrm{2}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)}\right]^{\mathrm{1}/\mathrm{3}} \\ $$$${but}\:{simply}\:\:{pq}=\frac{\mathrm{1}}{\mathrm{3}{k}}\:\:\:{hence} \\ $$$$\:{x}=\mathrm{2}\left({k}+\mathrm{1}\right){p}\:=\:\frac{\mathrm{2}\left({k}+\mathrm{1}\right)}{\:\sqrt{\mathrm{3}{k}}}\:\:. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${even}\:{for}\:\:{c}=\mathrm{1}\:\:{i}\:{dint}\:{get}\:{a} \\ $$$${correct}\:{answer},\:\:{please}\:{help} \\ $$$${error}-{freeing}\:{it}.\:\left(\mathcal{T}{hanks}!\right) \\ $$

Question Number 127968 Answers: 1 Comments: 0

How many xεR satisfy (x)^(1/7) −(x)^(1/5) = (x)^(1/3) −(√x)

$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{x}\epsilon\mathbb{R}\:\mathrm{satisfy}\:\sqrt[{\mathrm{7}}]{\mathrm{x}}\:−\sqrt[{\mathrm{5}}]{\mathrm{x}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\sqrt{\mathrm{x}}\: \\ $$

Question Number 127924 Answers: 0 Comments: 3

x^3 =x+c Solve for x, even when c<(2/(3(√3))).

$$\:\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Solve}\:{for}\:{x},\:{even}\:{when}\:{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}. \\ $$

Question Number 127908 Answers: 1 Comments: 0

(1/(1.3))+(2/(1.3.5))+(3/(1.3.5.7))+(4/(1.3.5.7.9))+...=?

$$\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}+\frac{\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}+...=? \\ $$

Question Number 127887 Answers: 0 Comments: 0