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Question Number 74196 Answers: 0 Comments: 1

solve for x: 4^x +6^x =9^x

$${solve}\:{for}\:{x}: \\ $$$$\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} \\ $$

Question Number 74164 Answers: 2 Comments: 0

factorize 6x^2 −11xy −10y^2

$${factorize} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} \:−\mathrm{11}{xy}\:−\mathrm{10}{y}^{\mathrm{2}} \\ $$

Question Number 74168 Answers: 3 Comments: 1

Question Number 74148 Answers: 1 Comments: 0

{ ((−x(√3)+2my(√2)=((√3)/3))),((2mx−3y(√6)=1)) :} help me solve it.

$$\begin{cases}{−{x}\sqrt{\mathrm{3}}+\mathrm{2}{my}\sqrt{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}\\{\mathrm{2}{mx}−\mathrm{3}{y}\sqrt{\mathrm{6}}=\mathrm{1}}\end{cases}\:\:\:\:\:\: \\ $$$$ \\ $$$${help}\:{me}\:{solve}\:{it}. \\ $$

Question Number 74129 Answers: 1 Comments: 0

2C_4 ^n = 35C_3 ^(n/2) ⇒ n = ?

$$\mathrm{2}\boldsymbol{{C}}_{\mathrm{4}} ^{\boldsymbol{{n}}} \:=\:\mathrm{35}\boldsymbol{{C}}_{\mathrm{3}} ^{\frac{\boldsymbol{{n}}}{\mathrm{2}}} \: \\ $$$$\Rightarrow\:\boldsymbol{{n}}\:=\:? \\ $$

Question Number 74121 Answers: 0 Comments: 1

Factor the polynomial ((c/2))x^2 +(b−((3c)/2))x+(c−b+a)

$$\mathrm{Factor}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\left(\frac{{c}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\left({b}−\frac{\mathrm{3}{c}}{\mathrm{2}}\right){x}+\left({c}−{b}+{a}\right) \\ $$

Question Number 74111 Answers: 1 Comments: 1

Question Number 74109 Answers: 1 Comments: 3

Question Number 74063 Answers: 0 Comments: 0

Question Number 74024 Answers: 1 Comments: 1

{ ((h^2 +y^2 +(k−z)^2 =s^2 )),((a^2 +(b−y)^2 +z^2 =s^2 )),((ah+y(y−b)+z(z−k)=0)),((((h+a)/2)+yz−(b−y)(k−z)=1)),((b+a(k−z)+hz=1)),((k+h(b−y)+ay=1)) :} Find s_(min) or at least express s=f(y) or g(z).

$$\begin{cases}{{h}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({k}−{z}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\left({b}−{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{ah}+{y}\left({y}−{b}\right)+{z}\left({z}−{k}\right)=\mathrm{0}}\\{\frac{{h}+{a}}{\mathrm{2}}+{yz}−\left({b}−{y}\right)\left({k}−{z}\right)=\mathrm{1}}\\{{b}+{a}\left({k}−{z}\right)+{hz}=\mathrm{1}}\\{{k}+{h}\left({b}−{y}\right)+{ay}=\mathrm{1}}\end{cases} \\ $$$${Find}\:\:{s}_{{min}} \:{or}\:{at}\:{least}\:{express} \\ $$$$\:{s}={f}\left({y}\right)\:{or}\:{g}\left({z}\right). \\ $$

Question Number 74006 Answers: 1 Comments: 1

If 3x^2 e^(log _x 27) =27000 then find x

$${If}\:\mathrm{3}{x}^{\mathrm{2}} {e}^{\mathrm{log}\:_{{x}} \mathrm{27}} =\mathrm{27000}\:{then}\:{find}\:{x} \\ $$

Question Number 74339 Answers: 0 Comments: 2

x^3 +ax^2 +bx+c=0 Let x=((pt+q)/(t+1)) p^3 t^3 +3p^2 qt^2 +3pq^2 t+q^3 +a(t+1)(p^2 t^2 +2pqt+q^2 ) +b(pt+q)(t^2 +2t+1) +c(t^3 +3t^2 +3t+1) = 0 ⇒ (p^3 +ap^2 +bp+c)t^3 +(3p^2 q+ap^2 +2apq+bq+2bp+3c)t^2 +(3q^2 p+aq^2 +2apq+bp+2bq+3c)t +(q^3 +aq^2 +bq+c) = 0 Let coeffs. of t^2 and t be zero. Subtracting and adding them 3pq+a(p+q)+b=0 & 3pq(p+q)+a{(p+q)^2 −2pq} +4apq+3b(p+q)+6c = 0 lets call pq=m , p+q=s ⇒ 3m+as+b=0 ....(i) 3ms+a(s^2 −2m)+4am+3bs+6c=0 ⇒ am+bs+3c=0 ....(ii) ⇒ s=((9c−ab)/(a^2 −3b)) ; m=((b^2 −3ac)/(a^2 −3b)) Now p,q are roots of eq. z^2 −sz+m=0 p,q = (s/2)±(√((s^2 /4)−m)) t^3 =−(((q^3 +aq^2 +bq+c)/(p^3 +ap^2 +bq+c))) (t≠−1) x=((pt+q)/(t+1)) .

$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${Let}\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$${p}^{\mathrm{3}} {t}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} {qt}^{\mathrm{2}} +\mathrm{3}{pq}^{\mathrm{2}} {t}+{q}^{\mathrm{3}} \\ $$$$+{a}\left({t}+\mathrm{1}\right)\left({p}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{2}{pqt}+{q}^{\mathrm{2}} \right) \\ $$$$+{b}\left({pt}+{q}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}\right) \\ $$$$+{c}\left({t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}\right)\:\:\:\:=\:\:\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\left({p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bp}+{c}\right){t}^{\mathrm{3}} \\ $$$$+\left(\mathrm{3}{p}^{\mathrm{2}} {q}+{ap}^{\mathrm{2}} +\mathrm{2}{apq}+{bq}+\mathrm{2}{bp}+\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$+\left(\mathrm{3}{q}^{\mathrm{2}} {p}+{aq}^{\mathrm{2}} +\mathrm{2}{apq}+{bp}+\mathrm{2}{bq}+\mathrm{3}{c}\right){t} \\ $$$$+\left({q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}\right)\:=\:\mathrm{0} \\ $$$${Let}\:{coeffs}.\:{of}\:{t}^{\mathrm{2}} \:{and}\:{t}\:{be}\:{zero}. \\ $$$${Subtracting}\:{and}\:{adding}\:{them} \\ $$$$\:\mathrm{3}{pq}+{a}\left({p}+{q}\right)+{b}=\mathrm{0}\:\:\:\& \\ $$$$\mathrm{3}{pq}\left({p}+{q}\right)+{a}\left\{\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq}\right\} \\ $$$$\:+\mathrm{4}{apq}+\mathrm{3}{b}\left({p}+{q}\right)+\mathrm{6}{c}\:=\:\mathrm{0} \\ $$$${lets}\:{call}\:\:{pq}={m}\:,\:\:{p}+{q}={s}\:\:\Rightarrow \\ $$$$\:\:\mathrm{3}{m}+{as}+{b}=\mathrm{0}\:\:\:\:....\left({i}\right) \\ $$$$\mathrm{3}{ms}+{a}\left({s}^{\mathrm{2}} −\mathrm{2}{m}\right)+\mathrm{4}{am}+\mathrm{3}{bs}+\mathrm{6}{c}=\mathrm{0} \\ $$$$\Rightarrow\:{am}+{bs}+\mathrm{3}{c}=\mathrm{0}\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{9}{c}−{ab}}{{a}^{\mathrm{2}} −\mathrm{3}{b}}\:\:\:;\:\:{m}=\frac{{b}^{\mathrm{2}} −\mathrm{3}{ac}}{{a}^{\mathrm{2}} −\mathrm{3}{b}} \\ $$$$\:{Now}\:\:{p},{q}\:\:{are}\:{roots}\:{of}\:{eq}. \\ $$$$\:\:\:{z}^{\mathrm{2}} −{sz}+{m}=\mathrm{0} \\ $$$$\:\:\:{p},{q}\:=\:\frac{{s}}{\mathrm{2}}\pm\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{4}}−{m}} \\ $$$$\:\:{t}^{\mathrm{3}} =−\left(\frac{{q}^{\mathrm{3}} +{aq}^{\mathrm{2}} +{bq}+{c}}{{p}^{\mathrm{3}} +{ap}^{\mathrm{2}} +{bq}+{c}}\right)\:\:\:\:\:\left({t}\neq−\mathrm{1}\right) \\ $$$$\:\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}}\:. \\ $$

Question Number 73775 Answers: 0 Comments: 0