Question and Answers Forum

AlgebraQuestion and Answers: Page 259

Question Number 96650 Answers: 1 Comments: 0

solve 2 ((2y−1))^(1/(3 )) = y^3 +1

$$\mathrm{solve}\:\mathrm{2}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{2y}−\mathrm{1}}\:=\:\mathrm{y}^{\mathrm{3}} +\mathrm{1} \\ $$

Question Number 96584 Answers: 1 Comments: 4

Question Number 96527 Answers: 2 Comments: 1

proof that 1^2 +2^2 +3^2 +....+n^2 =((n(2n+1)(n+1))/6)

$$\mathrm{proof}\:\mathrm{that}\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +....+\mathrm{n}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{2n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$

Question Number 96505 Answers: 2 Comments: 2

((√((8)^(1/(4 )) −(√((√2)+1))))/((√((8)^(1/(4 )) +(√((√2)−1))))−(√((8)^(1/(4 )) −(√((√2)−1)))))) ?

$$\frac{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}}{\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}+\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}−\sqrt{\sqrt[{\mathrm{4}\:\:}]{\mathrm{8}}−\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}}\:? \\ $$

Question Number 96460 Answers: 1 Comments: 8

Question Number 96407 Answers: 1 Comments: 4

Question Number 96390 Answers: 1 Comments: 0

Question Number 96340 Answers: 0 Comments: 4

The equations of two circles S_1 and S_2 are given by S_1 : x^2 + y^2 +2x +2y + 1 = 0 S_2 : x^2 + y^2 −4x + 2y +1 = 0. Show that S_1 and S_2 touch each other externally and obtain the equation of the common tangent T at the point of contact.

$$\mathrm{The}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{two}\:\mathrm{circles}\:{S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$$$\:{S}_{\mathrm{1}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}{y}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:{S}_{\mathrm{2}} :\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\:\mathrm{2}{y}\:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{S}_{\mathrm{1}} \:\mathrm{and}\:{S}_{\mathrm{2}} \:\mathrm{touch}\:\mathrm{each}\:\mathrm{other}\:\mathrm{externally}\:\mathrm{and}\:\mathrm{obtain} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common}\:\mathrm{tangent}\:{T}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact}. \\ $$

Question Number 96329 Answers: 0 Comments: 2

x⌊x⌊x⌊x⌋⌋⌋=88 x>0

$${x}\lfloor{x}\lfloor{x}\lfloor{x}\rfloor\rfloor\rfloor=\mathrm{88} \\ $$$${x}>\mathrm{0} \\ $$

Question Number 96321 Answers: 1 Comments: 1

It is given that x^2 =2^x . Find x.

$${It}\:{is}\:{given}\:{that}\:{x}^{\mathrm{2}} =\mathrm{2}^{{x}} .\:{Find}\:{x}. \\ $$

Question Number 96311 Answers: 1 Comments: 0

(4+(√(15)))^x + (4−(√(15)))^x = 62 x=?

$$\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{{x}} \:+\:\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{{x}} \:=\:\mathrm{62}\: \\ $$$${x}=? \\ $$

Question Number 96296 Answers: 1 Comments: 0

Question Number 96289 Answers: 0 Comments: 1

1010^x +2020^x =4040^x x=?

$$\mathrm{1010}^{{x}} +\mathrm{2020}^{{x}} =\mathrm{4040}^{{x}} \\ $$$${x}=? \\ $$

Question Number 96244 Answers: 1 Comments: 2

The line y = mx meets the parabola y = (x − a)(b − x) tangentially where 0 < a < b. Show that m = ((√b) − (√a))^2

$$ \\ $$$$\:\:\mathrm{The}\:\mathrm{line}\:{y}\:=\:{mx}\:\:\mathrm{meets}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\:\:{y}\:=\:\left({x}\:−\:{a}\right)\left({b}\:−\:{x}\right)\:\mathrm{tangentially}\:\mathrm{where} \\ $$$$\:\:\mathrm{0}\:<\:{a}\:<\:{b}.\:\mathrm{Show}\:\mathrm{that}\:{m}\:=\:\left(\sqrt{{b}}\:−\:\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 96189 Answers: 3 Comments: 0

find a common roots from the two quadratic eq 24x^2 +(p+4)x−1=0 and 6x^2 +11x+p+2=0

$$\mathrm{find}\:\mathrm{a}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq} \\ $$$$\mathrm{24x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{6x}^{\mathrm{2}} +\mathrm{11x}+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$

Question Number 96171 Answers: 0 Comments: 2

(4+(√(15)))^(3/2) −(4−(√(15)))^(3/2) = k(√6) find k

$$\left(\mathrm{4}+\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{4}−\sqrt{\mathrm{15}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\:\mathrm{k}\sqrt{\mathrm{6}} \\ $$$$\mathrm{find}\:\mathrm{k}\: \\ $$

Question Number 96131 Answers: 1 Comments: 0

(((x+4)^2 ))^(1/(3 )) + 4 (((x−3)^2 ))^(1/(3 )) + 5 ((x^2 +x−12))^(1/(3 )) = 0

$$\sqrt[{\mathrm{3}\:\:}]{\left({x}+\mathrm{4}\right)^{\mathrm{2}} }\:+\:\mathrm{4}\:\sqrt[{\mathrm{3}\:\:}]{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+\:\mathrm{5}\:\sqrt[{\mathrm{3}\:\:}]{{x}^{\mathrm{2}} +{x}−\mathrm{12}}\:=\:\mathrm{0} \\ $$

Question Number 96072 Answers: 1 Comments: 0

(x^2 +24x+24).(x^2 +x+24)= 24x^2

$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{24x}+\mathrm{24}\right).\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{24}\right)=\:\mathrm{24x}^{\mathrm{2}} \\ $$

Question Number 96021 Answers: 0 Comments: 1

if p and q are two complex number and p×q=m ,m is a real number . is there always exists a p^(1/3) and q^(1/3) (we know p^(1/3) and q^(1/3) each has actually 3 values) such that p^(1/3) ×q^(1/3) =m^(1/3) .where m^(1/3) is real .?? how to prove it?

Question Number 95964 Answers: 0 Comments: 0

{ ((x^2 +y^2 =13)),((2x^2 +3y=2xy^2 )) :}

$$\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{13}}\\{\mathrm{2x}^{\mathrm{2}} +\mathrm{3y}=\mathrm{2xy}^{\mathrm{2}} }\end{cases} \\ $$

Question Number 95967 Answers: 1 Comments: 1