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AlgebraQuestion and Answers: Page 345
Question Number 27983 Answers: 2 Comments: 0
$$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{two}\:\:\mathrm{factors}\:\mathrm{of}\:\mathrm{1000001}\:\mathrm{other}\:\mathrm{than}\:\mathrm{1}\:\mathrm{and}\:\mathrm{1000001} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\mathrm{1}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions}? \\ $$
Question Number 27977 Answers: 1 Comments: 0
$$\mid\mathrm{2}{x}+\mathrm{1}\mid\leqslant\mathrm{2} \\ $$
Question Number 27976 Answers: 1 Comments: 1
$${solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}<\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$
Question Number 27975 Answers: 1 Comments: 0
$${solve}\:{the}\:{inequality} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}>\mathrm{0} \\ $$
Question Number 27761 Answers: 1 Comments: 0
$$\mathrm{4}\left(\mathrm{2a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \\ $$
Question Number 27681 Answers: 0 Comments: 2
$${Find}\:{square}\:{root}\:{of}\:\mathrm{7}−\mathrm{30}\sqrt{\mathrm{2}}{i}\:. \\ $$
Question Number 27662 Answers: 0 Comments: 0
$${factorize}\:{in}\:{C}\left[{x}\right]\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\:+{z}^{\mathrm{2}} \:\:.\: \\ $$
Question Number 27587 Answers: 1 Comments: 1
$$\mathrm{divide}\:\mathrm{12x}\left(\mathrm{8x}−\mathrm{20}\right)\:\mathrm{by}\:\mathrm{4}\left(\mathrm{2x}−\mathrm{5}\right) \\ $$
Question Number 27559 Answers: 2 Comments: 1
$$\mathrm{Change}\:\mathrm{in}\:\mathrm{Q}#\mathrm{27507} \\ $$$$\mathrm{Solve}\:\mathrm{simultaneously}: \\ $$$$\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{y}=\mathrm{13} \\ $$$$\mathrm{x}+\mathrm{2}\sqrt{\mathrm{y}}=\mathrm{10} \\ $$
Question Number 27547 Answers: 0 Comments: 0
$${let}\:{give}\:{A}=\left(_{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}} ^{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}} \right)\:\:\:\:\:{find}\:{A}^{{n}} \:\:\:{and}\:\:{e}^{{A}} \:\:\:. \\ $$
Question Number 27507 Answers: 1 Comments: 4
$$\mathrm{2}\sqrt{{x}\:}+{y}=\mathrm{9}....\left(\mathrm{1}\right) \\ $$$${x}+\:\mathrm{2}\sqrt{{y}}=\mathrm{3}....\left(\mathrm{2}\right) \\ $$$$ \\ $$$${solve}\:{the}\:{simultaneous}\:{equation} \\ $$
Question Number 27503 Answers: 1 Comments: 0
$$\mathrm{If}\:\mathrm{x}=\mathrm{cy}+\mathrm{bz}\:,\mathrm{y}=\mathrm{az}+\mathrm{cx}\:\&\:\mathrm{z}=\mathrm{bx}+\mathrm{ay} \\ $$$$\mathrm{prove}\:\mathrm{that}\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{1}−\mathrm{b}^{\mathrm{2}} }=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:. \\ $$
Question Number 27486 Answers: 0 Comments: 0
Question Number 27449 Answers: 2 Comments: 0
$${factorise}\:{a}^{\mathrm{4}} −\left({b}+{c}\right)^{\mathrm{4}} \\ $$
Question Number 27469 Answers: 1 Comments: 3
Question Number 27419 Answers: 1 Comments: 0
$$\left({x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}\right)/\left({x}−\mathrm{1}\right) \\ $$
Question Number 27384 Answers: 1 Comments: 1
$${let}\:{give}\:\:{p}\left({x}\right)=\:\left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{n}} −\:\frac{\mathrm{1}+{itan}\alpha\:}{\mathrm{1}−{itan}\alpha}\:\:{factorize}\:{p}\left({x}\right)\:{inside} \\ $$$${C}\left[{x}\right]. \\ $$
Question Number 27382 Answers: 1 Comments: 0
$${resolve}\:{inside}\:{C}\:\:\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} +\left(\frac{{z}+{i}}{{z}−{i}}\right)^{{n}} =\:\mathrm{2}{cos}\theta\:{and}\mathrm{0}\:<\theta<\pi\:.{n}\:{integer}. \\ $$
Question Number 27376 Answers: 0 Comments: 2
Question Number 27300 Answers: 3 Comments: 1
Question Number 27213 Answers: 1 Comments: 1
$$\left[\left(\mathrm{16}{x}^{\mathrm{4}} −\mathrm{1}\right)\right]/\left[\mathrm{2}{x}−\mathrm{1}\right]\:{factorise}\:{it} \\ $$
Question Number 27112 Answers: 0 Comments: 2
Question Number 27103 Answers: 0 Comments: 1
$$\mathrm{the}\:\mathrm{intrest}\:\mathrm{on}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{money}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{6}.\mathrm{25}\:\mathrm{year}\:\mathrm{was}\:\frac{\mathrm{5}}{\mathrm{16}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{itself}.\mathrm{what} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{percent}? \\ $$
Question Number 27094 Answers: 1 Comments: 2
$${if}\:\mathrm{1}+{x}+{x}^{\mathrm{2}} =\mathrm{0}\:{find}\:{the}\:{value}\:{of}\: \\ $$$${A}=\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{6}} \:+\left(\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{6}} \:\:+...\:\left(\:\:{x}^{\mathrm{100}} +\frac{\mathrm{1}}{{x}^{\mathrm{100}} }\right)^{\mathrm{6}} \:. \\ $$
Question Number 27059 Answers: 0 Comments: 2
Question Number 27046 Answers: 0 Comments: 0
$${Considering}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}} \\ $$$${If}\:\:\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\Rightarrow\:\:\alpha^{\mathrm{2}} =−\frac{{p}}{\mathrm{3}} \\ $$$${if}\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\:\Rightarrow\:\beta^{\:\mathrm{3}} =\frac{{q}}{\mathrm{2}} \\ $$$${roots}\:{of}\:{the}\:{cubic}\:\:{eq}^{{n}} \:{are}: \\ $$$$\:\:\:\:{x}=\left[−\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \:. \\ $$$$\:{Why}\:{such}\:{a}\:{connection}? \\ $$$${If}\:{equation}\:{is}\:{quadratic}\:{even\_} \\ $$$$\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\:\Rightarrow\:\:\alpha=−\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\Rightarrow\:\beta^{\:\mathrm{2}} =\frac{{c}}{{a}} \\ $$$${roots}\:{of}\:{quadratic}\:{eq}.\:{are}: \\ $$$$\:\:\:\:{x}=\boldsymbol{\alpha}\pm\sqrt{\boldsymbol{\alpha}^{\mathrm{2}} −\boldsymbol{\beta}^{\:\mathrm{2}} }\: \\ $$$${why}\:{such}\:{a}\:{connection}\:?\: \\ $$
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