Menu Close

Question-189131




Question Number 189131 by Shrinava last updated on 12/Mar/23
Answered by cortano12 last updated on 12/Mar/23
(√(19+2(√(4×15)))) =2+(√(15))
$$\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{4}×\mathrm{15}}}\:=\mathrm{2}+\sqrt{\mathrm{15}} \\ $$
Answered by BaliramKumar last updated on 12/Mar/23
2. solution:−        x^2 +6x−4y^2 −16y−7  x^2 +2(x)(3)+3^2 _() −3^2 −(2y)^2 −2(2y)(4)−4^2 +4^2 −7_()   (x+3)^2 −(2y+4)^2   (x+2y+7)(x−2y−1)
$$\mathrm{2}.\:{solution}:− \\ $$$$\:\:\:\: \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{16}{y}−\mathrm{7} \\ $$$$\underset{} {\underbrace{{x}^{\mathrm{2}} +\mathrm{2}\left({x}\right)\left(\mathrm{3}\right)+\mathrm{3}^{\mathrm{2}} }}−\underset{} {\underbrace{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{y}\right)\left(\mathrm{4}\right)−\mathrm{4}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{7}}} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{2}{y}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{2}{y}+\mathrm{7}\right)\left({x}−\mathrm{2}{y}−\mathrm{1}\right) \\ $$
Commented by Shrinava last updated on 13/Mar/23
thankyou dearSer
$$\mathrm{thankyou}\:\mathrm{dearSer} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *