Menu Close

find-the-difference-of-the-roots-of-the-following-quadratic-equation-3-2-2-x-2-1-2-x-2-

Tinku Tara 0 Comments
Pin




Question Number 55658 by otchereabdullai@gmail.com last updated on 01/Mar/19
find the difference of the roots of the following quadratic equation (3+2(√(2 )))x^2 +(1+(√2))x =2
$$\mathrm{find}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{following}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:}\right)\mathrm{x}^{\mathrm{2}} \:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}\:=\mathrm{2} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19
α+β=−(((1+(√2) ))/(3+2(√2) )) αβ=((−2)/(3+2(√2) )) (α−β)^2 =(α+β)^2 −4αβ =((1+2(√2) +2)/((3+2(√2) )^2 ))−4(((−2)/(3+2(√2) ))) =((3+2(√2))/((3+2(√2) )^2 ))+(8/(3+2(√2) )) =(1/(3+2(√2) ))+(8/(3+2(√2))) =(9/(3+2(√2))) (α−β)=(√(9/(3+2(√2)))) =(3/( (√2) +1))
$$\alpha+\beta=−\frac{\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}\:\:\:\alpha\beta=\frac{−\mathrm{2}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:} \\ $$$$\left(\alpha−\beta\right)^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }−\mathrm{4}\left(\frac{−\mathrm{2}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:}+\frac{\mathrm{8}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{9}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\left(\alpha−\beta\right)=\sqrt{\frac{\mathrm{9}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}\:+\mathrm{1}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 01/Mar/19
Thank you Prof
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Prof} \\ $$
Commented by mr W last updated on 01/Mar/19
please check sir, should it not be: (3/( (√2)+1))=3((√2)−1) ?
$${please}\:{check}\:{sir},\:{should}\:{it}\:{not}\:{be}: \\ $$$$\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19
yes sir..in hurry...
$${yes}\:{sir}..{in}\:{hurry}… \\ $$
Commented by malwaan last updated on 02/Mar/19
(√((9(3−2(√2)))/(9−8)))=3(√(3−2(√2)))
$$\sqrt{\frac{\mathrm{9}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{9}−\mathrm{8}}}=\mathrm{3}\sqrt{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by mr W last updated on 02/Mar/19
and 3(√(3−2(√2)))=3(√(((√2)−1)^2 ))=3((√2)−1)
$${and}\:\mathrm{3}\sqrt{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}=\mathrm{3}\sqrt{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{3}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$
Commented by otchereabdullai@gmail.com last updated on 02/Mar/19
thanks Sir malwaan and Prof W for the correction
$$\:\:\mathrm{thanks}\:\mathrm{Sir}\:\mathrm{malwaan}\:\mathrm{and}\:\mathrm{Prof}\:\mathrm{W}\:\mathrm{for}\: \\ $$$$\mathrm{the}\:\mathrm{correction} \\ $$

Pin

Leave a Reply

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