Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

rps20190122_174705.jpg

Answers (3)

best_answer

@Deepak Garg

good approach.

since you should take assumption x,y are real numbers.

and using that assumption, since x^2-y^2=0  it implies x = y or x = -y.

also xy = -1/2 and since x,y are real it implies x = -y.

it means -\:x^2\:=\:\frac{1}{2}

so x\:=\:\frac{1}{\sqrt{2}}\:or\:x\:=\:-\:\frac{1}{\sqrt{2}}\:and\:x=\:-y

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

@Deepak Garg

(x+y)(x-y) = 0.

x and y are real numbers, it implies as xy = - 1/2 , x+y = 0 is only possiblity.

now x\:=\:\frac{1}{\sqrt{2}}\:or\:x\:=\:-\:\frac{1}{\sqrt{2}}\:and\:x=\:-y

Posted by

Kshitij

View full answer

so roots are (x,y ) = (\frac{1}{\sqrt{2}},-\:\frac{1}{\sqrt{2}}\:) or (-\:\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\:)

Posted by

Kshitij

View full answer

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question