![We\ are\ given\ that:\\x^2-y^2=15\\x+y=3 \sqrt{5}\\Now,\\We\ know\ that,\\If\ x\ and\ y\ are\ two\ real\ numbers,\\Then:\\(x+y)(x-y)=x^2-y^2\\Here,\\Substituting\ (x+y)=3\sqrt{5}\ in\ the\ LHS\ and\ (x^2-y^2)=15\ in\ the\ RHS,\\(x+y)(x-y)=x^2-y^2\\3\sqrt{5} (x-y)= 15\\Hence,\\(x-y)=\frac{15}{3\sqrt{5} } \\(x-y)=\frac{5*3}{3\sqrt{5} }\\(x-y)=\frac{5}{\sqrt{5} } [Cancelling\ 3\ from\ the\ numerator\ and\ denominator]\\(x-y)=\frac{\sqrt{5}*\sqrt{5} }{\sqrt{5} }\\(x-y)=\sqrt{5}](/tpl/images/1082/0687/67ee4.png)
X and y are two non-zero real numbers such that: x²-y²=15 and x+y=3√5. then x-y=
A. √15
B. √5<...
Mathematics, 01.02.2021 14:40 Angieleelee
X and y are two non-zero real numbers such that: x²-y²=15 and x+y=3√5. then x-y=
A. √15
B. √5
C. -3√5
Answers: 3
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