$$^{If\ m>n>0,\ x=m^2+n^2\ m,\ and\ y=2mn}$$ $$what\ is\ the\ value\ of\ \sqrt{x^2-y^2}in\ terms\ of\ mm\ and\ nn?$$
$$A. m^2+n^2$$
$$B. 2m^2+n^2$$
$$C. m^2+2n^2$$
$$D. n^2−m^2$$
$$E. m^2−n^2$$
OA is E
Which fast approach can i use to get here? can any expert help me out? Thanks so much
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$$^{If\ m>n>0,\ x=m^2+n^2\ m,\ and\ y=2mn}$$
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x + y = m^2 + n^2 + 2mnRoland2rule wrote:$$^{If\ m>n>0,\ x=m^2+n^2\ m,\ and\ y=2mn}$$ $$what\ is\ the\ value\ of\ \sqrt{x^2-y^2}in\ terms\ of\ mm\ and\ nn?$$
$$A. m^2+n^2$$
$$B. 2m^2+n^2$$
$$C. m^2+2n^2$$
$$D. n^2−m^2$$
$$E. m^2−n^2$$
OA is E
Which fast approach can i use to get here? can any expert help me out? Thanks so much
x + y = m^2 + 2mn + n^2
x + y = (m + n)^2
x - y = (m-n)^2
x^2 - y^2 = (x + y)(x - y)
x^2 - y^2 = [(m + n)^2]*[(m-n)^2]
[x^2 - y^2]^0.5 = (m + n)*(m-n)
[x^2 - y^2]^0.5 = (m + n)*(m-n)
[x^2 - y^2]^0.5 = m^2 - n^2
Answer is E
