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Tuesday, September 19, 2017

2017/024)If m is a root of A(x)=x^2−nx+10=0, and n is a root of B(x)=x−mx+20=0, what is the value of (\frac{n}{m} + \frac{m}{n} )^2

m is root of  A(x)=x^2−nx+10=0
so m^2-mn+10= 0

n is root of B(x)=x−mx+20=0 hence  n^2-mn+20 = 0

so we have m(m-n) = - 10 and n(m-n) = 20 or n = -2m

so \frac{n}{m} = -2 and \frac{m}{n} = -\frac{1}{2} or (\frac{n}{m} + \frac{m}{n})^2 = \frac{25}{4}

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