please solve this Share with your friends Share 0 Vijay Kumar Gupta answered this Dear Students, Consider the following equation. x2-6x+2=0Note that if α and β are roots of an equation ax2+bx+c=0, then sum of roots, α+β =-ba product of roots, α β=caComparing equation, x2-6x+2=0 with ax2+bx+c=0, we get a=1, b=-6, c=2Since m and n are the roots of equation x2-6x+2=0, using aboverelations, we get m+n=--61=6and mn=21=2This gives, m-n2=m2+n2-2mn =m2+n2-2mn+4mn-4mn =m2+n2+2mn-4mn =m+n2-4mn =62-42 =36-8 =28⇒m-n=±28=±27 This implies that, 1n- 1m=m-nnm=±272=±7 Regards 1 View Full Answer Jegannathan Anandaraman answered this Hello Dhruva, given quadratic equation x^2 - 6 x + 8 = 0 Here a = 1 b = -6 and c = 8 Sum of roots i.e m+n = -b/a = 6 Product of roots m * n = c/a = 8 Now 1/n - 1/m = (m-n)/ mn (m - n)^2 = (m + n)^2 - 4 m n = 36 - 32 = 4 So m - n = +/- 2 Hence required 1/n - 1/m = +/- 1/4 1
