Find the roots of 4x3 + 20x2 - 23x ...
Question

# Find the roots of 4x3 + 20x2 - 23x + 6 = 0 if two of its roots are equal.

JEE/Engineering Exams
Maths
Solution
130
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The equation: [ 4 x^{3}+20 x^{2}-23 x+6=0 ] has a double root. Find all roots. If the equation has a double root we must have a factor [ (x-r)^{2} ] for a root ( r ) since our equation has degree 3 and has real coefficients, the remaining root ( s ) must be real as well and we must have: [ 4 x^{3}+20 x^{2}-23 x+6=4(x-s)(x-r)^{2} ] Multiply the right hand side: [ 4 x^{3}+x^{2}(-8 r-4 s)+x(r(8 s+4 r))-4 r^{2} s ] We should have: [ begin{array}{ll} -8 r-4 s & =20 r(8 s+4 r) & =-23 -4 r^{2} s & =6 end{array} ] One could express ( r ) in ( s ) using the first equation and substitute in the second to obtain a quadratic equation in ( s ) : [ (-5-s)(3 s-5)=-23 ] Giving: [ begin{array}{ll} s=-6 & , r=1 / 2 s=8 / 3 & , r=-23 / 6 end{array} ] It is a simple exercise to show that the first combination generates the solution. On inspection we see that an easier method would have been to try the divisors of 6 to find ( s ).