Given that the equation x^2 +cx+d=0 and x^2 +3cx+4d=0 have a common root, express d in term of c a+b=-c, a+q=-3c ab=d aq=4d aq-ab=3d a(q-b)=3d...(1) q-b=-2c...(2) subst, a=-3d/2c a^2 +ca+d=0 (-3d/2c)^2 +c(-3d/2c)+d=0 d=(2/9)(c^2) am I right?
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Originally Posted by Trefoil2727 Given that the equation x^2 +cx+d=0 and x^2 +3cx+4d=0 have a common root, express d in term of c a+b=-c, a+q=-3c ab=d aq=4d aq-ab=3d a(q-b)=3d...(1) q-b=-2c...(2) subst, a=-3d/2c a^2 +ca+d=0 (-3d/2c)^2 +c(-3d/2c)+d=0 d=(2/9)(c^2) am I right? Put the value for d into your two quadratic equations and factor. They share a common solution x = -3c. -Dan
Substract de first ecuation from the second: The common root of the two quadratics is also the root of the last equation. So, the common root is Now replace x from any of the two quadratics with
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