If one of the roots of equation ${ x }^{ 2 }+ax+3=0$ is $3$ and one of the roots of the equation ${ x }^{ 2 }+ax+b=0$ is three times the other root, then the value of $b$ is

Substituting the root 3 into the first equation yields 9 + 3a + 3 = 0, so a is -4. For the second equation x squared - 4x + b = 0, let the roots be k and 3k; the sum of the roots gives k + 3k = 4, meaning k is 1. Since the roots are 1 and 3, their product gives the value of b as 3.

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