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# Root of Equation- B.Stat. (Hons.) Admission Test 2005 – Objective Problem 2

Try this beautiful problem of Algebra prticularly in cubic equation fromB.Stat. (Hons.) Admission Test 2005. You may use sequential hints to help you solve the problem.

## Competency in Focus: Root of Equation

This problem from Root of equation for B.Stat. (Hons.) Admission Test 2005 Objective Problem 2  is based on calculating a variable in a given equation.

## First look at the knowledge graph:- ## Next understand the problem

If $\sqrt{3}+1$ is a root of the equation $3x^{3}+ax^{2}+bx+12=0$ where a and b are rational numbers, then b is equal to (A) -6 (B) 2 (C) 6 (D) 10

### Algebra (Root of equation)

4/10
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Do you really need a hint? Try it first!
If the co effeicent of any polynomial equation is rational number and one of the root is a surd or coplex number, then the other root must be the conjugate.
Since our polynomial is of degree 3, there must be three roots of the equation. We already know the two of them so let the third one is $\gamma$.
we can use Vieta’s Theorem, that gives, the product of the roots is $\frac{c}{a}$ in our case $\frac{-12}{3}$ =$-4$. so we can say that $(1+\sqrt{3}) \times(1-\sqrt{3}) \times \gamma=-4$
Now we have calculated the value of gamma. Also we have from Vieta’s Theorem, that sum of products of the roots taken two roots at a time is $\frac{b}{3}$. So we can write,   $\frac{b}{3}=(1+\sqrt{3}) \times(1-\sqrt{3})+_{\gamma} \times\{(1+\sqrt{3})+(1-\sqrt{3})\}=2$.

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