## What are we learning ?

**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

##### Source of the problem

### B.Stat. (Hons.) Admission Test 2005 – Objective problem 2

##### Key Competency

### Algebra (Root of equation)

##### Difficulty Level

4/10

##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

## Start with hints

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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