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

## 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.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]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[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" _builder_version="4.3.1" inline_fonts="Aclonica" open="off"]

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

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### Algebra (Root of equation)

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.1" open="off"]

[/et_pb_text][et_pb_tabs _builder_version="4.3.1"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]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.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]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 $\gamma$.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]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 $(1+\sqrt{3}) \times(1-\sqrt{3}) \times \gamma=-4$[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]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,   $\frac{b}{3}=(1+\sqrt{3}) \times(1-\sqrt{3})+_{\gamma} \times\{(1+\sqrt{3})+(1-\sqrt{3})\}=2$.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" custom_margin="20px||20px||false|false" global_module="50753" saved_tabs="all" locked="off"][et_pb_fullwidth_header title="I.S.I. & C.M.I. Program" button_one_text="Learn more" button_one_url="https://www.cheenta.com/isicmientrance/" header_image_url="https://www.cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="4.2.2" title_level="h2" title_font="Acme||||||||" background_color="#220e58" custom_button_one="on" button_one_text_color="#1a0052" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

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