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Quadratic Equations

Section: Mathematics

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Question : 13 of 51

Marks: +1, -0

When the roots of

x^3+\alpha x^2+\beta x+6=0

are increased by 1 , if one of the resultant values is the least root of

x^4-6x^3+11x^2-6x=0

[AP EAPCET 22 May 2025 S2]

$\alpha-\beta+5=0$
$\alpha+\beta+7=0$
$2\alpha+\beta+7=0$
$2\alpha+3\beta-1=0$

Solution:

\;x(x^3-6x^2+11x-6)=0

\;\Rightarrow\;\;x(x-1)(x-2)(x-3)=0

\;\Rightarrow\;\;x=0,1,2,3

The least roots is 0 .

Let the roots of

x^3+\alpha x^2+\beta x+6=0

be

r_1, r_2, r_3

The roots of

x^3+\alpha x^2+\beta x+6=0

increased by 1 are

r_1+1, r_2+1, r_3+1

One of these roots is 0 , so

r_i+1=0

. for some

i^0

.

\Rightarrow r_i=-1

Since

r_i=-1

is root of

\;x^3+\alpha x^2+\beta x+6=0

\;\Rightarrow(-1)^3+\alpha(-1)^2+\beta(-1)+6=0

\;\Rightarrow-1+\alpha-\beta+6=0

\;\Rightarrow\alpha-\beta+5=0

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