If the expression x²-11 x+a and x²-14 x+2 a must have a common factor and a ≠0), then, the common factor is

Correct answer: C. (x-8) Here Let (x- α is the common factor then (x=α is root of the corresponding equation ( α^ 2-11 α+ a=0) (α^ 2-14 α+ 2 a= 0) Subtracting (3 α- a=0 α= a / 3) Hence ((a^2)/(9)- 11 (a)/(3)+a=0, a=0) or a=24 since (a ≠ 0, a= 24) ( ) the common factor of ((\ array l x^2-11 x+24 x²-14 x+48 array .) is clearly x -8

BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Mathematics
Chapter: Quadratic Equation

BITSAT 2009 · Mathematics · Quadratic Equation

If the expression x²-11 x+a and x²-14 x+2 a must have a common...

If the expression

x²-11 x+a

and

x²-14 x+2 a

must have a common factor and

\mathrm{a} \neq0), then, the common factor is

Options

A $(x-3)$
B $(x-6)$
C $(x-8)$
DNone of these

Correct answer

C. $(x-8)$

Step-by-step solution

Here Let (x-

\alpha is

the common factor then (\mathrm{x}=\alpha is root of

the corresponding equation (\therefore

\alpha^

2-11

\alpha+

a=0) (\alpha^

2-14

\alpha+

\mathrm{a}=

0) Subtracting (3

\alpha-

a=0 \Rightarrow

\alpha=

a / 3) Hence (\frac{a^2}{9}-

11

\frac{a}{3}+a=0, a=0)

or

a=24

since (\mathrm{a} \neq

\mathrm{a}=

24) (\therefore)

the

common factor of (\left(\

{

\begin{array}{l}x^2-11 x+24 \\ x²-14 x+48\end{array}

\right.) is clearly

x -8$

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