Question

3. Let 6a<4b<3c, then the equation (2x-a)(3x-b)+(3x-b)(4x-c)+(4x-c)(2x-a)=0
a) Both roots are Real
b) Both roots are equal
c) One Root lies in between (a/2, b/3)
d) One Root lies in between (b/3, c/4)​

Answer 1

Answer:

D. is the

correct answer

Step-by-step explanation:

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Answer 2

d) One Root that lies in between $(\[\frac{b}{3},\frac{c}{4}\])$ is the root of the equation $(2x-a)(3x-b)+(3x-b)(4x-c)+(4x-c)(2x-a)=0$

Given:

$(2x-a)(3x-b)+(3x-b)(4x-c)+(4x-c)(2x-a)=0$

To find:

The roots of the equation $(2x-a)(3x-b)+(3x-b)(4x-c)+(4x-c)(2x-a)=0$

Step-by-step explanation:

In the equation,

$(2x-a)$is common in two terms and take them out.

$(2x-a)[(3x-b)+(4x-c)]+(3x-b)(4x-c)=0$

$(3x-b)(4x-c)$is common in both terms so taking out that

$(3x-b)(4x-c)[(2x-a)-1]=0$

By solving$(3x-b)(4x-c)=0$

we get, $\[x=\frac{b}{3}\]$ ,$\[x=\frac{c}{4}\]$

By solving$(2x-a)-1=0$

we get,$\[x=\frac{-1+a}{2}\]$

Here the roots are,$\[\frac{b}{3},\frac{c}{4}\]$, $\frac{-1+a}{2}$

The one root lies between $(\[\frac{b}{3},\frac{c}{4}\])$

and the another root is $\frac{-1+a}{2}$

Hence, the roots are not real and equal