Question

the number of irrational roots of the equation (x-1)(x-2)(3x-2)(3x+1)=21 is

Answer 1

Answer:

answer is two irrational roots

Step-by-step explanation:

dont copy the whole solution in copy read it understand it and do it by itself

Answer 2

Given,

$\longrightarrow (x-1)(x-2)(3x-2)(3x+1)=21$

We see that,

$\longrightarrow(x-1)(3x-2)=3x^2-5x+2$

$\longrightarrow(x-2)(3x+1)=3x^2-5x-2$

Then our equation becomes,

$\longrightarrow(3x^2-5x+2)(3x^2-5x-2)=21$

$\longrightarrow(3x^2-5x)^2-4=21$

$\longrightarrow(3x^2-5x)^2=25$

$\longrightarrow3x^2-5x=\pm5$

$\longrightarrow3x^2-5x\pm5=0$

Taking the discriminant,

$\longrightarrow D=(-5)^2-(4\times3\times\pm5)$

$\longrightarrow D=25\pm60$

$\longrightarrow D=85\quad OR\quad D=-35$

This implies our equation has 2 irrational roots (since 85 is not a perfect square) and 2 non - real roots (since -35 is negative).

However, the equation has 2 irrational roots.