Answer:
The value of m is 6.
Step-by-step explanation:
Here, the given equation,
[tex]x^4-(3m+2)x^2+m^2=0[/tex]
[tex]x^4+0x^3-(3m+2)x^2+0x+m^2=0[/tex]
Let the roots of the equation are a-3b, a-b, a+b and a + 3b, ( they must be form an AP )
Thus, we can write,
[tex]a-3b+a-b+a+b+a+3b=\frac{\text{coefficient of }x^3}{\text{coefficient of }x^4}[/tex]
[tex]=\frac{0}{1}=0[/tex]
[tex]\implies a=0----(1)[/tex]
[tex](-3b)(-b)+(-b)(b)+(b)(3b)+(3b)(-3b)+(-b)(3b)+(-3b)(b)=\frac{\text{coefficient of }x^2}{\text{coefficient of }x^4}}[/tex]
[tex]=\frac{-3m-2}{1}[/tex]
[tex]3b^2-b^2+3b^2-9b^2-3b^2-3b^2=-3m-2[/tex]
[tex]-10b^2=-3m-2[/tex]
[tex]\implies b^2=\frac{3m+2}{10}-----(2)[/tex]
[tex](-3b)(-b)(b)(3b)=\frac{\text{Constant term}}{\text{coefficient of}x^4}= m^2[/tex]
[tex]9b^4=m^2[/tex]
[tex]9(\frac{3m+2}{10})^2=m^2[/tex]
[tex]9(\frac{9m^2+4+12m}{100})=m^2[/tex]
[tex]81m^2+36+108m=100m^2[/tex]
[tex]-19m^2+108m+36=0[/tex]
[tex]19m^2-108m-36=0[/tex]
[tex]19m^2-114m+6m-36=0[/tex]
[tex]19m(m-6)+6(m-6)=0[/tex]
[tex](19m+6)(m-6)=0[/tex]
[tex]\implies m=-\frac{6}{19}\text{ or }m=6[/tex]
But m > 0,
Hence, the value of m is 6.
