The given equation is `9x^(2)-9(a+b)x+(2a^(2)+5ab+2b^(2))=0.`
This os of the form `Ax^(2)+Bx+C=0,` where
`A=9,B=-9(a+b)" and "C=(2a^(2)+5ab+2b^(2)).`
`:." "D=(B^(2)-4AC)=81(a+b)^(2)-36(2a^(2)+5ab+2b^(2))`
`=81(a^(2)+b^(2)+2ab)-36(2a^(2)+5ab+2b^(2))`
`=9a^(2)+9b^(2)-18ab=9(a^(2)+b^(2)-2ab)`
`=9(a-b)^(2)ge0.`
So, the given equation has real roots.
Now, `sqrt(D)=sqrt(9(a-b)^(2))=3(a-b).`
`:." "alpha=(-B+sqrt(D))/(2A)=(9(a+b)+3(a-b))/(2xx9)=(6(2a+2b))/(18)=((2a+b))/(3),`
`:." "beta=(-B-sqrt(D))/(2A)=(9(a+b)-3(a-b))/(2xx9)=(6(a+2b))/(18)=((a+b))/(3)`
Hence, `((2a+b))/(3)` and `((a+2b))/(3)` are the roots of the given equation.
