Let α, β be the roots of the equation x^2 - 3x + r = 0, and α 2, 2β be the roots of the equation x^2 + 3x + r = 0. If the roots of the equation x^2 +…

From the first equation $x^2 - 3x + r = 0$x2−3x+r=0, the sum and product of the roots are:

$$\alpha + \beta = 3$$α+β=3

$$\alpha \beta = r$$αβ=r

From the second equation $x^2 + 3x + r = 0$x2+3x+r=0, the sum and product of the roots are:

$$\dfrac{\alpha}{2} + 2\beta = -3$$2α​+2β=−3

$$\left(\dfrac{\alpha}{2}\right)(2\beta) = r \Rightarrow \alpha \beta = r$$(2α​)(2β)=r⇒αβ=r

We have a system of linear equations in terms of $\alpha$α and $\beta$β:

$$\alpha + \beta = 3$$α+β=3

$$\alpha + 4\beta = -6$$α+4β=−6

Subtracting the first equation from the second gives:

$$3\beta = -9 \Rightarrow \beta = -3$$3β=−9⇒β=−3

Substituting $\beta = -3$β=−3 into $\alpha + \beta = 3$α+β=3 gives:

$$\alpha = 6$$α=6

Now, we can find $r$r:

$$r = \alpha \beta = 6 \times (-3) = -18$$r=αβ=6×(−3)=−18

The roots of the third equation $x^2 + 6x - m = 0$x2+6x−m=0 are given as $2\alpha + \beta + 2r$2α+β+2r and $\alpha - 2\beta - \dfrac{r}{2}$α−2β−2r​. Let us calculate their values:

$$R_1 = 2(6) + (-3) + 2(-18) = 12 - 3 - 36 = -27$$R1​=2(6)+(−3)+2(−18)=12−3−36=−27

$$R_2 = 6 - 2(-3) - \dfrac{-18}{2} = 6 + 6 + 9 = 21$$R2​=6−2(−3)−2−18​=6+6+9=21

The product of the roots of the equation $x^2 + 6x - m = 0$x2+6x−m=0 is $-m$−m. Therefore:

$$-m = R_1 R_2 = (-27)(21)$$−m=R1​R2​=(−27)(21)

$$-m = -567 \Rightarrow m = 567$$−m=−567⇒m=567

Answer: $567$567