a. Using Cramer's rule, solve: 3x + 2y + 9 = 0 , 2x 3y + 6 = 0 b. Express 3 + i in polar form. c. If one root of the equation ax^2 + bx + c = 0 be twice the other, show that 2b^2 = 9ac .

a. Solution using Cramer’s Rule

Given the system of equations: $$\{\begin{array}{l}3x+2y+9=0\\ 2x-3y+6=0\end{array}$$ Rewrite them in standard form: $$\{\begin{array}{l}3x+2y=-9\text{(1)}\\ 2x-3y=-6\text{(2)}\end{array}$$

The coefficient matrix $A$ and the constant matrix $B$ are: $$A=\left[\begin{array}{cc}3& 2\\ 2& -3\end{array}\right],B=\left[\begin{array}{c}-9\\ -6\end{array}\right]$$

Step 1: Compute the determinant of $A$ ($D$): $$D=\left|\begin{array}{cc}3& 2\\ 2& -3\end{array}\right|=(3)(-3)-(2)(2)=-9-4=-13$$

Step 2: Compute $D_x$ (replace the first column of $A$ with $B$): $$D_x=\left|\begin{array}{cc}-9& 2\\ -6& -3\end{array}\right|=(-9)(-3)-(2)(-6)=27+12=39$$

Step 3: Compute $D_y$ (replace the second column of $A$ with $B$): $$D_y=\left|\begin{array}{cc}3& -9\\ 2& -6\end{array}\right|=(3)(-6)-(-9)(2)=-18+18=0$$

Step 4: Solve for $x$ and $y$: $$x=\frac{D_x}{D}=\frac{39}{-13}=-3$$ $$y=\frac{D_y}{D}=\frac{0}{-13}=0$$

Final solution: $$\boxed{x=-3,y=0}$$

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b. Express $\sqrt{3}+i$ in polar form

A complex number $z=a+ib$ can be written in polar form as: $$z=r(\cos \theta +i\sin \theta )$$ where: $$r=\sqrt{a^2+b^2},\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$$

For $z=\sqrt{3}+i$:

• $a=\sqrt{3}$, $b=1$.

Step 1: Compute the modulus $r$: $$r=\sqrt{(\sqrt{3}{)}^2+(1{)}^2}=\sqrt{3+1}=\sqrt{4}=2$$

Step 2: Compute the argument $\theta$: $$\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ }\text{or}\frac{\pi }{6}\text{ radians}$$

Polar form: $$\boxed{2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)}$$

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c. If one root of $a{x}^2+bx+c=0$ is twice the other, show that $2b^2=9ac$

Let the roots be $\alpha$ and $2\alpha$.

Step 1: Use sum and product of roots: For a quadratic equation $a{x}^2+bx+c=0$: $$\alpha +2\alpha =-\frac{b}{a}\Rightarrow 3\alpha =-\frac{b}{a}\Rightarrow \alpha =-\frac{b}{3a}$$ $$\alpha \cdot 2\alpha =\frac{c}{a}\Rightarrow 2{\alpha }^2=\frac{c}{a}$$

Step 2: Substitute $\alpha$ into the product equation: $$2{\left(-\frac{b}{3a}\right)}^2=\frac{c}{a}$$ $$2\cdot \frac{b^2}{9a^2}=\frac{c}{a}$$ $$\frac{2b^2}{9a^2}=\frac{c}{a}$$

Step 3: Simplify to obtain the required relation: Multiply both sides by $9a^2$: $$2b^2=9ac$$

Thus, the required relation is proven: $$\boxed{2b^2=9ac}$$