Solving Equation $x^2-(2+i)x+(-1+7i)=0$

Question

$x^2-(2+i)x+(-1+7i)=0$

I tried to solve it and I got stuck here:

$x=(2+i)±\sqrt{\frac{7-24i}{2}}$

Answer 1

Firstly, you correctly found the discriminant: $$D = B^2 - 4AC = [-(2+i)]^2-4(-1+7i) = 7 - 24i.$$ Then: $$x_{1,2} = \frac{-B\pm \sqrt{D}}{2A}=\frac{2+i\pm\sqrt{7-24i}}{2}.\tag{1}$$ But: $$\sqrt{7-24i}=\pm(4-3i).$$ Why?

Let $\sqrt{7-24i} = z\implies z^2 = 7-24i$. If we let $z = a+bi \implies a^2-b^2 +2ab i = 7-24i $. Thus: $$\left\{ \begin{array}{l} a^2 - b^2 = 7\\ ab = -12 \end{array} \right. $$ Solving the above system in Reals, we get 2 pairs of solutions: $(a,b) = (4,-3)$ and $(a,b) = (-4,3)$. Both $z_1 = 4-3i$ and $z_2 = -4+3i$ satisfy the equation $z^2 = 7-24i$. No matter the choice we make for $\sqrt{27-4i}$ (either $4-3i$ or $-4+3i$), the solutions given by the quadratic formula will be the same, due to the "$\pm$ sign" in the numerator.

Apply this to $(1)$ and you will get the result.

Answer 2

HINT

Write $7 - 24\iota $ in the $r e^{\iota \theta}$ form, where $r = \sqrt{7^2 + 24^2}$ and $ \tan{\theta} = -\frac{24}{7} $. Note, $\theta$ will be in 4th quadrant.

Answer 3

By using factor method:

$$x^2-(2+i)x+(-1+7i)=0$$

• Rewrite the left hand side of the equation.

$$x^2-(2+i)x+(-1+7i)=(-1+7i)+(-2-i)x+x^{2}$$

• Factor the left hand side.

$$(( -3+i)+x)((1-2i)+x)=0$$ - Solve each term in the product separately.

$(-3+i)+x=0$ or $(1-2i)+x=0$

$$x=3-i \text{ or } x=-1+2i $$

Or we can Solve the quadratic equation by completing the square

indeed,

$$x^2-(2+i)x+(-1+7i)=0$$

Subtract $-1+7i$ from both sides:

$$x^2-(2+i)x=1-7i$$ Take one half of the coefficient of $x$ and square it, then add it to both sides

Add $\dfrac{3}{4}+i$ to both sides:

$$(\dfrac{3}{4}+i)+(-2-i)x+x^{2}=\dfrac{7}{4}-6i$$

• Factor the left hand side.

Write the left hand side as a square:

$$\left(x+(-1-\dfrac{i}{2})\right)^{2}=\dfrac{7}{4}-6i$$

• Eliminate the exponent on the left hand side.

Take the square root of both sides:

$$x+(-1-\dfrac{i}{2})=\sqrt{\dfrac{7}{4}-6i} \text{ Or } x+(-1-\dfrac{i}{2})=-\sqrt{\dfrac{7}{4}-6i}$$

add $1+\dfrac{i}{2}$ to both sides:

$$x=(1+\dfrac{i}{2})+\sqrt{\dfrac{7}{4}-6i} \text{ Or } x=(1+\dfrac{i}{2})-\sqrt{\dfrac{7}{4}-6i}$$

$$x=(1+\dfrac{i}{2})+\sqrt{4-6i-\dfrac{9}{4}} \text{ Or } x=(1+\dfrac{i}{2})-\sqrt{4-6i-\dfrac{9}{4}}$$

$$x=(1+\dfrac{i}{2})+\sqrt{\dfrac{(4-3i)^{2}}{4}} \text{ Or } x=(1+\dfrac{i}{2})-\sqrt{\dfrac{(4-3i)^{2}}{4}}$$

$$x=(1+\dfrac{i}{2})+\dfrac{(4-3i)}{2} \text{ Or } x=(1+\dfrac{i}{2})-\dfrac{(4-3i)}{2}$$

Answer 4

Some other answers explained how to get square roots of complex numbers, but it's also worth mentioning that there is an explicit formula you can get by following such methods in generality (leaving the variables in). The formula obtained is that the square root of $a + bi$ with nonnegative real part is given by

$$\sqrt{a + bi} = \sqrt{{a + \sqrt{a^2 + b^2} \over 2}} \pm \sqrt{{-a + \sqrt{a^2 + b^2} \over 2}}\,\,i$$

You choose the $\pm$ to ensure your square root is in the correct quadrant. So in your case

$$\sqrt{7 - 24i} = \sqrt{{7 + \sqrt{625} \over 2}} \pm \sqrt{{-7 + \sqrt{625} \over 2}}\,i$$ $$= 4 \pm 3i$$ Based on $7 - 24i$'s position in the fourth quadrant, you'd choose the $-$ in the $\pm$, so you'd have $$\sqrt{7 - 24i} = 4 - 3i$$ This is the square root with positive real part. $-4 + 3i$ is the square root with negative real part; you just multiply the other square root by $-1$. Due to the $\pm$ in the quadratic formula, it doesn't matter which square root you use when solving quadratic equations.