Find The Root Of The Answer

Since $\alpha$ is root of $x^2-4x+13=0$ .

$\implies \alpha^2-4\alpha+13=0$

$\Rightarrow \alpha^2=\color{#20A900}{4\alpha-13}$

We have to find:

$\implies \underbrace{\alpha^2}_{\color{#20A900}{4\alpha-13}}+4\beta+13$

$\Rightarrow 4\alpha-13+4\beta+13$

Using Vieta's Formula .

$\implies \color{#3D99F6}{\alpha+\beta}=4$

$\Rightarrow 4(\color{#3D99F6}{\alpha+\beta})=4×4=\color{#BA33D6}{\boxed{16}}$

From the equation we can find out that $\alpha + \beta = 4$ and $\alpha \cdot \beta = 13$ , so we have: $\alpha^2 +4\beta +13 = \alpha^2 +(\alpha + \beta)\beta + \alpha \cdot \beta$ $= \alpha^2 +2\alpha \cdot \beta + \beta^2 = (\alpha +\beta)^2 = 4^2=16$

First, find the roots of the equation $x^{2} - 4x + 13 = 0$ . In this problem, we can use the completing the square method.

$x^{2} - 4x + 13 = 0$

$x^{2} - 4x = -13$

$x^{2} - 4x + (\frac{4}{2})^{2} = -13 + (\frac{4}{2})^{2}$

$x^{2} - 4x + (\frac{2}{1})^{2} = -13 + (\frac{2}{1})^{2}$

$x^{2} - 4x + (2)^{2} = -13 + (2)^{2}$

$x^{2} - 4x + 4 = -13 + 4$

$(x - 2)^{2} = -9$

$x - 2 = \pm i\sqrt{9}$

$x = 2\pm 3i$

Then, set each of them equal to $\alpha$ and $\beta$ and substitute them to the expression $\alpha^{2} +4\beta + 13$ . The order of which root is equal to each does not matter.

Case 1:

$\alpha = 2 + 3i$ , $\beta = 2 - 3i$

$(2 + 3i)^{2} +4(2 - 3i) + 13$

$4+12i-9 +8-12i + 13$

$16$

Case 2:

$\alpha = 2 - 3i$ , $\beta = 2 + 3i$

$(2 - 3i)^{2} +4(2 + 3i) + 13$

$4-12i-9 +8+12i + 13$

$16$

$\therefore$ the answer is $\boxed{16}$ .

You can generalise this for any quadratic equation,

$p(x)=x^2-(\alpha + \beta)x+\alpha \beta$

$\alpha^2+(\alpha+\beta)\beta + \alpha\beta = \alpha^2 +\alpha\beta+\beta^2+\alpha\beta = (\alpha+\beta)^2$

Here $\alpha+\beta=4$ , so the answer is $4^2 = 16$

And if you never have never heard of Vieta's formula (like i had, so i am some knowledge richer), the roots are 2 +/- 3i. If you do the math right you will find 16 as answer too. I must say Vieta's makes it faster :)

From Vieta's Formula $\alpha + \beta = 4$
Now, since $\alpha$ is root of the equation; therefore;
$\alpha ^2 + 13 = 4\alpha$
Putting this in required equation; we get: $4(\alpha + \beta)$ $=4(4) = \boxed{16}$