$x^2+(a-3)x-4a-4=0$ Find square roots for the equation with a parameter

Question

Solve the following equation for $x$: $$x^2+(a-3)x-4a-4=0$$

The book I am reading says that the roots are: $x_1=4, x_2 = -a-1$.

I am stuck in showing that these are the roots.

My notes: $$D=a^2+10a+25$$ And I can use any $a$ for $D$, because there is no opportunity for $D<0$

Answer 1

$$x^{2}+(a-3)x-4a-4=0$$

Using Bashkara's formula: $$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$$

We have:

$$x=\frac{(-a+3)±\sqrt{(a-3)^{2}-4×(-4a-4)}}{2}$$ $$x = \frac{-a+3±\sqrt{a^{2}-6a+9+16a+16}}{2}$$ $$x=\frac{-a+3±\sqrt{a^{2}+10a+25}}{2}$$ $$x=\frac{-a+3±\sqrt{(a+5)^{2}}}{2}$$ $$x=\frac{-a+3±(a+5)}{2}$$

So we have that

$$x=\frac{-a+3+a+5}{2}$$ or $$x=\frac{-a+3-a-5}{2}$$

So we have that $x=4$ or $x=-a-1$.

Answer 2

Completing square no $D$: $$x^2+(a-3)x-4a-4=0$$ $$\tiny x^2+ (a-3)x+\left(\frac{a-3}{2}\right)^2-4a-4=\left(\frac{a-3}{2}\right)^2$$ $$\tiny\left(x+\frac{a-3}{2}\right)^2=4a+4+\frac{a^2-6a+9}{4}$$ $$\small\left(x+\frac{a-3}{2}\right)^2=\frac{a^2+10a+25}{4}$$ $$\left(x+\frac{a-3}{2}\right)^2=\left(\frac{a+5}{2}\right)^2$$ $$\left|x+\frac{a-3}{2}\right|=\left|\frac{a+5}{2}\right|$$ $$x+\frac{a-3}{2}=\pm\frac{a+5}{2}$$ $$x=\frac{-a+3\pm(a+5)}{2}$$ $$x_1=\frac{-a+3+a+5}{2}=4$$ $$\small x_2=\frac{-a+3-a-5}{2}=-a-1$$

Answer 3

The equation $x^2+(a-3)x-4a-4=0$ can be factored:

$$x^2+(a-3)x-4(a+1)=0$$ $$\Rightarrow (x-4)(x+(a+1))=0$$ $$\Rightarrow x-4=0~{\rm or~}x+(a+1)=0,$$ hence $$x\in \{4,-a-1\}.$$