What would the roots be for this quadratic equation $f(x)=2x^2-6x-8$?

Question

I am an 10 grade high school student. I really need help on this problem, my teacher was out today and i need help on completing this problem. Thank you.

Answer 1

To find the roots of $f$, solve $2x^2-6x-8=0$. First divide by $2$ to take away unnecessary coefficients to obtain $x^2-3x-4=0$, where we recognize that $x^2-3x-4=(x-4)(x+1)$ to see that the roots are $x=-1$ and $x=4$.

Answer 2

For an equation given $a$, $b$ and $c$ three real numbers.

$$ ax^2+bx+c=0 $$ Like it is here with $a=2$, $b=-6$ and $c=-8$ you first calculate what is called a discriminant $\Delta=b^2-4ac$. Just make the calculus. It has to be positive. $$ \Delta=36-4\times 2 \times \left(-8\right)=36+64=100 \geq 0 $$ Then there are two roots that are given by the formula $$ x_1=\frac{-b+\sqrt{\Delta}}{2a} \text{ and }x_2=\frac{-b-\sqrt{\Delta}}{2a} $$ It gives here with $\sqrt{\Delta}=\sqrt{100}=10$ $$ x_1=\frac{+6+10}{2\times 2}=\frac{16}{4}=4 \text{ and }x_2=\frac{6-10}{2 \times 2}=\frac{-4}{4}=-1 $$

There are two roots equal to $x_1=4$ and $x_2=-1$

And you can deduce by the way that $$ 2x^2-6x-8=2\left(x-4\right)\left(x+1\right) $$

Answer 3

In quadratic function $f(x)$, the term "root" generally refers to $x$ values such that $f(x) = 0$. So, in order to find it's roots, you need to find the solution set of the equation $$f(x) = 2x^2-6x-8 = 0$$ Notice that it is equivalent to $$2(x^2-3x-4) = 0 \implies x^2-3x-4 = 0$$ Now, you can either factorize it as $x^2-3x-4 = (x-4)(x+1) = 0$ to get to the roots (easier way when you can factorize the expression, which is not always the case) or you can find it's roots by using discriminant of the quadratic equation, which can be done as the following: $$x_{1,2} = \frac{-(-3) \pm \sqrt{(-3)^2-4 \cdot(-4)\cdot1}}{2 \cdot 1}$$ which yields $x_1 = 4$ and $x_2 = -1$ (harder way when you can factorize the expression but guarantees to find roots even if they are complex).