y=2x^2-x-3
Please help! Solve for the real roots of the equation and explain how you would use the graph of the equation to check the values for the real roots.

Question

y=2x^2-x-3
Please help! Solve for the real roots of the equation and explain how you would use the graph of the equation to check the values for the real roots.

myininaya · Answer

$$2x^2-x-3=0$$
You need to find two factors of a(c) that have sum b
You need to find two factors of 2(-3) that have sum -1
So 2(-3) is 2(-3) lol
and 2+(-3)=-1
So boom
Now we know 2x+(-3x)=-1x
So we are going to replace middle term with 2x-3x
or -3x+2x 
either one is fine
they mean the same thing
$$2x^2-3x+2x-3=0$$
Now factor by grouping shall be done
The first two terms have x in common so we can do 
$$x(2x-3)+2x-3=0$$
The last two terms have 1 in common so we can do
$$x(2x-3)+1(2x-3)=0$$
So now we have two terms and they both have the factor (2x-3) in common
So we can factor out (2x-3) like so
$$(2x-3)(x+1)=0$$

myininaya · Answer

Setting both factors =0 and solving for x will give us the two real solutions :)

myininaya · Answer

$$2x-3=0 \text{ or } x+1=0$$

anonymous · Answer

Wow, thank you so much myininaya!

myininaya · Answer

Now when you you finish getting these answers algebraically you can compare them to the answers by finding them graphically just by seeing where the graph crosses the x-axis (the horizontal axis)

anonymous · Answer

Great! Thanks!

myininaya · Answer

Np :)