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12:35 AM
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Q: How to solve the quadratic equation $x^2-1=2$?

steven Solve $x^2-1=2$ I have no idea how to do this can somebody please help me? I have tried working it out and I could never get the answer.

 
Move the -1 to the right-hand side. What does that give?
 
I got 3... so now what? @lemon
 
So $x^2=3$. What is the opposite of squaring?
 
Square root @lemon
 
Right, so take the square root of both sides of the equation. What do you get?
 
12:35 AM
x= positive/ negative 3? Can you work out the problem for me? @lemon
 
Almost, $x=\pm\sqrt{3}$. Does that make sense?
 
A little bit.. How would you solve (x+13)^2=7 @lemon
 
You want to get $x$ on its own. The first step is to get rid of the square (by square rooting both sides). And then move the 13 to the other side. What does that give you?
 
I do not know.. I am lost. @lemon
 
Well do the first step. What is the square root of $(x+13)^2$ and what is the square root of $7$?
 
12:35 AM
x+13? Could you please work it out? @lemon
 
We're not here to do your homework for you but we're happy to guide you. $x+13$ is correct. And the sqrt of 7 is simply $\sqrt{7}$. So you have $x+13=\sqrt{7}$. Now move the 13 to the other side. What do you get?
 
i am not asking you to do my homework...! I just need an explanation so i can fully understand... @lemon
 
@steven witty though dfg is (+1), he's winding you up.
Do you understand how we got from $(x+13)^2=7$ to $x+13=\sqrt{7}$?
 
Could somebody please work this out for me please? @lemon
yes @lemon I do
 
So now move the 13 to the other side of the equal sign. What does that give you? (Take a shot...)
 
12:35 AM
@lemon-20?????????
 
No. When you move a positive number from one side of an equation to the other, what happens to it?
 
negative? Could u please help me
 
I am helping you! That's right. So the +13 moves to the other side to become -13. Therefore, $x+13=\sqrt{7}$ becomes $x=\sqrt{7}-13$. Right?
 
Ye that is right @lemon
 
And that's the final answer ($x=\sqrt{7}-13$). You can't simplify that any further.
If you have any more such questions then let me know here
 

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