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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.

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I feel from the comments that you lack some understanding of equations in general, not just quadratic equation since this problem should be quite simple. Let's try and fix that!

First off, what does it mean to solve the equation $x^2-1=2$? It means to find the value(s) that we can plug into the $x$ such that the equality holds. For instance plugging $x=1$ into the expression on the left hand side yields $x^2-1=1^2-1=0$, so this is not a solution since it does not equal $2$ as we wished. Instead of just guessing for various $x$-values, one often uses operations such as addition, multiplication, etc on both sides of the equality to try and isolate $x$.

Let's try to isolate $x$ in the equation using this method.

$$x^2-1=2$$

We want $x$ to alone on one side so first off we want to get rid of the $-1$, we do this by adding $1$ to each side.

\begin{align}x^2-1+1&=2+1 \implies\\ x^2&=3 \end{align}

Next off we want to get rid of the square. We use the opposite operation, square root:

\begin{align}\sqrt{x^2}&=\sqrt{3} \implies\\ x&=\pm\sqrt{3} \end{align}

Now $x$ is isolated and we are left with the solution. Either $x=\sqrt{3}$ or $x=-\sqrt{3}$.

Let's do a quick calculation of the other example.

\begin{align}8x^2-200&=0 &\text{ Add 200 to each side}\\ 8x^2 &=200 &\text{ Divide by 8}\\ x^2 &= \frac{200}{8} &\text{ Simplify}\\ x^2 &= 25 &\text{ Square root}\\ x &= \pm\sqrt{25}&\text{ Simplify}\\ x &= \pm 5. \end{align} And that is the solution.

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$$x^2-1=2 \Rightarrow x^2=1+2 \Rightarrow x^2=3 \Rightarrow \sqrt{x^2}=\sqrt{3} \Rightarrow |x|=\sqrt{3} \Rightarrow x=\pm \sqrt{3}$$

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$ x^2 - 1 = 2 \implies x^2 = 3 \implies x = \pm \sqrt{3} $

We have two solutions because both solutions result in the same square. We simply added $1$ to both sides and took the square root. Both steps are correct, because we are making the same transformation to both sides of the equation. So if they are equal before the transformation, they have to be equal after the transformation.

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$$\begin{array}{llll} x^2-1&=&2 &(\text{given})\\ (x^2-1)+1&=&2+1 &(\text{additive axiom of equality})\\ x^2+(-1+1)&=&3 &(\text{associative field axiom})\\ x^2+0&=&3 &(\text{additive inverse field axiom})\\ x^2&=&3 &(\text{additive identity}) \end{array}$$

now I'm stuck. :(

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$x^2-1=2$ Initial Problem
$x^2=3$ Add 1 to both sides.
$x= \sqrt{3} , -\sqrt{3}$ Square root both sides and you have your two solutions

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You could also graph this to find what the 2 possible values for $x$ are. To do this, you can rewrite it using the following method:

$$x^2 - 1 = 2$$

Subtract 2 from both sides:

$$x^2 - 1 - 2 = 2 - 2$$

To make it equal to $0$: $$x^2 - 3 = 0$$

Then graph it (I used Desmos Graphing Calculator):

Graph

You will see that the 2 $x$ intercepts (where the red line goes over the green and blue lines, and $y = 0$) are $-\sqrt{3}$ (or $-1.73205$) and $\sqrt{3}$ (or $1.73205$)

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