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How can I find out what X equals in this?

$$x^2 - 2x - 3 = 117$$

How would I get started? I'm truly stuck.

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What is a quadratic equation? –  copper.hat Jun 10 '13 at 20:35
    
Do you know, how to solve quadratic equation $x^2-2x-120=0$? –  mcihak Jun 10 '13 at 20:36

4 Answers 4

Hint: Use the quadratic formula! Rewrite your equation as:

$$x^{2} - 2x - 120 = 0$$

The quadratic formula says that for a quadratic equation of the form $ax^{2} + bx + c = 0$, the roots of the equation (i.e., values of $x$ satisfying the equation), we have:

$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

Using this information, can you solve your equation?

(Note: ordinarily, I would prove the quoted results, but the quadratic formula is a ubiquitous result, information about which can be found just about anywhere on the internet. For proof of the quadratic formula, in particular, André Nicolas has provided a particularly nice one here.)

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As answered by others, you can use the quadratic formula to find the roots of any quadratic polynomial. But this one is a special case where you can simply factor the polynomial. $$ x^2 - 2x - 3 = 117 \\ x^2 - 2x - 120 = 0 \\ x^2 + 10x - 12x - 120 = 0 \\ x(x + 10) - 12(x + 10) = 0 \\ (x + 10)(x - 12) = 0 $$ So the two solutions are $x = -10$ and $x = 12$.

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A different way if you have not seen the quadratic formula yet.

Recall that $(x-1)^2 = x^2-2x+1$ and so

$$ x^2-2x-3 = \left(x^2-2x+1\right)-4=(x-1)^2-4 $$ and your equation $120=x^2-2x-3$ becomes equivalent to $$ 117=(x-1)^2-4 $$ so $(x-1)^2 = 121 = 11^2$. Therefore, $x-1 = 11$ or $x-1 = -11$.

In the first case, $x = 11+1 = 12$.

In the second case, $x = -11+1 = -10$.

Check

In the first case, $x=12$ so $x^2-2x-3 = 144-24-3 = 117$.

In the second case, $x=-10$, so $x^2-2x-3 = 100+20-3 = 117$.

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Hint:1.$$ax^2 +bx +c=0 \to D=b^2-4ac\ge0 $$$$\ \to\begin{cases} \color{green}{x_1=\frac{-b+\sqrt{D}}{2a}} \\ \color{red}{x_2=\frac{-b-\sqrt{D}}{2a}} \\ \end{cases} $$$$$$ 2. $$x^2 +bx +c=(x+\frac{b}{2})^2=\frac{b^2}{4}-c\ge0\quad \text{then} x=\pm\sqrt{\frac{b^2}{4}-c}-\frac{b}{2}$$ 3. find $x_1$and $x_2 $by solving following system \begin{cases} x_1+x_2=\frac{-b}{a} \\ x_1x_2=\frac{c}{a} \\ \end{cases}

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