How do you solve the equation $x^{2} - 2 x - 1 = 0$ $x^2-2x-1=0$ by graphing?

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How do you solve the equation $x^{2} - 2 x - 1 = 0$ $x^2-2x-1=0$ by graphing?

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Answer 1 · 2016-12-20T17:17:51

Solutions are $- \frac{1}{2}$ $-1/2$ and $2 \frac{1}{2}$ $2 1/2$

Explanation:

When we draw graph of $y = x^{2} - 2 x - 1$ $y=x^2-2x-1$ ,

as a point on $x$ $x$ -axis means $y = 0$ $y=0$ ,

intercepts on $x$ $x$ -axis, give the solution of equation $x^{2} - 2 x - 1 = 0$ $x^2-2x-1=0$

Now, if $x = 2$ $x=2$ , $y = - 1$ $y=-1$

if $x = 0$ $x=0$ , $y = - 1$ $y=-1$

if $x = - 2$ $x=-2$ , $y = 7$ $y=7$

if $x = 1$ $x=1$ , $y = - 2$ $y=-2$

if $x = 3$ $x=3$ , $y = 2$ $y=2$ and if $x = - 3$ $x=-3$ , $y = 14$ $y=14$

So joining points ${(2, - 1), (0, - 1), (- 2, 7), (1, - 2), (3, 2), (- 3, 14)}$ ${(2,-1),(0,-1),(-2,7),(1,-2),(3,2),(-3,14)}$ we get the graph
graph{x^2-2x-1 [-10, 10, -5, 5]}

It shows that intercepts on $x$ $x$ -axis are $- \frac{1}{2}$ $-1/2$ and $2 \frac{1}{2}$ $2 1/2$

Hence, these are the solutions.

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How do you solve the equation $x^{2} - 2 x - 1 = 0$ $x^2-2x-1=0$ by graphing?

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Explanation:

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