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In $9-14 :$ a Sketch the graph of each function. b. From the graph, estimate the roots of the function to the nearest tenth. c. Find the exact irrational roots in simplest radical form.$$f(x)=x^{2}-2 x-1$$
a) graphb) The roots of $f$ ROUNDED TO THE NEAREST TENTH are the $x=-0.4$ and $x=2.4$ as shown in the pic.c) $x=1+\sqrt{2}$ and $x=1-\sqrt{2}$
Algebra
Chapter 5
QUADRATIC FUNCTIONS AND COMPLEX NUMBERS
Section 1
Real Roots of a Quadratic Equation
Equations and Inequalities
Quadratic Functions
Complex Numbers
Polynomials
McMaster University
Harvey Mudd College
Baylor University
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Okay, This question asks us about the roots to this quadratic equation. So the first thing it wants us to do is try and find these routes graphically, which I have. The graph over here ends in black. Those points are where f of X equals zero. So these air, where we have solutions to the equation f of X equals zero so we can approximate the X values that will give us zero. So we get X values of approximately negative zero point for and 2.4 when we do this. So now we want to solve it. Exactly. So we sent our equation f of X equals zero, and then we're going to complete the square. So completing the square turns a quadratic X squared plus B x plus C into a new quadratic a Times X plus d squared plus e where de is equal to be over to a and E is equal to b squared, see minus B squared over for a So now let's find D and E for our quadratic. So de is equal to be over to a which is negative two divided by two or negative one and then e is equal to C minus. D squared over for a and that's just equal to negative one minus or sorry. This should be a B squared one minus B squared, which is four divided by for a and that is just equal to one minus 4/4 or one minus one for negative one minus one, which is negative. Two. So now are transformed. Equation is X minus one quantity squared minus two is equal to zero. So now we can add to to both sides to get X minus. One squared is equal to two. And then if we square root both sides, we get X minus. One is equal to plus or minus the square root of two. And now we can split up our solutions. We have X minus one is equal to the positive squared of to and X minus one is equal to the negative squared too. So in the first case, we have X equals one plus the squared of to giving us our positive solution. And then we have X equals one minus squared of two, which gives us our negative solution. And that's all right
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