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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.$$f(x)=x^{2}+4 x+2$$

a) graphb) The roots of $f$ ROUNDED TO THE NEAREST TENTH are the $x=-3.4$ and $x=-0.6$ as shown in the pic.c) $x=-2+\sqrt{2}$ and $x=-2-\sqrt{2}$

Algebra

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 1

Real Roots of a Quadratic Equation

Equations and Inequalities

Quadratic Functions

Complex Numbers

Polynomials

Baylor University

University of Michigan - Ann Arbor

Idaho State University

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01:32

In mathematics, the absolu…

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In $9-14 :$ a Sketch the g…

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In $9-14$ : a. For each gi…

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Okay, This question would like us to find the solutions to this quadratic equation. First, it wants us to find a graphical estimate for these solutions. So I grabbed this in Dismas here, and we're looking for the points where F of X is equal to zero. And that's just where it's crossing the X axis. So our solutions air here and here and zooming in the approximate X values of these solutions Negative 0.6 for ah, one more to the right and then negative 3.4 for the left. But now we want to complete the square to find the exact values of thes roots. So we want to set our quadratic equation equals zero. And now, to solve this, we're gonna use the complete the square method where we convert X squared plus B X plus C into the form a times X plus d squared plus e and de is a constant that's be over to a and E is a constant that c minus b squared over four a. So now we just need to find the DND values for the specific quadratic so d is equal to be over to a which is four over to purchased two and then e is equal to C minus. B squared over for a which is tu minus. Be square to 16 over two A. So we get to minus eight, which is negative. Six. So now if we do this you or on my bed this should be a four on the bottom, which is negative. Two. So now we can rewrite this in our completed square form of X plus d quantity squared plus e, which is minus two. And now this is much easier to solve. So now we have X plus two quantity squared is equal to two when we add to double sites. So taking the square root we get X Plus two is equal to plus or minus the square root of two or splitting our solutions we get X Plus two equals square root of two and X plus two equals minus square root of two and then solving we get X equals negative two plus route to and X equals negative too minus root too. And these are the exact forms of our approximations we found from the graph

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