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To solve the equation given in step 2 of Example 2 , Sarah multiplied each side of the equation by 8 and then added 1 to each side to complete the square. Show that $16 x^{2}-8 x-8$ is the square of a binomial and will lead to the correct solution of $0=2 x^{2}-x-1$

$x=1, x=-\frac{1}{2}$

Algebra

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 1

Real Roots of a Quadratic Equation

Equations and Inequalities

Quadratic Functions

Complex Numbers

Polynomials

Campbell University

McMaster University

Lectures

01:32

In mathematics, the absolu…

01:11

01:31

Solve the equation by comp…

04:14

Use completing the square …

02:06

Solve each quadratic equat…

04:36

01:25

Solve each equation by com…

02:15

Solve by completing the sq…

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04:06

Solve the quadratic equati…

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When solving the equation …

Okay. This question wants us to solve the following quadratic by completing the square. So the first thing we can do to make our lives easier is pull out a factor of eight from everything. Because all three terms here too visible by eight. And after doing this, we see that we just pulled an eight out of the whole thing and that really is it going to affect our solution? His eight can never make this left hand side zero so we can just divide by it on each side. And now we're left with a simpler quadratic two X squared minus X minus one equals zero. Okay, this question wants us to solve the following quadratic by completing the square. So what we can do? Okay, this question wants us to solve the following quadratic by completing the square. So the first thing we can do to make our lives easier is pull out a factor of eight from everything. Because all three terms here, too visible by eight turns quadratic of the form X squared. Must be experts see into a Times X plus d squared plus e where d is equal to be over to a and E is equal to C minus. B squared over four. A. So now all they have to do is find D and E. So in our case, de is equal to be, which is negative. One over two times A and A is two, so we have four on the bottom, and then e is equal to see negative one minus B squared over for a so completing the square turns quadratic of the form X squared. Must be experts see into a Times X plus D squared plus e, where D is equal. To be over to a and E is equal to C minus. B squared over four a or we have a negative one minus 1/8 so negative 9/8 so now are quadratic can be turned into two times X minus 1/4 quantity squared minus 9/8 equals zero or adding night eighths. We get two minus X minus V. So now all they have to do is find d and E. So in our case, de is equal to be, which is negative. One over two times a day is too. So we have four on the bottom and then e is equal to see negative one minus B squared over for a fourth squared equals 9/8 for X minus 1/4 squared is equal to 9/16 and then we can take our square room and get X minus of fourth is equal to plus or minus the square root of 9/16. So now we have our two cases we have, or we have a negative one minus 1/8 so negative 9/8 so now are quadratic can be turned into two times X minus 1/4 quantity squared minus 9/8 equals zero or adding night eighths. We get two minus X minus. Affects minus 1/4 equals positive. 3/4 because the square root of 9/16 is 3/4 and then we have X minus 1/4 equals negative 3/4 the negative group. So our first equation has solution 4/4 which is one and our second equation has solution negative 2/4 which is just minus 1/2. Fourth squared equals 9/8 or X minus 1/4 squared is equal to 9/16 and then we can take our square room and get X minus of fourth is equal to plus or minus the square root of 9/16. So now we have our two cases. We have that.

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