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Use the quadratic formula to solve the equation, negative x

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squared plus 8x is equal to 1.

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Now, in order to really use the quadratic equation, or to

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figure out what our a's, b's and c's are, we have to have

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our equation in the form, ax squared plus bx plus c is

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equal to 0.

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And then, if we know our a's, b's, and c's, we will say that

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the solutions to this equation are x is equal to negative b

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plus or minus the square root of b squared minus 4ac-- all

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of that over 2a.

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So the first thing we have to do for this equation right

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here is to put it in this form.

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And on one side of this equation, we have a negative x

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squared plus 8x, so that looks like the first two terms. But

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our constant is on the other side.

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So let's get the constant on the left hand side and get a 0

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here on the right hand side.

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So let's subtract 1 from both sides of this equation.

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The left hand side of the equation will become negative

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x squared plus 8x minus 1.

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And then the right hand side, 1 minus 1 is 0.

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Now we have it in that form.

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We have ax squared a is negative 1.

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So let me write this down. a is equal to negative 1.

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a is equal to negative 1.

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It's implicit there, you could put a 1 here if you like.

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A negative 1.

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Negative x squared is the same thing as negative 1x squared.

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b is equal to 8.

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So b is equal to 8, that's the 8 right there.

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And c is equal to negative 1.

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That's the negative 1 right there.

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So now we can just apply the quadratic formula.

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The solutions to this equation are x is equal to negative b.

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Plus or minus the square root of b squared, of 8 squared,

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minus 4ac-- let me do it in that green color --minus 4,

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the green is the part of the formula.

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The colored parts are the things that we're substituting

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into the formula.

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Minus 4 times a, which is negative 1, times negative 1,

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times c, which is also negative 1.

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And then all of that-- let me extend the square root sign a

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little bit further --all of that is going to be

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over 2 times a.

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In this case a is negative 1.

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So let's simplify this.

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So this becomes negative 8, this is negative 8, plus or

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minus the square root of 8 squared is 64.

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And then you have a negative 1 times a negative 1, these just

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cancel out just to be a 1.

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So it's 64 minus is 4.

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That's just that 4 over there.

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All of that over negative 2.

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So this is equal to negative 8 plus or minus the

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square root of 60.

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All of that over negative 2.

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And let's see if we can simplify the radical

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expression here, the square root of 60.

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Let's see, 60 is equal to 2 times 30.

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30 is equal to 2 times 15.

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And then 15 is 3 times 5.

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So we do have a perfect square here.

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We do have a 2 times 2 in there.

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It is 2 times 2 times 15, or 4 times 15.

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So we could write, the square root of 60 is equal to the

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square root of 4 times the square root of 15, right?

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The square root of 4 times the square root of 15,

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that's what 60 is.

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4 times 15.

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And so this is equal to-- square root of 4 is 2 times

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the square of 15.

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So we can rewrite this expression, right here, as

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being equal to negative 8 plus or minus 2 times the square

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root of 15, all of that over negative 2.

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Now both of these terms right here are divisible by either 2

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or negative 2.

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So let's divide it.

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So we have negative 8 divided by negative 2, which is

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positive 4.

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So let me write it over here.

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Negative 8 divided by negative 2 is positive 4.

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And then you have this weird thing.

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Plus or minus 2 divided by negative 2.

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And really what we have here is 2 expressions.

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But if we're plus 2 and we divide by negative 2, it will

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be negative 1.

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And if we take negative 2 and divide by negative 2, we're

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going to have positive 1.

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So instead of plus or minus, you could imagine it is going

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to be minus or plus.

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But it's really the same thing.

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Right?

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It's really now minus or plus.

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If it was plus, it's now going to be a minus.

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If it was a minus, it's now going to be a plus.

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Minus or plus 2 times the square root of 15.

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Or another way to view it is that the two solutions here

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are 4 minus two roots of 15, and 4 plus two roots of 15.

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These are both values of x that'll satisfy this equation.

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And if this confuses you, what I did, turning a plus or minus

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into minus plus.

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Let me just take a little bit of an aside there.

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I could write this expression up here as two expressions.

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That's what the plus or minus really is.

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There's a negative 8 plus 2 roots of 15 over negative 2.

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And then there's a negative 8 minus 2 roots of 15 over

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negative 2.

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This one simplifies to-- negative 8 divided by

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negative 2 is 4.

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2 divided by negative 2 is negative 1.

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2 times a 4 minus the square root of 15.

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And then over here you have negative 8 divided by negative

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2, which is 4.

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And then negative 2 divided by negative 2, which is plus the

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square of 15.

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And I just realized I made a mistake up here.

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When we're dividing a 2 divided by negative 2, we

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don't have this 2 over here.

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This is just a plus or minus the root of 15.

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We just saw that when I did it out here.

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So this is minus the square root of 15.

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And this is plus the square root of 15.

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So the two solutions for this equation-- It's good that I

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took that little hiatus there, that little aside there.

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The two solutions could be 4 minus the square root of 15,

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or x, or and, x could be 4 plus the square root of 15.

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Either of those values of x will satisfy this original

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quadratic equation.

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