![]() The videos on factoring if you find that confusing. ![]() We see on the left hand side simplifies to, this If I add them I get 10 and when I multiply Side, I also have to do to the right hand side. Maintain the equality, anything I do the left hand We have a constant term that is the square of half of theĬoefficient on the first degree term. On the left hand side, it will be a perfect square if The left hand side, not the last term, this expression Perfect square trinomial, is that this last And the way we can do that,Īnd saw this in the last video where we constructed a So I want to add something toīoth sides of this equation so that this left hand sideīecomes a perfect square. ![]() I want to add something toīoth sides of this equation. So all I did is add 75 toīoth sides of this equation. To add something here to complete the square Leave some space here, because we're going And so we get x squared plusġ0x, and then negative 75 plus 75. Sides to get rid of the 75 from the left hand The right hand side just so it kind of clears The left hand side into a perfect square. This term right here, this 10, half of this 10 is 5. Is not a complete square, or this is not a perfect This, just the way it's written, you might try to factor And then 300 dividedīy 4 is what? That is 75. To the left hand side, I also do the rightĬontinue to be valid. And I can obviouslyĭo that, because as long as whatever I do So let's just divideīy 4, this divided by 4, that divided by 4,Īnd the 0 divided by 4. Step here, I don't like having this 4 outįront as a coefficient on the x squared term. Help?Ĭomplete the square to solve 4x squared plusĤ0x minus 300 is equal to 0. Why didn't they divide the 2 term by 2 in the beginning? And why did they times the added term by 2 at the end? Looking back at it, I'm thinking they multiplied the last term by 2 to make it even with the equation in the paratheses, but I've also seen equations when the term isn't multiplied by the leading coeffiecient. ![]() Then divide the middle term to get 3/4, then I subtract that term squared from -1 to get -1 - 9/16, to which I got 25/16 = (x+3/4)^2 or 2(x+3/4)^2 - 25/16īut the hint for the equation showed this process instead: The explanations suck as to why you do this and not that, so can someone help me out please? Sometimes you divided everything by the leading coefficient, sometimes you don't divide the last term by the leading coefficient, sometimes you multiple the squared middle term by the leading coefficient. As you can see, quadratic equations that look like fractions are just as simple to rearrange and solve.So the practice after this video only managed to completely confuse me. Plug the discriminant part into the rest of the general equation to find the two solutions for x.įinally, these are the two possible answers for x. As you can see, it is a positive number, so there should be two real roots. This is the bit under the square root sign. Using the general formula for solving quadratic equations, we usually calculate the discriminant part first. Solve the equation using the general formula. Usually, mathematicians do not like to see too many negative numbers in an equation. Multiplying throughout by -1 removes the negatives on the first two coefficients, but it turns the +8 into a -8 obviously. The -4x terms add up to give -8x, and the pair of 4s add up to give 8. Moving all the terms to the LHS helps to simplify a little more. Here you just expand out the brackets on the LHS, and simplify a little. The terms rearrange like this and as if by magic, the fraction is gone and you have something more sensible that you can work with. Rearrange the fraction to look like a quadratic equation. Here is an algebraic fraction that is a quadratic in disguise. Quadratic equations involving fractions are common, and one usually uses the cross multiplication method to form the equation.
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