For example, solve x²+6x=-2 by manipulating it into (x+3)²=7 and then taking the square root.
Log in Emilyy Jade 8 years agoPosted 8 years ago. Direct link to Emilyy Jade's post “how did you get 25/4 and ...” how did you get 25/4 and how did you equal it to 7 • (18 votes) Megu 8 years agoPosted 8 years ago. Direct link to Megu's post “The 25/4 and 7 is the res...” The 25/4 and 7 is the result of completing the square method. To factor the equation, you need to first follow this equation: x^2 + 2ax + a^2. To figure out the a, you need to take the 5 and divide it by 2 (because 2ax), which becomes 5/2. a=5/2. Hope this helps! :3 (50 votes) Federico 8 years agoPosted 8 years ago. Direct link to Federico's post “How was 7 added at the 6 ...” How was 7 added at the 6 paragraph? Where it shows the steps on how to complete the square? • (8 votes) Kim Seidel 8 years agoPosted 8 years ago. Direct link to Kim Seidel's post “When Sal adds 9 to the le...” When Sal adds 9 to the left side, he must also add 9 to the right side. The right side was "-2". (26 votes) Santos Gonzalez 8 years agoPosted 8 years ago. Direct link to Santos Gonzalez's post “what does what does it me...” what does what does it mean when there is a letter i In the solution of the equation? • (4 votes) Miican Lin 8 years agoPosted 8 years ago. Direct link to Miican Lin's post “An "i" means the answer i...” An "i" means the answer is the square root of a negative number. Since that doesn't work in the normal everyday world - but does have uses elsewhere - the "i" is used to make it easier to simplify the answers (and confuse the people ~_^). "i" is defined as the square root of negative 1, and can be factored out. Like an "x" or other variable, terms with "i" can only be added to or subtracted from other terms containing "i". However, "i" squared = -1. It becomes a regular number and can be added to regular numbers. This is important to remember when checking your answers. (11 votes) Annika Breen 7 years agoPosted 7 years ago. Direct link to Annika Breen's post “How do you solve 5x^2 -10...” How do you solve 5x^2 -10x =23 • (3 votes) Kim Seidel 7 years agoPosted 7 years ago. Direct link to Kim Seidel's post “I'm going to assume you w...” I'm going to assume you want to solve by completing the square. Hope this helps. (8 votes) STEM 6 years agoPosted 6 years ago. Direct link to STEM's post “Does each quadratic equat...” Does each quadratic equation always have a square and those missing parts of the square; or is it (this method) just for the perfect square equations? • (4 votes) Monil Mehta 2 years agoPosted 2 years ago. Direct link to Monil Mehta's post “Not every quadratic equat...” Not every quadratic equation always has a square. It may have a square, missing parts for a square, or even both, in which case you could use the completing the square method. But no, for the most part, each quadratic function won't necessarily have squares or missing parts. It's possible, but not common. But since every number is a square and has a square root, you can still do it, though it would be much more painful. (2 votes) LoganLandonLiam 8 years agoPosted 8 years ago. Direct link to LoganLandonLiam's post “How do I do a quadratic e...” How do I do a quadratic equation using completing the square if a or x^2 have a number in front of it ( ie. 4x^2) because I have tried many things with it and it doesn't add up or subtract out. • (1 vote) doctorfoxphd 8 years agoPosted 8 years ago. Direct link to doctorfoxphd's post “This would be the same as...” This would be the same as rule 2 (and everything after that) in the article above. This leaves Or, you can divide EVERY term by 4 to get (8 votes) lew.lehmann 4 years agoPosted 4 years ago. Direct link to lew.lehmann's post “Often when coming across ...” Often when coming across quadratics with non-zero leading co-efficients I'm not sure whether to pull out a common factor, or to divide by the leading co-efficient. For example in the above question in my first attempt I factored out a 4 to get 4(x^2 + 5x)-3=0. However, when attempting to solve from there, I end up at a dead end, or with a different result to if I'd divided out the leading co-efficient. Are these 2 methods interchangeable, and I'm just doing something wrong? Or, is there a preferred approach for some reason? • (3 votes) A/V 4 years agoPosted 4 years ago. Direct link to A/V's post “I wouldn't say that both ...” I wouldn't say that both methods are exactly interchangable; however, it's best to factor out. Factoring it out preserves the equation whilst dividing the equation by, for example it's GCF, removes that aspect of the equation. As a result from that, there might be some loss of solutions. Your completing the square method is exactly on point with the first half. Just remember when you find (b/2)², you must add that result in the parenthesis, and subtract it out and multiply it by the 4 by the number outside. Less abstractly: 4(x²+5x+(25/4)) - 3(4)(25/4) = 0 And solve by then. Hopefully that gives some insight (3 votes) Marioland 8 years agoPosted 8 years ago. Direct link to Marioland's post “This method is only avail...” This method is only available if factoring does not help correct? • (0 votes) Daniel Li 4 years agoPosted 4 years ago. Direct link to Daniel Li's post “No, completing the square...” No, completing the square can be used to solve any quadratic equation whether or not factoring works. (3 votes) braydonives 6 years agoPosted 6 years ago. Direct link to braydonives's post “What if my x^2 value has ...” What if my x^2 value has a coefficient in front. What do I do with it? • (2 votes) Kim Seidel 6 years agoPosted 6 years ago. Direct link to Kim Seidel's post “That is covered in a late...” That is covered in a later lesson: https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-by-completing-the-square/v/completing-the-square-to-solve-quadratic-equations You can find more examples at this link: https://www.purplemath.com/modules/sqrquad.htm (2 votes) maral abdulaziz 4 years agoPosted 4 years ago. Direct link to maral abdulaziz's post “Even after reading the ar...” Even after reading the article, I still don't know how to complete the square for this equation -0.4x^2+8x • (2 votes) Kim Seidel 4 years agoPosted 4 years ago. Direct link to Kim Seidel's post “!) Factor out -0.4 so th...” !) Factor out -0.4 so the x^2 term has a coefficient of 1 Hope this helps. (3 votes)Want to join the conversation?
In x^2 +5x = 3/4, The a^2 is missing.
Then you need to square it, (because a^2) which becomes 5^2/2^2.
5x5 is 25, and 2x2 is 4, so the a^2 is 25/4.
When you add 25/4 on the left side, you have to add 25/4 on the right side as well.
Remember, the 3/4 is still there, so add them: x^2 + 5x + 25/4 = 25/4 + 3/4
25/4 + 3/4 = 28/4. 28 divided by 4 is 7.
So, when he add the 9, he gets: "- 2 + 9 = 7"
Hope this helps.
1) Divide the entire equation by 5: x^2 - 2x = 23/5
2) Complete the square: -2/2 = -1. (-1)^2 = +1. Add +1 to both sides: x^2 - 2x + 1 = 23/5 + 1
3) Rewrite the left side as a binomial squared, and add the fractions on the right: (x-1)^2 = 28/5
4) Take square root of both sides: sqrt(x-1)^2 = +/- sqrt(28/5)
5) Simplify both radicals: x-1 = +/- sqrt(4*7/5) x-1 = +/- 2 sqrt(7/5) * sqrt(5/5) x-1 = +/- 2 sqrt(35) / 5
6) Add 1 to both sides: x = 1 +/- 2 sqrt(35) / 5 x = 5/5 +/- 2 sqrt(35) / 5 x = [5 +/- 2 sqrt(35)] / 5
You are correct that you cannot get rid of it by adding or subtracting it out. As shown in rule 2, you have to divide by the value of a (which is 4 in your case). In the example following rule 2 that we were supposed to try, the coefficient of x² is 4. So, we have to divide the x² AND the x terms by 4 to bring the coefficient of x² down to 1.
Let's use the example they gave us:
4x² + 20x -3 = 0
move the constant term4x² + 20x = 3
divide through the x² term and x term by 4 to factor it out
4(4x²/4 + 20x/4 ) = 34( x² + 5x ) = 3x² + 5x = 3/4 → I prefer this way of doing it
Now we complete the square by dividing the x-term by 2 and adding the square of that to both sides of the equation. That is 5/2 which is 25/4 when it is squaredx² + 5x +25/4 = 3/4 + 25/4 → simplify the right sidex² + 5x +25/4 = 28/4 → Hey, that is equal to 7
Now rewrite the perfect square trinomial as the square of the two binomial factors
We especially designed this trinomial to be a perfect square so that this step would work:x² + 5x +25/4 = (x + 5/2)²
If you get stuck on the fractions, the right-hand term in the parentheses will be half of the x-term.
(x + 5/2)² = 7
Take the square root of both sides, remembering to take both the positive and negative square root of the number on the right
√ (x + 5/2)² = ± √ 7
x + 5/2 = ± √ 7
Isolate the x by subtracting away the constant that is still with it
x = - 5/2 ± √ 7
That gives us our two correct solutions for xx = - 5/2 + √ 7 x = - 5/2 - √ 7
-0.4(x^2-20x)
2) Now complete the square by taking [-20/2]^2 = (-10)^2 = 100. Add 100 inside the parentheses and subtract -0.4(100) outside the parentheses.
-0.4(x^2-20x+100)- (-0.4)(100)
-0.4(x^2-20x+100)+40
3) Factor the trinomial: -0.4(x-10)^2+40
