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# Solving quadratics by completing the square: no solution

Sal solves the equation 4x^2+40x+280=0 by completing the square, only to find there's no solution for this equation. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• At Sal divides everything by 4 because they are divisible by 4. what do we divide by if there is no common factor? for example: 0 = x^2 -3x -12
• In that case you don't divide at all. Sal just did it to simplify, which isn't always possible. Instead you go on to the next step and fill in the square. In the case of the one you gave, because the coefficient of x is -3, you will get (x - 3/2)^2 as the square, so you need to add and subtract (-3/2)^2. But sometimes you might want to divide anyway, for example if x^2 has a coefficient, like in 3x^2 + 2x + 1 = 0. It'd be easier without that 3 there, so one way to go is to divide and get x^2 + 2x/3 + 1/3 = 0, and then you can add and subtract (1/3)^2 in order to fill in the square to get (x + 1/3)^2. It isn't necessary, it depends on whether you find it makes it easier.
• Does the square root always have to equal zero
• When finding the roots, solutions, or zeros (all the same thing) to a quadratic function, yes, you set them equal to zero, and solve for the independent variable (x).
• At Sal factors out a 4 to simplify the equation, which I understand. I also understand that, in the end, this equation doesn't have any real roots because we can't take the square of a negative. However, wouldn't the final answer be -45 = 4(x+5)^2 instead of -45 = (x+5)^2. The 4 that he factored out seems to have disappeared into the ether. Am I missing something?
• John,
Rather that "factor out the 4", Sal divided both sides of the equation by 4. On the left was a zero, and 0/4 = 0, so yes, the 4 did look like it evaporated into thin air.
But if he had just factored the 4, the answer would be
-180 = 4(x+5)². And we could still then divide both sides by 4 and the answer would be -45 = (x+5)²

0=4x² + 40x +280 Let's just factor out the 4
0=4(x² + 10x + 70) And and and subtract the 25 inside the parenthesis.
0=4(x² + 10x + 25 - 25 + 70) Now convert the perfect square
0=4((x+5)² -25+70) subtract the 25 from 70
0=4((x+5)² + 45) Distribute the 4
0=4(x+5)² + 4*45 and multiply 4*45
0=4(x+5)² + 180 Now subtract 180 from each side
-180 = 4(x+5)² And now you could divide each side by 4
-45 = (x+5)²

So, as you see the 4 didn't just disappear. It is that when one side was 0, when both sides were divided by 4 it just looked like it was factored out and disappeared.

I hope that makes it click for you.
• What would the complex roots be?
• Hello gatogreensleeves,
the complex roots would be x = -5 +j*(squareroot(45)) and x = -5 -j*(squareroot(45)).
Regards!
• why can't you use x = -b + or - sqrt b squared - 4ac/2a for this?
• You certainly can use that formula (which is called the quadratic formula). However, Sal is trying to explain HOW to find the solutions of the equation so that people will understand how equations are manipulated. Simply plugging numbers into a formula certainly works, but it doesn't help with understanding and it makes quite a boring video.
• at what is a complex number?
• Complex numbers are a new class of numbers that support roots of negative numbers.
They are written in the form a + bi where i is sqroot(-1). Note that complex numbers include the real numbers; you could think of a real number as being a complex number where b=0.
In the UK they are introduced at A-level, and used extensively in engineering degree calculations. At GCSE it is sufficient to state that there are no real roots. If you were to sketch the graph, it would never intersect the x-axis (i.e. there is no real value of x such that y=0).
• At , Sal said we can add a billion and subtract a billion without changing the equation. I agree completely with that.
But can we add infinite and subtract infinite without changing the equation?
• No, because you can't actually add or subtract infinity. It is not a number, it is an abstract concept.

Infinity minus infinity is not zero. Similarly, infinity divided by infinity is not one. Therefore, you cannot apply the same rules of inverse operations to infinity as you would to an actual number.
• My professor said to find the vertex of the parabola y=x^2-4x+6. When you complete the square you end up getting an answer that has no solution as Sal showed above. But then why is the vertex (2,2)?

Thank You!
• The vertex is different than the solution. The vertex is the maximum/minimum point of the line, and the solution(s) is where the line crosses the x-axis. The vertex can be found using this equation: x=(-b)/2a. This will get you the x value of the vertex. Plug that into your original equation to find y, and you have the coordinates of the vertex. :)
• When talking about the complex roots would you need a 3 dimensional graph to plot the roots? With x, y and i axes? Or is this not possible? If not why not?
• Generally complex numbers are graphed on a slightly modified coordinate plane with real numbers represented by what we usually think of as the x-axis and imaginary numbers represented by what is usually thought of as the y-axis. For example, the number 3 + 4i on the complex plane would be in the same place as the point (3, 4) on the x-y plane.

That said, I don't see why what you propose wouldn't be a reasonable representation. It'd certainly be interesting to see how the graphs look with both real and complex roots showing!