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## Algebra (all content)

### Course: Algebra (all content)>Unit 9

Lesson 11: Systems of quadratic equations

# Quadratic system with no solutions

Sal solves a system of two quadratic equations algebraically and finds the system has no solutions. He then graphs the equations to show that this is true. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Is there a video on how to put an equation in the vertex form? It's appears like easy manipulation however, is there something that could help me better understand how Sal played those equations so eloquently?
• 18 / 0.12= 150. How do you get that?
• an easy way to think about it is that you want to clear away the decimal. so by moving the decimal from .12 two places to the right (or multiply by 100). and because you moved it to the right for .12 you also want to do that to the 18 to make it 1800. 1800/12=150.
• Would the two parabolas intersect when using an imaginary axis?
• The imaginary axis is a completely different dimension than the Cartesian Plane. It is used to visualize an imaginary point, so technically you can't graph a parabola on an imaginary axis. The two parabolas from the video won't intersect on the real axis, which means they won't intersect at all.
I hope that helped! 🙂
• @ Isn't it incorrect to say there is no solution at all? Is there not a solution that just involves i?
• Yeah, but he wants to find all the real solutions. There are solutions that involve i, but they are not real solutions.
• can someone please explain how 2(x-4)^2+3 turns into 2(x^2-8x+16)+3?
• (x-4)^2 =(x-4)(x-4)
You need to foil or multiply both together.
(x-4)(x-4)
Whenever you see two numbers in parenthesis being added or subtracted, you need to multiply the terms together:
Ex:
(x+3)^2=(x+3)(x+3)
(4x-2)^2=(4x-2)(4x-2)

Foiling
(x-4)(x-4)
Multiply 1st terms (x*x)
x^2
Multiply 1st term to last term
(x)(-4)
-4x
Multiply middle terms
(-4)(x)
Multiply last terms
(-4)(-4)=16
x^2-4x-4x+16
x^2-8x+16
You can also just distribute the terms
(x-4)(x-4)
Distribute either one. I will distribute the x-4 to the other x-4

x(x-4) - 4(x-4)
x^2-4x - 4x+16
Combine like terms
x^2-4x-4x+16
x^2-8x+16
• so should i have studied quadratic equations before i got to this? it looks like that's a later topic in the algebra play list. and where do i find the vertex form/formula?

for instance at sal says "this is in vertex form" referring to y = 2 (x-4)^2 + 3 "where x = 4 and y = 3"... and then he plots that point. i don't knw what that means or where i can look for it in the playlist.

at i think this might be completing the square?
• Yes, you need to know quadratics before doing these videos -- they seem to have been put too early on this list. It is possible to do some simple non-linear equations before mastering quadratics, but I would not recommend it. Try going to the Functions videos and then coming back to these videos once you're completely done with quadratics.
• What if in one or both of the equations y and x are squared?
• You'd have to simplify until the equation is either linear or quadratic.
Think of it like this
ax^2+bx+c = a ( Substitute ^2 ) ^2 + b ( Substitute ^2 ) + c

If an equation is just y^2 = ax^2+bx+c
convert it to y = sqrt(a) * x + sqrt (b) sqrt ( x ) + sqrt(c) and then insert this into the other equation and solve.
How can you manipulated?
• You get one solution when the discriminant b²-4ac=0. The graph of the parabola would have the vertex on the x-axis, so that it only touches at one point.
• Hello, I have a question about Quadratic system with a wrong solution.
Equation 1: ax^2 = y
Equation 2: bx = y
for any real number a & b.
It will yield:
ax^2 = bx
ax^2 - bx = 0
x(ax-b) = 0
Hence, x = 0 & y = 0 Or x = b/a & y = b^2/a
Although we can simply find x = 0 & y = 0 is the true answer (by plotting a graph), what is the reason/meaning behind of the equation ax-b = 0 (why does it cause)? Is this mathematic fault in the quadratic system unevitable? Sorry for bothering and many thanks.
• What leads you to believe that x = b/a & y = b^2/a is not a solution? It is a valid solution. If you substitute it back into both equations, the ordered pair works in both equations.
• At , why did he add the "plus 1 and minus 1" there? Won't it be cancelled out and therefore he would be adding nothing to the equation? Is there something I'm missing or not?