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# Quadratic equations word problem: triangle dimensions

Sal solves a geometry problem using a quadratic equation. Created by Sal Khan and Monterey Institute for Technology and Education.

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• At when he says "let me make it clear... lets do it this way"
How did he know to multiply the 1/2 with the b before distributing?
Is that part of the order of operations? Are you supposed to do any multiplications that come before the parentheses first before distributing always?
• Communative property of multiplication- You can do it in any order.
• How do you do factoring by grouping?
• I'll explain how to do factoring by grouping but first be aware that.....
The quadratic equation is in the form of...........
Ax²+Bx+C=0
Where:
(note:coefficient of means the number beside)
A=the coefficient of x²
B=the coefficient of x
C=the constant term(or the term that doesn't have an x beside it)
And we want to find two coefficients of x (q and w) to replace B which when you.......
multiply them they equal = A*C =q*w
add them they equal = A+B =q*w

so from..... Ax²+Bx+C=0
you'll get.... Ax²+qx+wx+C=0
you'll group them to.. (Ax²+qx)+(wx+C)=0
you'll further simplify it with real numbers...

Okay lets have an example so that we can understand it even more.
Lets say we want to factor an equation the form of Ax²+Bx+C=0
like 2x²+13x+20=0
We know that our A=2, B=13 and C=20
So we want to find two coefficients of x (q and w) to replace B which when you.......
multiply them they equal = A*C =q*w= 2*20=40
add them they equal = A+B =13=q*w =8*5
Now we know that our q and w are..... 8 and 5 because the satisfy all the conditions we are looking for.
Lets just plug in their values to..
2x²+qx+wx+20=0
2x²+8x+5x+20=0
(2x²+8x)+(5x+20)=0 ( I grouped them apropriately so that I can factor out)
2x(x+4)+5(x+4)=0 ( I factored out a 2 and 5)
(x+4) (2x+5)=0 ( I factored out the (x+4) from 2x and 5)
• What I am still so confused on how to solve problems using quadratic equations?
• Can someone give me a website please? Where I can practice more examples like the one on the video.
• So are all measurement/geometry problems using quadratics like the one in the video going to have a negative answer and a positive answer? If not, how will you know which one is the correct answer? Will one give you a different answer from the other when you put the value in the original equation?
• With respect to Geometry, the right answer will always be positive, since it's impossible to have negative area, length, volume, etc. So if there's one positive answer and one negative answer, the correct one will be positive.

The triangle problem that Sal was doing in the video will always have one positive and one negative answer. This is because the area will always be positive making "c" (ax^2+bx+c) always negative. When "c" is negative there is only one positive solution and one negative solution.

It's also important to remember that sometimes the quadratic equation will give two positive answers or two negative answers (if you plug in x^2-5x+6 to the quadratic equation, you get an example of two positive answers, while x^2+5x+6 gives you an example of two negative answers.)

Hope that helps
• At , Sal says the area is 30 inches SQUARED equal to 1/2 the base times height, yet he wrote it as:

30 = 1/2*b(b-4)

Why didn't he write it as 30^2 like the problem said?:

30^2 = 1/2*b(b-4)
• What if you had 35n^2 + 22n + 3 = 0?
What should I do?
• 3 only has two factors, so it is inevitable that the solution will look something like this...
(_n + 3)(_n + 1)
Now, all we have to deal with are the factors of 35 and there are four possibilities (1 & 35, 35 & 1, 5 & 7, and 7 & 5), so it shouldn't be too hard to check each one.
1.) (1n + 3)(35n + 1)
35n^2 + n + 105n + 3
35n^2 + 106n + 3 =/= 35n^2 + 22n + 3
This does not work, they are not equal
2.) (35n + 3)(n + 1)
35n^2 + 35n + 3n + 3
35n^2 + 38n + 3 =/= 35n^2 + 22n + 3
This does not work
3.) (5n + 3)(7n + 1)
35n^2 + 5n + 21n + 3
35n^2 + 26n + 3 =/= 35n^2 + 22n + 3
This does not work, so it must be the last possibility
4.) (7n + 3)(5n + 1)
35n^2 + 7n + 15n + 3
35n^2 +22n + 3 = 35n^2 + 22n + 3
They match so (7n + 3)(5n + 1) is the correct factorization.

To finish the problem...
(7n + 3)(5n +1) = 0
This means that either 7n + 3 = 0 or 5n + 1 = 0
7n + 3 = 0
7n = -3
n = -3/7
5n + 1 = 0
5n = -1
n = -1/5
n = -3/7 or -1/5
• why is the best way to solve a quadratic to have all the terms on one side and equal to zero?
• By putting it equal to 0, you are then finding the roots (or zeroes or x intercepts or solutions) where the quadratic crosses the x axis. If you keep a 60 on the other side, you have to find where b^2-4b crosses the y=60 line which is okay if you are completing the square and solving by taking the square root, but not for factoring which finds the x intercepts. Even the quadratic formula requires you to have it =0 so you know a, b, and c.
• are there any quadratics that can't be factored?
• Try x^2 + 2x + 2.

Any quadratic of the form ax^2 + bx + c can be solved using the formula ( -b +/- sqrt(D) )/2a with D= b^2-4*a*c. However, if D is less than zero it cannot be solved regularly. This requires the introduction if the imaginary number i = sqrt(-1). I think this allows us to factor all quadratics.
• At - , why is this the case? Why is the best way to solve a quadratic is to set it equal to 0?