Some questions about quadratics, please explain?

Question

1) I've factorised an equation to
8 (x - 9) (x - 4).
How do I solve for x, can I just ignore the 8 before the brackets and say that x is 9 or 4?

2) 16t^2 - 70t + 76
How do I solve this? I usually work out a number that multiplies to give 76 + 16 and that adds to give - 70, but I can't work that out in my head, how do I do it in a simpler way?

3) .003v^2 + .03v + 12.2 = 0
How do I do this?


P.S I don't just want answers, I want explanations
thanks a lot!

Answer 1

Solving for x

8 (x - 9)(x - 4) = 0

Roots

8 = 0

- - - - - -

Roots

x - 9 = 0

x - 9 + 9 = 0 + 9

x = 9

- - - - -

Roots

x - 4 = 0

x - 4 + 4 = 0 + 4

x = 4

- - - - - -s-

Answer 2

I'll tell you how I'm learned to solve these...In your country you do it different however ;)

1) x^2 - 4x - 9x + 36 = 0 (I opened the brackets)
x^2 - 13x + 36 = 0
D = b^2 - 4ac (where b = 13 ,a = 1 and c = 36)
D = 13^2 - 4.1.36
D = 169 - 144
D = 25 ; 25^-2 = 5

x = - b - D^-2 /2a = 13-5/2.1 = 13-5/2 = 8/2 = 4
x = -b + D^-2 /2a = 13+5/2 = 18/2 = 9

2) 16t^2 - 70t +76 = 0
D = 70^2 - 4.16.76
D = 4900 - 4.1216
D = 4900 - 4864
D = 36 ; 36^-2 = 6

x = -b - D^-2/2a = 70 -6/2.16 =64/32 = 2
x = -b + D^-2 /2a = 70 + 6 /32 = 76/32 = 2 ,375

I don't get the third one ...what is ".003v^2 + .03v + 12.2 = 0" ... the "." before the 003v^2 means that it's "0.003v^2" or what ??..

Edit : Just saw the answer above...Now I get it ..

0.003v^2 - 0.03v = 12.2 = 0 / .1000
3v^2 - 30v + 12 200 = 0 (as explained above ,multiply by 1000)

D = b^2 - 4ac (I think that I'm writing this for 857 time :-D )
D = 900 - 3.12 200 .4
D = 900 - .....
D = - ..... (the number doesn't matter ... no decision)

Hope that helped ... I wrote it the most simple way

Answer 3

1. For solving the equation, you do ignore the 8. That is a separate factor, and as it does not involve x, it cannot be equated to zero or give an answer.

2.
The technique you describe for finding factors needs to be extended a bit. It is more complicated when the coefficient of x^2 (or t^2 in this equation) is not 1.

Notice first that you can extract a factor of 2, and reduce the numbers.
2(8t^2 - 35t + 36) = 0
Now map the equation in brackets on to the general quadratic at^2 + bt + c = 0.
That gives a = 8, b = -35, c = 36.
Evaluate the discriminant, which is b^2 - 4ac = 35^2 - 4*8*36.
That comes to 73. As 73 is not a perfect square, you can abandon factorisation by trial and error. Use the formula:
t = ( -b +/- sqrt(d) ) / 2a, where d is the discriminant you have just worked out. Hope you can carry on now.

3.
The greatest number of decimal places you have here is 3, in the first term. Multiply the equation by 1000 to get rid of them:
3v^2 + 30v + 12200 = 0
Now continue as in question 2.

Answer 4

1) What you wrote here is an expression, not an equation. Where's the rest of the equation? Was it what you wrote, set equal to zero? If that's what you have, then yes, x=4,9 are obviously the only solutions to 8(x-9)(x-4) = 0, because the only way to make the product of three numbers zero is to have at least one of them equal zero.

2) Again, where's the rest of the equation? We can't "solve" anything unless there's an equation. Is it that expression, set equal to zero? If so, then you can factor out a 2 to get
2(8t^2 - 35t + 38). The only integer factors for 38 are (1, 38) and (2,19). Testing these in different combinations with (1,8) and (2,4) gives 2(t - 2)(8t - 19) as the factoring. You could also resort to using a method like completing the square or the quadratic formula, and get something equivalent.

3) First multiply both sides by 1000 to clear away the decimal. You get 3v^2 + 30v + 12200 = 0. But if you apply one of the techniques I mentioned in #2, you'll find that it has no solution. For example, you can rewrite this as
v^2 + 10v + 12200/3 = 0
v^2 + 10v + 25 + 12125/3 = 0
(v + 5)^2 = -12125/3
But you can't have a square number equal a negative number.

Answer 5

1) Yes. U can just ignore the 8 before the brackets. It's just a common factor that has been taken out. So yes, the answer for x is 4 or 9.

2) 16t^2 - 70t + 76 = 0
2(8t^2 - 35t + 38) = 0 (Take out the 2 as a common factor, u may ignore it from now onwards)

By using x={-B+-[sqrt(B^2-4AC)]}/2A where Ax^2+Bx+C=0
t = (-(-35)+{sqrt[(-35)^2 - 4(8)(38)]}) /2(8)
t = [-35+sqrt(1225-1216)]/16
t = [-35+sqrt(9)]/16
t = (-35+3)/16
t = (-32)/16
t = -2

or

t = (-(-35)-{sqrt[(-35)^2 - 4(8)(38)]}) /2(8)
t = [-35-sqrt(1225-1216)]/16
t = [-35-sqrt(9)]/16
t = (-35-3)/16
t = (-38)/16
t = -19/8

3) 0.003v^2 + 0.03v + 12.2 = 0
(multiply by 1000, especially if u don't like to work with decimals):
3v^2 + 30v + 12200=0
By using x={-B+-[sqrt(B^2-4AC)]}/2A where Ax^2+Bx+C=0
v = (-30+-{sqrt[30^2 - 4(3)(12200)]}) /2(3)
v = [-30+-sqrt(900-146400)]/6
v = [-30+-sqrt(-145500)]/6

v is not solvable becos there is no answer for sqrt(-145500), i.e. sqrt cannot be done for -ve numbers.

Hope that my explanations are clear enough for u.

Answer 6

1) i don't think you can just ignore the 8...

2)16t^2 - 70t +76
i do the same thing... i don't know if there's a simpler way.
(16t - 38) (t - 2) = 0

3)

Answer 7

For the first one, yea... x = +9 or +4.

For the others, the quadratic equation might be what you're after.

http://en.wikipedia.org/wiki/Quadratic_equation