quadratic equation?

Question

Please help with this problem---x squared minus 13x + 36=0... our teacher @ school is REALLY bad with explinations and counldn't really teach this. Once I know how to work one I can go on with the rest of my work...or you can IM me @ biznitchil@yahoo and walk me through it

Answer 1

You need to try to think of two numbers that multiply to 36 (the third number) and sum to -13 (the second number). Once you've found them, the factored form is (x+__)(x+__), and the x-values making either term 0 are your answers.

Note that this works for negative numbers as well. If there is a coefficient for x^2, these get a little trickier, and there are a couple of ways to tackle those. You can either factor out the coefficient, or find numbers multiplying to the coefficient and mixing the numbers to get the middle term.

Answer 2

You can solve this by factoring the polynomial at the left side of the equation. You need to find two numbers that multiply to 36 and add to -13. -9 and -4 are your numbers. -9*-4 = 36 and -9+-4 = -13.

So the factored form of x^2-13x+36 is (x-9)(x-4)

To solve, set each of the factors equal to zero and solve each factor for x. You will get two answers.

Hope this helps. Good Luck.

Answer 3

there are many ways to solve this trinomial. because its quadratic, one way is take first and last coefficient.. multiply.. in this case, you get 36..
now, find the factors of 36 that will add up to the middle coefficient, in this case, 13..
lets see..making a list.
1*36
2*18
3*12
4*9
6*6

but the one that stands out is 4 and 9.. because, -9 and -4 is 13..

and, (-9)(-4) is 36..

now.. just write in parenthesis as
(x-4)(x-9).. because you multiplied by one in the start, there is nothing else to do..

however, if you had a factor of 2 in the squared variable, you would multiply by 2, but you would have to divide by 2 at the end... which undoes the multiplying by to .... try this..


there is a name to this method, don't remember it right now.. but my students call it, the X method.. cause you place your product on top of the x, and middle coefficient at the bottom of X, and factors you find place on the left, and right of X.... its hard to explain without pictures, and a board..

Math Teacher..

Answer 4

x^2 - 13x + 36 = 0

you want it in the form such that:
(x+a)(x+b) = 0

expand the above
x^2 +ax + bx + ab = 0
x^2 +(a+b)x + ab = 0

the original equation
x^2 + (-13)x + 36 = 0

therefore:
a+b = -13
ab = 36

solve for a & b, a = -4, b = -9

Finally: (x-4)(x-9)=0

Answer 5

What are you trying to solve it for? you trying to factor? if you're trying to solve for x it's x^2-13x= -36, x^2 = 13x-36, puzzle it out...somewhere between 9 and 10, if you have a TI-83+ calculator go to math, bottom option on the first screen called Solver, punch in the equation and it gives you what x equals.

Answer 6

ax^2+bx+c=0
a=1
b=-13
c=36

x1=(-b+sqrt(b^2-4ac)/(2a)
=(13+sqrt(169-144))/2=(13+5)/2=9

x2=(13-5)/2=8/2=4

solution set={4,9}

OR

The given equation can be writen as
(x-4)(x-9)=0
0.0=0
x-4=0 x1=4
x-9=0 x2=9

Answer 7

b^2-4ac=169-144=25
x=-b(+,-)sqrtofb^2-4ac=13(+,-)5/2=9and4