How do you solve an quadratic equation?

Question

Give at least 1 example for the method

Answer 1

x=

-b +/- SQRT (bSQ-4ac
-------------------------------
2a

whereas: a(x)SQ+b(x)+c=number

problem: xSQ+2x+3=6

a=1
b=2
c=3

-2 +/- SQRT (2SQ+4*1*3)
-----------------------------------
2*1

2SQ=4
4*1*3=12
2*1=2
4+12=16
SQRT16=4

-2+4
------
2

-2+4=2

2/2=1

-2-4
------
2

-2-4= -6

-6/2= -3

x= 1 AND -3

Answer 2

To start you need to get the whole equation on one side so it is equal to zero (x^2(x squared)=6x-8 ------->x^2-6x+8=0). Then you can factor it, separating it into two terms (x^2-6x+8=0 -------> (x-2)(x-4)=0). If you don't know how to factor, you will need to learn this first and it is two complicated to explain by just typing. Most every Algebra 1 book explains factoring. When you have the equation down to (x-2)(x-4)=0, you know that for that to equal 0, one of the terms has to be equal to 0 because any number times 0 equals 0. So, set both x-2 and x-4 equal to 0 (x-2=0, x-4=0) and solve for x.

x-2=0
+2 +2
x=2

x-4=0
+4 +4
x=4

A quadratic equation will give you either two answers, like above, one answer, when both answers are the same, or no answers when there is no real number that will fit the equation. If an equation will not factor, you can use the quadratic equation which is too complicated to type, google it. You can also solve by graphing the equation and finding where it crosses the x-axis.

Answer 3

Quadratic equations require that you factor out a polynomial and set each factor equal to zero to find the roots of the equation:

x^2 + 4x +4 = 0

Find the factors of the last term 4 that when manipulated will give you the middle term, 4. 2 and 2 are factors of 4 AND when you add them together you get 4. Assuming that you know the form in which to write the factored terms, the equation becomes:

(x+2)(x+2) = 0

Use the FOIL method to check your factoring: First, Outer, Inner, Last.

Now set each term equal to 0 to determine the roots of the equation.

x+2 = 0; x = -2
x-2 = 0; x = 2

If the problem is asking for a length of something or something that requires a positive value, you can almost always eliminate the negative value of x that is obtained. In this case, -2 would be a root but not part of the solution.

Answer 4

You may use the quadratic equation which verbally stated is {-b + or - (the square root of ((b^2)-4ac)) / 2a} where ax^2 +bx +c is the quadratic you want to solve. Or by the way of factoring the eqation. x^2 - 4 would factor down to (x+2)*(x-2). Another example of this would be x^2 - 5x +6 would break down to (x-2)*(x-3). To solve for x set the broken down components to zero. Say for the first example, x =2, -2. For the second x= 2,3. Quadratics can also become more complex in the manner that 3x^2 - 11x +5 may be given the solutions for this would be fractions approximately .531625 and 3.13504

Answer 5

you could do the slip and slide method where you take the first part and times to the last and then when you separate what is left over divide it. You should then have the correct answer. The method you will prefer once you know math really well is to just guess and check. 3x^2+9x+6

Which equals 3(x^2+3x+2)
then equals...{3(x+1)(x+2)}


or the slip and slide...3x^2+9x+6

(x^2+9x+18)
[(x+6)(x+3)}
[3(x+2)(x+1)]