How to solve quadratic equation by completingthe square

give me an equation
x equared + 6x=-8 thanks po
welcome kiddo :) mas tama yata ung isang sumagot?
wala pa pong tamang sagot eh
why? completing the square ung ginawa ko. ang hirap lang ipakita kasi mas okay na isulat




Solve (x – 3)(x – 4) = 0.Okay, this one is already factored for me. But how do I solve this?Think: If I multiply two things together and the result is zero, what can I say about those two things? I can say that at least one of them must also be zero. That is, the only way to multiply and get zero is to multiply by zero. (This is sometimes called "The Zero Factor Property" or "Rule" or "Principle".)

Warning: You cannot make this statement about any other number! You can only make the conclusion about the factors ("one of them must equal zero") if the product itself equals zero. If the above product of factors had been equal to, say, 4, then we would still have no idea what was the value of either of the factors; we would not have been able (we would not have been mathematically "justified") in making anyclaim about the values of the factors. Because you can only make the conclusion ("one of the factors must have equalled zero") if the product equals zero, you must always have the equation in the form "(quadratic) equals (zero)" before you can attempt to solve it.

The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I'll set them eachequal to zero:x – 3 = 0   or   x – 4 = 0This gives me simple linear equations, and they're easy to solve:x = 3  or  x = 4And this is the solution they're looking for:  x = 3, 4

Note that "x = 3, 4" means the same thing as "x = 3  or  x = 4"; the only difference is the formatting. The "x = 3, 4" format is more-typically used.

One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic equations may be verified by plugging them back into the original equation, and making sure that they work, that they result in a true statement. For the above example, we would do the following:


Checking x = 3 in (x – 3)(x – 4) = 0:

([3] – 3)([3] – 4) ?=? 0 
     (3 – 3)(3 – 4) ?=? 0
               (0)(–1) ?=? 0
                       0   =   0

Checking x = 4 in (x – 3)(x – 4) = 0:

([4] – 3)([4] – 4) ?=? 0 
     (4 – 3)(4 – 4) ?=? 0
                 (1)(0) ?=? 0
                       0   =   0

So both solutions "check" and are thus verified as being correc

 x^{2} +6x=-8 . put two blanks after 6x (the sign is positive always) and -8
x squared +6x +[blank]=-8+[blank]
b/2= 6/2=3, square the 3 then it will be put on the blanks
x squared + 6x + 9=-8+9
x squared + 6x +9-1=0
x squared +6x + 8, then factor it. it will become
(x+4)(x+2)=0 then solve the value of x
x=-4 and x=-2
hope it helped :)