(x+8)(x+8)=0

x^{2}+12x+40=4

I don't understand how to find out how many sol'ns there are

raindropsfurrever
Mar 21, 2018

#1**+3 **

A polynomial equation may have as many solutions as its degree, but sometimes some of the solutions are the same. We can find how many solutions each equation has by finding those solutions.

First one:

(x + 8)(x + 8) = 0 Set each factor equal to zero.

x + 8 = 0 or x + 8 = 0

x = -8 or x = -8

Notice that these two solutions are exactly the same, so really the equation just has one solution:

x = -8

Here is another way to work it:

(x + 8)(x + 8) = 0

(x + 8)^{2} = 0 Take the ± square root of both sides.

x + 8 = ±√0

x + 8 = ±0 Zero is neither positive nor negative, so we can drop the ± .

x + 8 = 0

x = -8

Second one:

x^{2} + 12x + 40 = 4 Subtract 4 from both sides of the equation.

x^{2} + 12x + 36 = 0 Now factor the left side of the equation...

(x + 6)(x + 6) = 0 This is just like the last problem now. Set each factor equal to zero.

x + 6 = 0 or x + 6 = 0

x = -6 or x = -6

These two solutions are identical, so really there is just one solution. It is:

x = -6

hectictar
Mar 21, 2018