Question 203173: Solving Quadratics
2x2 - 4x + 1 = 0
8. 3x2 + 2x = 7
9. x2 + 8x + 9 = -7
Answer by jejess(7)   (Show Source):
You can put this solution on YOUR website! This is all about factoring
for the first one we have 2x^2 -4x +1 =0
You need to find two number that multiply 2 but add to negative 4
-2, -2
now the question will read like this
2x^2-2x-2x+1=0
To factor this one we group the first two terms together and the last two terms together.
(2x^2-2x)(-2x+1)=0
Now we factor out what we can from the barkets
In the first one (2x^2-2x) What is common about the two terms? 2 and x
so we remove them
2x(x-1)
now do the same for the second set of barkets (-2x+1)
-1 is common
-(2x-1)
Now the 2x-1 is common so we rewrite the question as this
(2x-1)(x-1)=0
Solve for x by splitting the barkets up
2x-1 = 0 OR x-1=0
slove by rearranging
2x-1=0 or x-1=0
2x=1 x=1
x=1/2
If you do not understand factoring and rearraging please say so in next question.
I will answer the third one and you can try the second your own if you don't understand it please re-post with where you seem to go wrong. ie; factoring.
The next one is a little tricky because the question is equal to something (-7)
We must re-write the question then to make it equal zero in order to factor it.
x^2+8x+9=-7
now will read this
x^2+8x+9+7=0 REMEMBER when you move a number or term it changes signs a postive become a negative and vis versa.
Collect like terms; so you add 9 and 7 to get 16
x^2+8x+16=0
Now you need to find two numbers that add to 8 but mult. to 16
4 and 4 and you split the middle term (8x) into these two numbers so the question reads like this now.
x^2+4x+4x+16=0
(x^2 +4x)(4x+16)=0
x(x+4)4(x+4)=0
(x+4)(x+4)=0
or (x+4)^2
x must be = to -4
x+4=0
x=-4
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