Question 4743: solve each equation for x using factoring. please show work.
x2-4x+1=13
x2+2x+48=15x
5x2+7=52
3x2=4-11x
2x3+4x=6x2
2x2-4=7x
Answer by guapa(62)   (Show Source):
You can put this solution on YOUR website! 1)x²-4x+1=13 In order to factor the right side has to be 0. So, we substract 13 from both sides.
x²-4x+1-13=13-13
x²-4x-12=0 Now we need 2 numbers whose sum is -4 and whose factor is -12. To find them we express 12 in its prime factors. 2*2*3 By grouping these factors in various ways we find that 6(2*3), 2(2)and 2*6=12 Since we need -12 one of these factors has to be negative. -6*2=-12 and -6+2=-4 Now we can express the middle terms as -6x+2x. Therefore, we can complete the factoring as follows:
x²-6x+2x-12=0 Factor by grouping(x from first two terms and 2 from last two terms)
x(x-6)+2(x-6)=0 Factor common binominal factor of (x-6) from both terms.
(x-6)(x+2)=0 For all real numbers: ab=0 if and only if a=0 or b=0
Let's use this technique:
a=0, (x-6)=0 , x=6 ->(6-6)=0
b=0, (x+2)=0 , x=-2 ->(-2+2)=0
The solution set is (-2,6)
Check:
6²-4(6)+1=13
36-24+1=13
13=13
(-2)²-4(-2)+1=13
4-(-8)+1=13
4+8+1=13
13=13
Use this process for all following problems
2)x²+2x+48=15x
x²-13x+48=0
48 expressed in its prime factors: 2*2*2*2*3. The solution to this problem is not factorable since there are no two numbers whose sum is -13 and whose product is 48.
3)5x²+7=52
5x²+7-52=52-52
5x²-45=0 Divide both sides by 5
(5x²-45)/5=0/5
x²-9=0 The difference of two squares {a²-b²= (a-b)(a+b)}
(x-3)(x+3)=0
x-3=0, x=3
x+3=0, x=-3 Solution set is (-3,3)
4)3x²=4-11x
3x²+11x-4=0 (2 numbers whose sum is 11 and whose product is -12(3*-4)
3x²+12x-x-4=0
3x(x+4)-1(x+4)=0
(x+4)(3x-1)=0
x+4=0, x=-4
3x-1=0, x= 1/3 Solution set is (-4,1/3)
5)2x³+4x=6x²
2x³-6x²+4x=0 Divide both sides by 2
x³-3x²+2x=0 Factor out x
x(x²-3x+2)=0
x(x²-2x-x+2)=0
x(x-2)(x-1)=0
x-2=0, x=2
x-1=0, x=1 Solution set is (0,1,2)
6)2x²-4=7x
2x²-7x-4=0
2x²-8x+x-4=0
2x(x-4)+1(x-4)=0
(x-4)(2x+1)=0
x-4=0, x=4
2x+1=0, x= -(1/2) Solution set is (-1/2,4)
I hope this helps
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