SOLUTION: solve the questions give below by factorization
(1)3x2+x=1 (2)4x2+9=12x (3)5x(x+2)=4(3x=1) (4)2x+7=x2 (5)x2-3x=3x=7 (6)3(4x-1)=x2
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Question 2598: solve the questions give below by factorization
(1)3x2+x=1 (2)4x2+9=12x (3)5x(x+2)=4(3x=1) (4)2x+7=x2 (5)x2-3x=3x=7 (6)3(4x-1)=x2
Found 2 solutions by longjonsilver, kiru_khandelwal:
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
by pure factorisation?
1) cannot be factorised easily. You would have to do the "complete the square" method.
2)
(2x - 3)(2x - 3) = 0
so 2x-3 = 0
2x = 3
x = 3/2
3) 5x(x+2) = 4(3x+1) --> + or - on the last term, seeing as you have "="?
this does not factorise easily. Again, do you want the "complete the square" method?
4)
Same as 3).
5) Again you have 2 "=" signs..which is wrong?
6)
Again this does not factorise easily.
jon.
Answer by kiru_khandelwal(79) (Show Source): You can put this solution on YOUR website!
4x2 + 9 = 12x
=> 4x2 - 12x + 9 = 0
Now the equation is of the form
ax2 + bx + c = 0
Multiply a and c.....and factorize the product 'ac' such that the sum of the factors is b
In our case a=4 b = -12 and c = 9
a*c = 4*9 = 36
Now we have to split 36 into factors whose sum is -12
so 36 = (-6)*(-6)
and sum of -6 and -6 is -12
therefore,
4x2 -12x + 9 =0 can be written as
4x2 -6x -6x + 9 = 0
=> 2x(2x-3)-3(2x-3)=0
=> Taking the common factore (2x-3) out
=> (2x-3)(2x-3)=0
According to the zero product factor....
2x-3 = 0
=> x = 3/2
Similarly others can be solved
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