what are the factors for 7x^2-63=0 and for 6x^2=13x+5
2007-02-01 07:24:33 · 7 answers · asked by Jessica R 1 in Science & Mathematics ➔ Mathematics
7x² - 63 = 0
7x²- 63 + 63 = 0 + 63
7x² = 63
7x²/ 7 = 63/7
x² = 63/7
x² = 9
√x² = √9
x = ± 3
The answer is x = ± 3
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method 2
7x² - 63 = 0
7(x² - 9) = 0
7(x + 3)(x - 3)
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Roots
7 = 0
x + 3 = 0
x + 3 - 3 = 0 - 3
x = - 3
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Roots
x - 3 = 0
x - 3 + 3 = 0 + 3
x = 3
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6x² = 13x = 5
6x² - 13x - 5 = 0
6x² - 2x - 15x - 5 = 0
2x(3x+ 1) - 5(3x + 1)
(2x - 5)(3x + 1)
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2007-02-01 07:53:39 · answer #1 · answered by SAMUEL D 7 · 1⤊ 0⤋
7x^2-63=0
transp. -63 to the other side.
7x^2=63 or
x^2=63/7 or x^2=9
Trans. 9 to the other side:
x^2-9=0
x*x-3*3=0
the factors are: (x-3)(x+3)=0
6x^2=13x+5
Transp. 13x+5 to the other side:
6x^2-13x-5=0
Form :ax^2-bx-c+0
with a=6 ;b=13 and c=8
equation of degre second.
Solve for x' and x''.
x'=(-b +or- square rootB) /2a
with B=b^2-4ac
x'=-1/3 and x''=5/2
the factors are:
(3x+1)(2x-5)=0
2007-02-01 07:48:20 · answer #2 · answered by Johnny 2 · 1⤊ 0⤋
obviously 7 can be divided evenly into both terms thus
7x^2-63 = 7(x^2-9)
the term in parentheses factors as (x-3)*(x+3) therefore the factors are 7, (x-3), and (x+3)
6x^2=13x +5
subtract 13x+5 from both sides of equation:
6x^2 -13x -5 =0 this can be done by inspection or trial and error
we want two factors in the form (nx +/ a) and (nx +/- b)
the first part is easy: factor the 6x^2 as 3x and 2x and factor 5 as 1 and 5 and place them in the brackets thus:
(3x + 1)(2x - 5) I do these by trial and error. the minus sign and the factors of 5 must be placed so that the sum of 3x*-5 +2x*1 =13
2007-02-01 07:48:10 · answer #3 · answered by bignose68 4 · 1⤊ 0⤋
7x^2-63=0
what do you notice? 63/7 = 9...handy! you can now factor out the 7 as 7 is divisible by all the terms in the equation...giving:
7(x^2-9)=0
6x^2=13x+5
move everything to one side and keep x^2 positive...giving:
6x^2-13x-5=0
since 6 has factors of : 1, 2, 3, 6...then we need to find the perfect combination for x^2...
we also need to find factors of 5 (since its prime number...we can only use 5 and 1
we need numbers that add up to -13 when 1 number is multiplied with the corresponding number and the other multiplied to the other corresponding number and then the two products added.
now for trial and error:
say we use 1 and 6 as x^2's coefficient:
6x5=30...that is way out of the range...
1x5=5...closer but not close enough if added by one...
now we try 2 and 3
2x5=10 close...
3x1=3
10+3=13...but then...notice a problem? yes...one of them has to be negative...so it will not add up to 13 afterall!
how bout:
3x5=15
2x1=2 <-- make this one negative...
add them together:
15-2=13 <-- wow!
hence...we will get the equation as
(2x-5)(3x-1) = 0
we can also solve for x...
x=5/2 or x=1/3!
2007-02-05 06:54:39 · answer #4 · answered by ChristopheraX 4 · 0⤊ 0⤋
a) 7 will go into both terms so factor it out:
7(x^2 - 9) = 0
(x^2 - 9) is a difference of squares = (x + 3)(x - 3), so
7(x^2 - 9) = 7(x + 3)(x - 3)
Therefore, if 7(x + 3)(x - 3) = 0, then x = 3 or - 3.
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(b.) 6x^2=13x+5, so 6x^2 -13x - 5 = 0
Therefore, (3x + 1)( 2x - 5) = 0 ...
Which means 3x + 1 = 0, so x = -1/3
or 2x - 5 = 0, so x = 5/2
2007-02-01 07:35:20 · answer #5 · answered by Pythagoras 7 · 1⤊ 0⤋
factor out 7
x^2-9=0
x+3 and x-3
6x^2-13x-5=0
6x^2-15x+2x-5
2x-5 or 3x+1
x = 5/2 or -1/3
2007-02-01 07:26:38 · answer #6 · answered by Anonymous · 1⤊ 0⤋
7(x^2-9)=0
7(x+3)(x-3)=0
x=0, -3, and 3
6x^2-13x-5=0
(2x-5)(3x+1)=0
x=5/2 and -1/3
2007-02-01 08:14:44 · answer #7 · answered by Anonymous · 0⤊ 0⤋