1. -10x^2 - 7x +3
2. y-6=y^2
3. 2y^2 +y =-11y
4. 2(3x^2-1) = 6x +5x
2006-09-02 15:52:58 · 8 answers · asked by avenger 2 in Science & Mathematics ➔ Mathematics
They are all quadratics. Use the quadratic formula to solve for the two roots of each equation.
x1 = (-b + sqrt(b^2 - 4ac)) / 2a
x2 = (-b - sqrt(b^2 - 4ac)) / 2a
1) a = -10, b= -7, c = 3
2) a = 1, b = -1, c = 6
3) a = 2, b = 13, c = 0
4) a = 6, b = -11, c = -2
although 4) seems odd in that 6x + 5x may be something else.
Once you have calculated the two roots x1,x2, then that factored form is
a(x-x1)(x-x2)
roots x1,x2 may be
case 1) real & distingct (b^2 - 4ac) > 0
case 2) double (b^2 - 4ac) = 0
case 3) conjugate pair (b^2 - 4ac) < 0
2006-09-02 15:55:29 · answer #1 · answered by none2perdy 4 · 0⤊ 0⤋
-10x^2 - 7x +3 =
-10x^2 - 10x +3x + 3 =
-10x(x + 1) + 3(x + 1) =
(-10x + 3)(x + 1)
At 2, 3 and 4 the question is peculiar.
2006-09-03 06:06:29 · answer #2 · answered by Thermo 6 · 0⤊ 0⤋
1. (-10x + 3) (x + 1)
2. not factorable
3. 2y (y + 6)
4. (6x + 1) (x - 2)
2006-09-03 00:47:55 · answer #3 · answered by Lin 2 · 0⤊ 0⤋
1. -10x² - 7x +3
-10x² -10x +3x +3
(-10x² - 10x) + (3x + 3)
-10x(x+1)+3(x+1)
(x+1)(3-10x)
2. y²-y+6=0
This is not factorable anymore.
3. 2y²+12y=0
2y(y+6)=0
4. 6x²-2=11x
6x²-11x-2=0
6x²-12x+x-2=0
(6x²-12x)+(x-2)=0
6x(x-2)+(x-2)=0
(x-2)(6x+1)=0
2006-09-02 23:12:24 · answer #4 · answered by fictitiousness ;-) 2 · 0⤊ 0⤋
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2006-09-03 00:36:55 · answer #5 · answered by Kit 3 · 0⤊ 0⤋
1 (5x-2)(2x-1)
is this your homework lol
2006-09-02 23:01:56 · answer #6 · answered by Jun 2 · 0⤊ 0⤋
You need to g back and really **learn** how
(ax+b)*(cx+d) = a*c*x² + (a*d+b*c)*x +b*d and why it works that way.
Doug
2006-09-02 23:07:38 · answer #7 · answered by doug_donaghue 7 · 0⤊ 0⤋
Don't know the problem sorry.
2006-09-02 22:58:43 · answer #8 · answered by ? 6 · 0⤊ 0⤋