Solving quadratic equations by factoring
1) 9x^2 - 81= 0
2)y^2 + 24y = -144
3) 20y^2 -13y + 2 =0
4) 6x^2 + 7x = -2
5) y^3 - 15y = -2y^2
6) 4x^2 - 64 = 0
2006-10-15 08:08:07 · 8 answers · asked by Stephanie B 1 in Science & Mathematics ➔ Mathematics
Set the factors to equal zero after you are done factoring it out.
I hope this helps.
1)9x^2-81=0
9x^2=81
x^2=9
x=+/-3
2) y^2+24y=-144
y^2+24y+144=0
(y+12)(y+12)=0
y= -12
3)20y^2-13y+2=0
(5y -2)(4y-1)
y= (2/5, 1/4)
4)6x^2+7x=-2
6x^2+7x+2=0
(3x+ 2)(2x+1)=0
x= (-3/2, -1/2)
5)y^3-15y= -2y^2
y^3+2y^2-15y=0
(y^2-3y)(y+5)=0
y=(-3,-5)
6) 4x^2-64=0
4x^2=64
x^2=16
x= (-4, 4)
2006-10-15 08:23:11 · answer #1 · answered by Answerer17 6 · 0⤊ 0⤋
1) 9x^2 - 81
(3x-9)(3x+9)
3x - 9 = 0 and 3x+9 = 0
3x = 9 and 3x = -9
So, x = 3 and x = -3
2) y^2 + 24y = -144
y^2 + 24y +144 = 0
(y+12)(y+12) = 0
y = -12
3) 20y^2 -13y +2 = 0
(5y-2)(4y-1) = 0
5y-2 = 0 and 4y-1 = 0
So
y = 2/5 and y = 1/4
4) 6x^2 + 7x=-2
6x^2+7x+2=0
(3x+2)(2x+1)=0
3x+2 = 0 and 2x+1 =0
x = -2/3 and x= -1/2
5) y^3-2y^2-15y = 0
(y^2-5y)(y+3) = 0
y^2-5y=0 and y+3 = 0
We still have to factor the first equation
(y)(y-5)=0 and y+3=0
y=0, y=5 and y=-3
6) 4x^2-64 =0
x^2 = 16
x = 4 and -4
Regards,
Mysstere
PS--The person who answered before me needs to check their answers. Their 4th answer is incorrect b/c if you multiply the first terms with the FOIL method, they only get 3x^2, and not the 6x^2 as in your question.
2006-10-15 15:21:06 · answer #2 · answered by mysstere 5 · 0⤊ 0⤋
1) 9x^2-81=0 9x^2-81+81=0+81 9x^2=81
9x^2/9=81/9 x^2=9 sqrt(x^2)=sqrt(9) x=3
2) y^2+24y= -144 y^2+24y+144=0
(y+12)(y+12)=0 y=-12
3) 20y^2 -13y+2=0 (5y-2)(4y-1)=0 (5y-2)=0
y=2/5 (4y-1)=0 y=1/4 so y=2/5 or 1/4
4) 6x^2+7x=-2 6x^2+7x+2=0 (3x+1)(2x+2)=0
(3x+1)=0 x=-1/3 (2x+2)=0 x=-1 so
x=-1/3 or -1
5) y^3-15y=-2y^2 y^3+2y^2-15y=0 y(y^2+2y-15)
=0 y(y-3)(y+5)=0 so y=0 or 3 or -5
6) 4x^2-64=0 (4x^2-64)/4=0/4 x^2-16=0
(x+4)(x-4)=0 so x= -4 or 4
2006-10-15 15:24:28 · answer #3 · answered by Rex 4 · 0⤊ 0⤋
1) x= +3 or -3
2) y= -144 or -168
3) not clear
4) x= 2 or -5/6
5) y= square root of -15 or 2y^2
6) x= +4 or -4
2006-10-15 15:41:37 · answer #4 · answered by obinna p 1 · 0⤊ 0⤋
1) (3x+9)(3x-9)= 0 x=3,-3
2) (y+12)^2= 0 y=-12
3)(5y - 2)(4y -1)=0 y=2/5,1/4
4)(3x + 1)(x + 2)=0 x=-1/3,-2
5)y(y^2-2y-15)= 0 y(y + 3)(y- 5)= 0 y=0,-3,5
6)(2x - 8)(2x-8) = 0 x=4,-4
#4 below is correct as pointed out
2006-10-15 15:20:31 · answer #5 · answered by want2no 5 · 0⤊ 0⤋
1) X=+ or - 3
2) y=-12
3) y=8 and 5
4) x+-4 and -3
5) y=0 and -5 and 3
6) x = + or - 4
2006-10-15 15:12:17 · answer #6 · answered by Faraz S 3 · 0⤊ 0⤋
9x*2 - 81 = 0 now add 81 to both sides of the equation
9x*2=81 divide both sides by 9
x*2=9 now take the square root of both sides
x=3
See its' that simple, I am sure you can do the others now!
2006-10-15 15:17:24 · answer #7 · answered by mystic_golfer 3 · 0⤊ 0⤋
i know how to do them but too many
2006-10-15 15:16:12 · answer #8 · answered by 7 · 0⤊ 0⤋