x2 - 17x + 60 = 0
x2 - 16x + 64 = 0
x2 + 4x + 0 = 0
x2 + 8x + 15 = 0
Four algebra problems. . .?
2009-07-13 3:48 pm
回答 (8)
2009-07-13 4:05 pm
✔ 最佳答案
x^2 - 17x + 60 = 0 (x-5)(x-12)=0
x-5=0 x-12=0
x=5 x=12
x=5, 12 answer//
x^2 - 16x + 64 = 0
(x-8)(x-8)=0
x-8=0 x-8=0
x=8 x=8
x=8,8 answer//
x^2 + 4x + 0 = 0
x^2+4x=0
x(x+4)=0
x=0 x+4=0
........x=-4
x=0,-4 answer//
x^2+8x+15 = 0
(x+3)(x+5)=0
x+3=0 x+5=0
x=-3 x=-5
x=-3,-5 answer//
2009-07-13 4:26 pm
x² - 17x + 60 = 0
x² - 17/2x = - 30 + (- 17/2)²
x² - 17/2x = - 120/4 + 289/4
(x - 17/2)² = 169/2
x - 17/2 = +/- 13/2
x = 13/2 + 17/2, x = 30/2, x = 30
x = - 13/2 + 17/2, x = 4/2, x = 2
x² - 16x + 64 = 0
x² - 8x = - 64 + (- 8)²
x² - 8x = - 64 + 64
(x - 8)² = 0
x - 8 = 0
x = 8
x² + 4x + 0 = 0
x² + 2x = 2²
x² + 2x = 4
(x + 2)² = 4
x + 2 = +/- 2
x = 2 - 2, x = 0
x = - 2 - 2, x = - 4
x² + 8x + 15 = 0
x² + 4x = - 15 + 4²
x² + 4x = - 15 + 16
(x + 4)² = 1
x + 4 = 1
x = 1 - 4, x = - 3
x = - 1 - 4, x = - 5
x² - 17/2x = - 30 + (- 17/2)²
x² - 17/2x = - 120/4 + 289/4
(x - 17/2)² = 169/2
x - 17/2 = +/- 13/2
x = 13/2 + 17/2, x = 30/2, x = 30
x = - 13/2 + 17/2, x = 4/2, x = 2
x² - 16x + 64 = 0
x² - 8x = - 64 + (- 8)²
x² - 8x = - 64 + 64
(x - 8)² = 0
x - 8 = 0
x = 8
x² + 4x + 0 = 0
x² + 2x = 2²
x² + 2x = 4
(x + 2)² = 4
x + 2 = +/- 2
x = 2 - 2, x = 0
x = - 2 - 2, x = - 4
x² + 8x + 15 = 0
x² + 4x = - 15 + 4²
x² + 4x = - 15 + 16
(x + 4)² = 1
x + 4 = 1
x = 1 - 4, x = - 3
x = - 1 - 4, x = - 5
2016-11-06 10:16 pm
4. Divide and rationalize the denominator for sqrt49/sqrt500 A. 7 sqrt 5 OVER 2 B. 7 sqrt 5 OVER 50 C. 7 OVER 2 sqrt5 D. 35/10 sqrt{40 9} = 7 sqrt{500} = 10(sqrt{5}) we've right here: [sqrt{40 9} = 7]/[10(sqrt{5})] we are able to now rationalize the denominator. to attain this, multiply the precise and backside via [10(sqrt{5})]. very final answer is: determination B or 7 sqrt 5 OVER 50
2009-07-13 5:30 pm
Solve for x.
1)
x^2 - 17x + 60 = 0
x^2 - 5x - 12x + 60 = 0
(x^2 - 5x) - (12x - 60) = 0
x(x - 5) - 12(x - 5) = 0
(x - 5)(x - 12) = 0
x - 5 = 0
x = 5
x - 12 = 0
x = 12
∴ x = 5, 12
__________________________
2)
x^2 - 16x + 64 = 0
x^2 - 8x - 8x + 64 = 0
(x^2 - 8x) - (8x - 64) = 0
x(x - 8) - 8(x - 8) = 0
(x - 8)(x - 8) = 0
x - 8 = 0
x = 8
∴ x = 8
__________________________
3)
x^2 + 4x + 0 = 0
x^2 + 4x = 0
x(x + 4) = 0
x = 0
x + 4 = 0
x = -4
∴ x = -4, 0
__________________________
4)
x^2 + 8x + 15 = 0
x^2 + 5x + 3x + 15 = 0
(x^2 + 5x) + (3x + 15) = 0
x(x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
∴ x = -5, -3
1)
x^2 - 17x + 60 = 0
x^2 - 5x - 12x + 60 = 0
(x^2 - 5x) - (12x - 60) = 0
x(x - 5) - 12(x - 5) = 0
(x - 5)(x - 12) = 0
x - 5 = 0
x = 5
x - 12 = 0
x = 12
∴ x = 5, 12
__________________________
2)
x^2 - 16x + 64 = 0
x^2 - 8x - 8x + 64 = 0
(x^2 - 8x) - (8x - 64) = 0
x(x - 8) - 8(x - 8) = 0
(x - 8)(x - 8) = 0
x - 8 = 0
x = 8
∴ x = 8
__________________________
3)
x^2 + 4x + 0 = 0
x^2 + 4x = 0
x(x + 4) = 0
x = 0
x + 4 = 0
x = -4
∴ x = -4, 0
__________________________
4)
x^2 + 8x + 15 = 0
x^2 + 5x + 3x + 15 = 0
(x^2 + 5x) + (3x + 15) = 0
x(x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
∴ x = -5, -3
2009-07-13 3:57 pm
q1. x= 5 or 12 factorise or use the quadratic formula
q2. x= 8 (a repeated root), complete the square is quickest method
q3. x= 0 or -4 Factor x out and solve the linear equation
q4. x= -5 or - 3 same method as q1
q2. x= 8 (a repeated root), complete the square is quickest method
q3. x= 0 or -4 Factor x out and solve the linear equation
q4. x= -5 or - 3 same method as q1
2009-07-13 3:57 pm
All four of these quadratic equations can be easily factored. Have you tried ?
x^2 + bx + c = 0
Look for factors of 'c' that will combine to 'b'
Be careful with signs.
Have you used the "zero-product property" of multiplication ?
Zero - Product Property states that if the product of two or more factors is zero, then at least one of the factors must be zero.
Go for it. You can do it.
x^2 + bx + c = 0
Look for factors of 'c' that will combine to 'b'
Be careful with signs.
Have you used the "zero-product property" of multiplication ?
Zero - Product Property states that if the product of two or more factors is zero, then at least one of the factors must be zero.
Go for it. You can do it.
2009-07-13 3:54 pm
use the discriminant method to find all roots...
if D<0..then find out the complex conjugate roots...
if D=0..then both roots equal...
if D>0..then both real n distinct...
hope it helps...
if D<0..then find out the complex conjugate roots...
if D=0..then both roots equal...
if D>0..then both real n distinct...
hope it helps...
2009-07-13 3:53 pm
1)
( x - 12)( x- 5 ) = 0 x=12 or x=5
2)
(x-8)^2=0 x=8 twice root
3)
x (x+4)=0 x= 0 or x=-4
4)
(x+5)(x+3)=0 x=-3 or x= -5
( x - 12)( x- 5 ) = 0 x=12 or x=5
2)
(x-8)^2=0 x=8 twice root
3)
x (x+4)=0 x= 0 or x=-4
4)
(x+5)(x+3)=0 x=-3 or x= -5
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