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secondary 2 | Maths
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hii, i need help with part (ii)
thanks!
Date Posted: 3 years ago
Views: 175
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x² + px + 8 = 0
Since x = -2 is a solution, sub x = -2 into the equation
(-2)² + p(-2) + 8 = 0
4 - 2p + 8 = 0
12 = 2p
p = 6
So, x² + 6x + 8 = 0
Since x = -2 is a solution, (x+2) must be a factor.
Factorise,
(x + 2)(x + 4) = 0
x + 2 = 0 or x + 4 = 0
x = -2 or x = -4
So, the other solution is x = -4
Alternatively,
x² + 6x + 8 = 0
Now we complete the square.
(x² + 2(3)x + 3²) + 8 - 3² = 0
(x + 3)² - 1 = 0
(x + 3)² = 1
x + 3 = ±√1
x + 3 = 1 or x + 3 = -1
x = -2 or x = -4
Since x = -2 is a solution, sub x = -2 into the equation
(-2)² + p(-2) + 8 = 0
4 - 2p + 8 = 0
12 = 2p
p = 6
So, x² + 6x + 8 = 0
Since x = -2 is a solution, (x+2) must be a factor.
Factorise,
(x + 2)(x + 4) = 0
x + 2 = 0 or x + 4 = 0
x = -2 or x = -4
So, the other solution is x = -4
Alternatively,
x² + 6x + 8 = 0
Now we complete the square.
(x² + 2(3)x + 3²) + 8 - 3² = 0
(x + 3)² - 1 = 0
(x + 3)² = 1
x + 3 = ±√1
x + 3 = 1 or x + 3 = -1
x = -2 or x = -4
Or, you can 'factor out' the (x + 2) first.
x² + 6x + 8 = 0
x² + 2x + 4x + 8 = 0
x(x + 2) + 4(x + 2) = 0
(x + 2)(x + 4) = 0
x + 2 = 0 or x + 4 = 0
x = -2 or x = -4
x² + 6x + 8 = 0
x² + 2x + 4x + 8 = 0
x(x + 2) + 4(x + 2) = 0
(x + 2)(x + 4) = 0
x + 2 = 0 or x + 4 = 0
x = -2 or x = -4
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