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secondary 2 | Maths
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Date Posted: 3 years ago
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Alternatively, since x = 5 is a solution, (x - 5) is a factor of x² - qx + 10.
Let the other solution be x = p
(x - 5)(x - p) = 0
x² - (p+5)x + 5p = 0
Comparing coefficients,
5p = 10
p = 2
The other solution is x = 2.
Comparing coefficients,
p+5 = q
q = 2 + 5 = 7
Let the other solution be x = p
(x - 5)(x - p) = 0
x² - (p+5)x + 5p = 0
Comparing coefficients,
5p = 10
p = 2
The other solution is x = 2.
Comparing coefficients,
p+5 = q
q = 2 + 5 = 7
See 1 Answer
Substitute x = 5 into the equation,
5² - q(5) + 10 = 0
25 - 5q + 10 = 0
35 - 5q = 0
5q = 35
q = 35 ÷ 5 = 7
Next, recall that sum of roots = -b/a
Equation is x² - 7x + 10 = 0
Sum of roots = - (-7)/1 = 7
So, 5 + (the other root) = 7
The other root = 7 - 5 = 2
5² - q(5) + 10 = 0
25 - 5q + 10 = 0
35 - 5q = 0
5q = 35
q = 35 ÷ 5 = 7
Next, recall that sum of roots = -b/a
Equation is x² - 7x + 10 = 0
Sum of roots = - (-7)/1 = 7
So, 5 + (the other root) = 7
The other root = 7 - 5 = 2
Otherwise, you can also factorise the obtained equation.
x = 5 is a solution so (x - 5) must be one of the factors.
x² - 7x + 10 = 0
(x - 5)(x - 2) = 0 (do your cross factorisation)
x - 5 = 0 or x - 2 = 0
x = 5 or x = 2
x = 5 is a solution so (x - 5) must be one of the factors.
x² - 7x + 10 = 0
(x - 5)(x - 2) = 0 (do your cross factorisation)
x - 5 = 0 or x - 2 = 0
x = 5 or x = 2
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