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secondary 4 | A Maths
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help pls
Date Posted: 1 month ago
Views: 56
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done
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clear
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can someone check my working pls
Date Posted:
1 month ago
1. you are trying to compare coefficients of ...
5x^2-20x+15-k = 0 ...... (1)
to
x^2-(2ab+1)x+(ab)^2+ab = 0 ...... (2)
note that coefficients of x^2 in (1) & (2) are different, so first you need to divide (1) by 5 throughout, & make coefficients of x^2 the same, before you can compare the other coefficients.
2. why do you want to use 'ab (alpha-beta)' as your variable name? single-letter variable names are less confusing.
3. overall, the method you are using is quite tedious. a simpler way is to remember that quadratic curves are always symmetrical. as k varies, the line of symmetry will remain the same.
consider k=0,
so 5(x-1)(x-3) = 0.
the 2 roots are x=1 & x=3,
line of symmetry is at x=2.
since you are looking for 2 roots that differ by 1, they will be x=1.5 & x=2.5 (0.5 units from line x=1)
substitute x=1.5 or x=2.5 into 5(x-1)(x-3) = k,
gives k = -3.75.
5x^2-20x+15-k = 0 ...... (1)
to
x^2-(2ab+1)x+(ab)^2+ab = 0 ...... (2)
note that coefficients of x^2 in (1) & (2) are different, so first you need to divide (1) by 5 throughout, & make coefficients of x^2 the same, before you can compare the other coefficients.
2. why do you want to use 'ab (alpha-beta)' as your variable name? single-letter variable names are less confusing.
3. overall, the method you are using is quite tedious. a simpler way is to remember that quadratic curves are always symmetrical. as k varies, the line of symmetry will remain the same.
consider k=0,
so 5(x-1)(x-3) = 0.
the 2 roots are x=1 & x=3,
line of symmetry is at x=2.
since you are looking for 2 roots that differ by 1, they will be x=1.5 & x=2.5 (0.5 units from line x=1)
substitute x=1.5 or x=2.5 into 5(x-1)(x-3) = k,
gives k = -3.75.
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