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secondary 3 | A Maths
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Anyone can help me w this ques? Thankuu
Date Posted: 3 years ago
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x^2 + kx = 3-k
x^2 + kx + (k-3) = 0
For real roots/solutions,
b^2 - 4ac >= 0
k^2 - 4(k-3)(1) >= 0
k^2 - 4k + 12 >= 0
By observation, k^2 - 4k + 12 is never less than 0 for all k, hence the equation x^2 + kx = 3-k has real roots for all k.
x^2 + kx + (k-3) = 0
For real roots/solutions,
b^2 - 4ac >= 0
k^2 - 4(k-3)(1) >= 0
k^2 - 4k + 12 >= 0
By observation, k^2 - 4k + 12 is never less than 0 for all k, hence the equation x^2 + kx = 3-k has real roots for all k.
Thankuu
Presentation flaw.
At the b2 - 4ac line, we should not immediately put “>=“ symbol as this question asks us to prove it. By placing it on that line, we are already assuming it to be true.
If the marker is strict, one mark may be deducted for the presentation.
At the b2 - 4ac line, we should not immediately put “>=“ symbol as this question asks us to prove it. By placing it on that line, we are already assuming it to be true.
If the marker is strict, one mark may be deducted for the presentation.
That issue is only the minor one.
More importantly, merely stating 'by observation' will result in loss of marks as well since it is not immediately obvious that the discriminant ≥ 0 for all real k
(It actually IS > 0 for all real k)
Completing the square should be shown to justify this.
x² + kx = 3 - k
x² + kx + k - 3 = 0
Discriminant (b² - 4ac)
= k² - 4(1)(k - 3)
= k² - 4k + 12
= k² - 2(k)(2) + 2² + 8
= (k - 2)² + 8
Since (k - 2)² ≥ 0 for all real values of k (the square of any real value is always 0 or positive) ,
Then (k - 2)² + 8 ≥ 8 for all real values of k.
This in turn means that (k - 2)² + 8 > 0 for all real values of k.
Since the discriminant is always positive for all real values of k, the roots/solutions of the quadratic equation are always real for all real values of k.
Side note : student, realise that we have obtained the vertex form of a parabola equation whereby :
① The coefficient of x² is positive (upward sloping curve, U-shaped)
② The vertex, a minimum point in this case, has coordinates (2,8).
So this means that the minimum value of the discriminant is 8, which is > 0
More importantly, merely stating 'by observation' will result in loss of marks as well since it is not immediately obvious that the discriminant ≥ 0 for all real k
(It actually IS > 0 for all real k)
Completing the square should be shown to justify this.
x² + kx = 3 - k
x² + kx + k - 3 = 0
Discriminant (b² - 4ac)
= k² - 4(1)(k - 3)
= k² - 4k + 12
= k² - 2(k)(2) + 2² + 8
= (k - 2)² + 8
Since (k - 2)² ≥ 0 for all real values of k (the square of any real value is always 0 or positive) ,
Then (k - 2)² + 8 ≥ 8 for all real values of k.
This in turn means that (k - 2)² + 8 > 0 for all real values of k.
Since the discriminant is always positive for all real values of k, the roots/solutions of the quadratic equation are always real for all real values of k.
Side note : student, realise that we have obtained the vertex form of a parabola equation whereby :
① The coefficient of x² is positive (upward sloping curve, U-shaped)
② The vertex, a minimum point in this case, has coordinates (2,8).
So this means that the minimum value of the discriminant is 8, which is > 0
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