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secondary 3 | E Maths
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The answer for the 3rd one is ‘all real numbers’. Please explain to how that is the answer. Thank you
Date Posted: 2 years ago
Views: 213
In the quadratic formula x = -b +-..., the term inside the square root is
b² - 4ac.
The square root is actually the culprit to why some equations do not have solutions. This is because the square root of negative numbers do not have real values.
Now, in this case, the equation 3x² + kx - 1 = 0 takes the form ax² + bx + c = 0 where a = 3. b = k and c = -1. Inside the square root, the value of the expression (which is b² - 4ac) is
k² - 4(3)(-1)
which simplifies to
k² + 12.
In short, the term inside the square root is k² + 12. Upon closer inspection, we notice that k² is a squared expression, and its property is such that its lowest value is 0 (found when k = 0). Since the lowest possible value of k² is 0 (when k = 0), the lowest possible value of k² + 12 is correspondingly 12. And in fact, this is only the minimum value; it can go higher than 12.
As such, the term inside the square root will be positive, and the value of the square root will therefore be positive as well. Then, the "+-" in the formula will give rise to two different values for x. Hence, we always have two solutions to the equation 3x² + kx - 1 = 0.
b² - 4ac.
The square root is actually the culprit to why some equations do not have solutions. This is because the square root of negative numbers do not have real values.
Now, in this case, the equation 3x² + kx - 1 = 0 takes the form ax² + bx + c = 0 where a = 3. b = k and c = -1. Inside the square root, the value of the expression (which is b² - 4ac) is
k² - 4(3)(-1)
which simplifies to
k² + 12.
In short, the term inside the square root is k² + 12. Upon closer inspection, we notice that k² is a squared expression, and its property is such that its lowest value is 0 (found when k = 0). Since the lowest possible value of k² is 0 (when k = 0), the lowest possible value of k² + 12 is correspondingly 12. And in fact, this is only the minimum value; it can go higher than 12.
As such, the term inside the square root will be positive, and the value of the square root will therefore be positive as well. Then, the "+-" in the formula will give rise to two different values for x. Hence, we always have two solutions to the equation 3x² + kx - 1 = 0.
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If you take A-math, you can easily use discriminant to solve. For E-math, you have to realise that this quadratic curve is U-shaped and therefore has a minimum point. If the minimum point lies above the x-axis, then it means that there is no solution, if it lies on the x-axis -- 1 solution, and if it lies below the x-axis -- 2 solutions. So the solution is just to complete the square and get the coordinates of the min point and prove where it lies on the xy-plane.
Date Posted:
2 years ago
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