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secondary 3 | E Maths
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How to find the solution for x^2-2x-5
Date Posted: 3 years ago
Views: 248
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In your working you have solved x² - 2x - 3 = 0
That's the intersection between the curve y = x² - 2x - 3 and horizontal line y = 0
Now you want to solve x² - 2x - 5 = 0
x² - 2x - 5 = 0
x² - 2x - 5 + 2 = 0 + 2
x² - 2x - 3 = 2
This is the same as solving for the intersection between the curve y = x² - 2x - 3 and horizontal line y = 2
So draw the line y = 2 and find the intersection points between it and the curve
You should get x = -1.45 or x = 3.45 (3s.f)
(Exact answer is x = 1 - √6 or x = 1 + √6 when quadratic formula is used)
That's the intersection between the curve y = x² - 2x - 3 and horizontal line y = 0
Now you want to solve x² - 2x - 5 = 0
x² - 2x - 5 = 0
x² - 2x - 5 + 2 = 0 + 2
x² - 2x - 3 = 2
This is the same as solving for the intersection between the curve y = x² - 2x - 3 and horizontal line y = 2
So draw the line y = 2 and find the intersection points between it and the curve
You should get x = -1.45 or x = 3.45 (3s.f)
(Exact answer is x = 1 - √6 or x = 1 + √6 when quadratic formula is used)
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