Answer:
- 20. The vertex is (2/3, 14/3) | p = 3, q = -2/3 and r = 14/3
- 21. 20x² + 2x - 3 = 0
Explanation:
20.
Given
To find
- The least value of the y and the corresponding value of x
- Constants p, q and r such that 3x² - 4x + 6 = p(x + q)² + r
Solution
The given is the parabola with positive a coefficient, so it opens up and the minimum point its vertex.
The vertex has x = -b/2a and corresponding y- coordinate is found below:
- x = - (- 4)/2*3 = 2/3, and
- y = 3(2/3)² - 4(2/3) + 6 = 4/3 - 8/3 + 6 = 14/3
- So the vertex is (2/3, 14/3)
The vertex form of the line has the equation:
- y = a(x - h)² + k, where (h, k) is the vertex
Plugging in the values:
Comparing with p(x + q)² + r, to find out that:
- p = 3, q = -2/3 and r = 14/3
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21.
(i) α and β are the roots of: ax² + bx + c = 0
Show that:
- α + β = -b/a and αβ = c/a
Solution
Knowing the roots, put the equation as:
- (x - α)(x - β) = 0
- x² - αx - βx + αβ = 0
- x² - (α+β)x + αβ = 0
Comparing this with the standard form:
Divide by a to make the constants of x² same:
Now comparing the constants:
- - (α+β) = b/a ⇒ α+β = - b/a
- αβ = c/a
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(ii)
Given
- α and β are the roots of: 3x² - x - 5 = 0
To Find
- The equation with roots 1/2α and 1/2β
Solution
The sum and the product of the roots:
- α + β = -b/a = 1/3
- αβ = c/a = -5/3
The equation is:
- (x - 1/2α)(x - 1/2β) = 0
- x² - (1/2α + 1/2β)x + 1/(2α)(2β) = 0
- x² - (α + β)/(2αβ)x + 1/4αβ = 0
- x² - (1/3)/(2(-5/3))x + 1/(4(-5/3)) = 0
- x² + 1/10x - 3/20 = 0
- 20x² + 2x - 3 = 0