1) a) Solve the quadratic equation ( 3 - y/4 ) ( 3 - y/4 ) = 26, leaving the radical
sign ' √ ' in the answer .
b) Form a quadratic equation in x whose roots are half the values of the roots
of ( 3 - y/4 ) ( 3 - y/4 ) = 26 respectively .
2) If α and β are the real roots of the equation 3x^2 + ( k - 1 )x - 12 = 0 and
α^2 + β^2 = 9 , find all the possible values of k .
3) Given that α and β are the roots of the equation 2x^2 - 4x - 5 = 0, where α > β.
Find the value of each of the following expressions without solving the
equation. (leave the radical sign ' √ ' in the answer if necessary .)
a) ( β/( 1-α ) ) ( α/( 1-β ) )
b) α - β
c) ( α-2β )/α + ( β - 2α )/β
4) Given that α and β are the roots of the equation x^2 + 3x - 8 = 0 .
From a quadratic equation whose roots are
a) 2 + (1/α) and 2 + (1/β)
b) α^2 + 1 and β^2 +1
c) α^2 - β and β^2 - α
5) If one root of the equation x^2 - (2k - 5)x + k^2 - 5k + 56/9 = 0 , where k is a
constant , is twice the other root,
a) prove that one root is (2k - 5)/3
b) find the two possible values of k
F4 Maths (Quadratic) β/α
2012-10-26 5:44 am
回答 (1)
2012-10-27 9:48 am
✔ 最佳答案
1a) ( 3 - y/4 ) ( 3 - y/4 ) = 263 - y/4 = ± √26
y/4 = 3 ± √26
y = 12 ± 4√26b) x = y/2 , i.e. y = 2x The required equation is
( 3 - 2x/4 ) ( 3 - 2x/4 ) = 26
(3 - x/2)² - 26 = 0
x²/4 - 3x + 9 - 26 = 0
x² - 12x - 68 = 0
2) α² + β² = 9
(α + β)² - 2αβ = 9
( - (k - 1) / 3 )² - 2(- 12 / 3) = 9
(k - 1)² / 9 + 8 = 9
(k - 1)² - 9 = 0
(k - 1 - 3)(k - 1 + 3) = 0
k = 4 or k = - 2
3a) ( β / (1 - α) ) ( α / (1 - β) )
= αβ / ( 1 - (α + β) + αβ )
= (-5/2) / ( 1 - (4/2) + (-5/2) )
= 5 / 7b)α - β
= √( α - β )² for α > β.
= √( (α + β)² - 4αβ )
= √( (4/2)² - 4(-5/2) )
= √14c)(α - 2β) / α + (β - 2α) / β
= 1 - 2β/α + 1 - 2α /β
= 2 ( 1 - (β/α + α /β) )
= 2 ( 1 - (α² + β²) / (αβ) )
= 2 ( 1 - ((α + β)² - 2αβ) / (αβ) )
= 2 ( 1 - ( (4/2)² - 2(-5/2) ) / (-5/2) )
= 46/5
4a) Sum of roots
= 2 + 1/α + 2 + 1/β
= 4 + (α + β) / (αβ)
= 4 + (- 3) / (- 8)
= 35/8Product of roots
= (2 + 1/α) (2 + 1/β)
= 4 + 2(1/α + 1/β) + 1/(αβ)
= 4 + 2(3/8) + 1/(-8)
= 37/8The required equation is x² - (35/8)x + 37/8 = 0 ,
i.e. 8x² - 35x + 37 = 0
b)Sum of roots
= α² + 1 + β² + 1
= (α + β)² - 2αβ + 2
= (- 3)² - 2(- 8) + 2
= 27Product of roots
= (α² + 1) (β² + 1)
= (αβ)² + α² + β² + 1
= (αβ)² + (α + β)² - 2αβ + 1
= (- 8)² + (- 3)² - 2(- 8) + 1
= 90The required equation is x² - 27x + 90 = 0.
c)Sum of roots
= α² - β + β² - α
= (α + β)² - 2αβ - (α + β)
= (- 3)² - 2(- 8) - (- 3)
= 28 Product of roots
= (α² - β) (β² - α)
= (αβ)² - (α³ + β³) + αβ
= (αβ)² - (α + β)(α² - αβ + β²) + αβ
= (αβ)² - (α + β)( (α + β)² - 3αβ ) + αβ
= (- 8)² - ( - 3)( ( - 3)² - 3(- 8) ) + (- 8)
= 155The required equation is x² - 28x + 155 = 0.
5a) Let 2α = β :Sum of roots
α + 2α = 2k - 5
α = (2k - 5)/3b)Product of roots
(2k - 5)/3 * 2(2k - 5)/3 = k² - 5k + 56/9
2(2k - 5)² = 9k² - 45k + 56
8k² - 40k + 50 = 9k² - 45k + 56
k² - 5k + 6 = 0
(k - 2) (k - 3) = 0
k = 2 or k = 3.
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