Question 1 :
Construct a quadratic equation with roots 7 and −3.
Solution :
Let α and β be the roots of the required quadratic equation
α = 7, β = -3
General form of a quadratic quadratic equation :
x2 - (α + β)x + α β = 0
x2 - (7 + (-3))x + 7(-3) = 0
x2 - 4x - 21 = 0
Question 2 :
A quadratic polynomial has one of its zeros 1 + √5 and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution :
Let α = 1 + √5, β be the roots of the required quadratic equation
p(x) = x2 - (1 + √5 + β)x + (1 + √5)β
p(1) = 12 - (1 + √5 + β)1 + (1 + √5)β
2 = 1 - 1 - √5 - β + β + √5 β
2 + √5 = √5 β
β = (2 + √5)/√5
α + β = (1 + √5) + (2 + √5)/√5
= [√5(1 + √5) + (2 + √5)]/√5
= (√5 + 5 + 2 + √5)/√5
= (7 + 2√5)/√5
= (7√5 + 10)/5
αβ = (1 + √5) ⋅ (2 + √5)/√5
Question 3 :
If α and β are the roots of the quadratic equation x2 + √2x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β .
Solution :
General form of a quadratic quadratic when α and β are the roots of the equation :
x2 - (α + β)x + α β = 0
Now, let us find sum and product of roots of the quadratic equation
x2 + √2x + 3 = 0
α + β = -√2/1 = - √2
α β = 3/1 = 3
here α = 1/α and β = 1/β
x2 - (1/α + 1/β)x + (1/α)(1/β) = 0
x2 - ((α + β)/α β)x + (1/αβ) = 0
x2 - ((- √2)/3)x + (1/3) = 0
3x2 + √2x + 1 = 0
Hence the required quadratic equation is 3x2 + √2x + 1 = 0.
Question 4 :
If one root of k(x − 1)2 = 5x − 7 is double the other root, show that k = 2 or −25.
Solution :
k(x − 1)2 = 5x − 7
k(x2 - 2x + 1) = 5x − 7
kx2- 2kx - 5x + k + 7 = 0
kx2- x(2k + 5) + (k + 7) = 0
If one root is α, then the other root β = 2α
α + β = -b/a α + 2α = (2k + 5)/k 3α = (2k + 5)/k α = (2k + 5)/3k ----(1) |
αβ = c/a α (2α) = (k + 7)/k 2α2 = (k + 7)/k --(2) |
Applying (1) in the second equation, we get
2((2k + 5)/3k)2 = (k + 7)/k
(8K2 + 40K + 50) / 9k2 = (k + 7)/k
(8K2 + 40K + 50)/9K = (k + 7)
8K2 + 40K + 50 = 9K(K + 7)
8K2 + 40K + 50 = 9K2 + 63k
9K2 - 8K2 + 63k - 40k - 50 = 0
K2+ 23k - 50 = 0
(k + 25) (k - 2) = 0
k + 25 = 0 k - 2 = 0
k = -25 k = 2
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