Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
α/β + 2(1/α + 1/β) + 3αβ = 8
Hence the value of α/β + 2(1/α + 1/β) + 3αβ is 8.
Question 13. If the squared
difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, find the value of p.
Solution:
Let as assume that the two zeroes of the polynomial are α and β.
Given that,
f(x) = x2 + px + 45
Now,
Sum of the zeroes = α + β = – p
Product of the zeroes = α × β = 45
therefore,
(α + β)2 – 4αβ = (-p)2 – 4 x 45 = 144
(-p)2
= 144 + 180 = 324
p = √324
Hence the value of p will be either 18 or -18.
Question 14. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – px + q, prove that [(α2 / β2) + (β2 / α2)] = [p4/q2] – [4p2/q] + 2
Solution:
Given that,
α and β are the roots of the quadratic polynomial.
f(x) =
x2 – px + q
Now,
Sum of the zeroes = p = α + β
Product of the zeroes = q = α × β
therefore,
LHS = [(α2 / β2) + (β2 / α2)]
= [(α^4 + β4) / α2.β2]
= [((α+ β)^2 – 2αβ)2 + 2(αβ)2] / (αβ)2
= [((p)2 – 2q)2 + 2(q)2] / (q)2
= [(p4 + 4q2 –
4pq2) – 2q2] / q2
= (p4 + 2q2 – 4pq2) / q2 = (p/q)2 – (4p2/q) + 2
LHS = RHS
Hence, proved.
Question 15. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.
Solution:
Given that,
α and β are the zeroes
of the quadratic polynomial
f(x) = x2 – p(x + 1)– c
Now,
Sum of the zeroes = α + β = p
Product of the zeroes = α × β = (- p – c)
therefore,
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
= (− p – c) + p + 1
= 1 – c = RHS
therefore, LHS = RHS
Hence proved.
Question 16. If α and β are the zeroes of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic
polynomial having α and β as its zeroes.
Solution:
Given that,
α + β = 24 ——(i)
α – β = 8 ——(ii)
By solving the above two equations, we will get
2α = 32
α = 16
put the value of α in any of the equation.
Let we substitute it in (ii) and we will get,
β = 16 – 8
β = 8
Now,
Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24
Product of the zeroes = αβ = 16 × 8 = 128
Then, The quadratic polynomial = x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128
Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128
Question 17. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 1, find a quadratic polynomial whose zeroes are 2α/β and 2β/α.
Solution:
Given that,
f(x) = x2 – 1
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = – 1
therefore,
Sum of the zeroes of the new polynomial
= [(2α2 + 2β2)] / αβ
= [2(α2 + β2)] / αβ
= [2((α + β)2 – 2αβ)] / αβ = 4/(-1)
After substituting the value of the sum and products of the zeroes we will get,
As given in the question,
Product of the zeroes
= (2α)(2β) / αβ = 4
Hence, the quadratic polynomial is
x2 – (sum of the zeroes)x + (product of the zeroes)
= kx2 – (−4)x + 4x2 –(−4)x + 4
Hence, the required quadratic polynomial is f(x) = x2 + 4x + 4
Question 18. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 3x – 2, find a quadratic polynomial whose zeroes are 1/(2α + β) and 1/(2β + α).
Solution:
Given that,
f(x) = x2 – 3x – 2
Sum of the zeroes = α + β = 3
Product of the zeroes = αβ = – 2
therefore,
Sum of the zeroes of the new polynomial
= 1/(2α + β) + 1/(2β + α)
= (2α + β + 2β + α) / (2α + β)(2β + α)
= (3α + 3β) / (2(α2 + β2) + 5αβ)
= (3 x 3) / 2[2(α + β)2 – 2αβ + 5 x (-2)]
= 9 / 2[9-(-4)]-10 = 9/16
Product of zeroes = 1/(2α + β) x 1/(2β + α)
= 1 / (4αβ + 2α2 + 2β2
+ αβ)
= 1 / [5αβ + 2((α + β)2 – 2αβ)]
= 1 / [5 x (-2) + 2((3)2 – 2 x (-2))] = 1/16
therefore, the quadratic polynomial is,
x2– (sum of the zeroes)x + (product of the zeroes)
= (x2 + (9/16)x +(1/16))
Hence, the required quadratic polynomial is (x2 + (9/16)x +(1/16)).
Question 19. If α and β are the zeroes of the quadratic polynomial f(x) = x2
+ px + q, form a polynomial whose zeroes are (α + β)2 and (α – β)2.
Solution:
Given that,
f(x) = x2 + px + q
Sum of the zeroes = α + β = -p
Product of the zeroes = αβ = q
therefore,
Sum of the zeroes of new polynomial = (α + β)2 + (α – β)2
= (α + β)2 + α2 + β2 – 2αβ
= (α + β)2 + (α + β)2 – 2αβ –
2αβ
= (- p)2 + (- p)2 – 2 × q – 2 × q
= p2 + p2 – 4q = p2 – 4q
Product of the zeroes of new polynomial = (α + β)2 x (α – β)2
= (- p)2((- p)2 – 4q)
= p2 (p2–4q)
therefore, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – (2p2 – 4q)x +
p2(p2 – 4q)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2p2 –4q) x + p2(p2 – 4q)).
Question 20. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 2x + 3, find a polynomial whose roots are:
(i) α + 2, β + 2
(ii) [α-1] / [α+1], [β-1] / [β+1]
Solution:
Given that,
f(x) = x2 – 2x + 3
Sum of the zeroes = α + β = 2
Product of the zeroes = αβ = 3
(i) Sum of the zeroes of new polynomial = (α + 2) + (β + 2)
= α + β + 4 = 2 + 4 = 6
Product of the zeroes of new polynomial = (α + 1)(β + 1)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4 = 3 + 2(2) + 4 = 11
therefore, quadratic polynomial is :
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – 6x +11
Hence, the required quadratic polynomial is f(x) = k(x2 – 6x + 11).
(ii) Sum of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= [(α-1)(β+1) + (β-1)(α+1)] / (α+1)(β+1)
= [αβ + α – β – 1 + αβ – α + β – 1)] / (α+1)(β+1)
= (3-1+3-1) / (3+1+2) = 2/3
Product of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= 26 = 13(2/6) = 1/3
therefore, the quadratic
polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – (2/3)x + (1/3)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2/3)x + (1/3))
Question 21. If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate:
(i) α – β
(ii) 1/α – 1/β
(iii) 1/α + 1/β
– 2αβ
(iv) α2β + αβ2
(v) α4 + β4
(vi) 1/(aα + b) + 1/(aβ + b)
(vii) β/(aα + b) + α/(aβ + b)
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
Solution:
Given that,
f(x) = ax2 + bx + c
Sum of the zeroes of polynomial = α + β = -b/a
Product of
zeroes of polynomial = αβ = c/a
Since, α + β are the zeroes of the given polynomial therefore,