问题11。如果α和β是二次多项式f(x)= 6x 2 + x – 2的零,则求出(α/β)+(β/α)的值
解决方案:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = -1/6,
Product of the zeroes =α × β = -1/3.
Now,
(α/β) +(β/α) = (α2 + β2) – 2αβ / αβ
Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
= -25/12
Hence the value of (α/β) +(β/α) is -25/12.
问题12。如果α和β是二次多项式f(x)= 6x 2 + x – 2的零,则求出α/β+ 2(1 /α+ 1 /β)+3αβ的值
解决方案:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = 6/3
Product of the zeroes = α × β = 4/3
Now,
α/β + 2(1/α + 1/β) + 3αβ = [(α2 + β2) / αβ] + 2(1/α + 1/β) + 3αβ
[ ((α + β)2 – 2αβ) / αβ] + 2(1/α + 1/β) + 3αβ
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.
问题13。如果二次多项式f(x)= x 2 + px + 45的零的平方差等于144,则找到p的值。
解决方案:
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.
问题14.如果α和β是二次多项式f(x)的= X 2的零- PX + Q,证明[(α2 /β2)+(β2 /α2)] = [P 4 / q 2 ] – [4p 2 / q] + 2
解决方案:
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.
问题15.如果α和β是二次多项式f(x)= x 2 – p(x + 1)– c的零,则表明(α+ 1)(β+ 1)= 1 – c。
解决方案:
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.
问题16。如果α和β是二次多项式的零,使得α+β= 24并且α–β= 8,则找到一个以α和β为零的二次多项式。
解决方案:
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
问题17 。如果α和β是二次多项式f(x)= x 2 – 1的零,则找到一个零为2α/β和2β/α的二次多项式。
解决方案:
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
问题18:如果α和β是二次多项式的零,则f(x)= x 2 – 3x – 2,找到一个零次为1 /(2α+β)和1 /(2β+α)的二次多项式。
解决方案:
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)).
问题19. i fα和β是二次多项式f(x)= x 2 + px + q的零,形成一个零为(α+β) 2和(α–β) 2的多项式。
解决方案:
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)).
问题20.如果fα和β是二次多项式f(x)= x 2 – 2x + 3的零,则求根的多项式为:
(i)α+ 2,β+ 2
(ii)[α-1] / [α+ 1],[β-1] / [β+ 1]
解决方案:
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))
问题21.如果α和β是二次多项式f(x)= ax 2 + bx + c的零,则求值:
(i)α–β
(ii)1 /α– 1 /β
(iii)1 /α+ 1 /β–2αβ
(ⅳ)α2 +β2αβ
(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/α]
解决方案:
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,
(i) α – β
The two zeroes of the polynomials are :
= [√-b+b2-4ac]/2a – ([-b+√(b2-4ac)]/2a)
= [-b+√(b2-4ac) + b+√(b2-4ac)] / 2a
= √(b2-4ac) / a
(ii) 1/α – 1/β
= (β-1) / αβ = -(α-β)/αβ ——-(1)
From above question as we know that,
α-β = √(b2-4ac) / a
and,
αβ = c/a
Put the values in (i) and we will get,
= -[(√(b2-4ac))/c]
(iii) (1/α) + (1/β) – 2αβ
= (α+β)/αβ – 2αβ ———- (i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= (-b/a x a/c – 2c/a) = -[b/c + 2c/a]
(iv) α2β + αβ2
= αβ(α + β) ——–(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/ a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= c/a(-b/a) = -bc/a^2
(v) α4 + β4
= (α2 + β2)2 – 2α2β2
= ((α + β)2 – 2αβ)2 – (2αβ)2 ———(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it in (i), we will get
= [(-b/a) -2(c/a)]2 – [2(c/a)2]
= [(b2 -2(ac)) / a2]2 – [2(c/a)2]
= [(b2 – 2ac)2 – 2a2 c2] / a4
(vi) 1/(aα + b) + 1/(aβ + b)
= (aβ + b + aα + b) / (aα + b)(aβ + b)
= (a(α + β) + 2b) / (a2 x αβ + abα + abβ + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= b / (ac – b2 + b2) = b/ac
(vii) β/(aα + b) + α/(aβ + b)
= [β(aβ + b) + α(aα + b)] / (aβ + b)(aα + b)
= [aα2 + bβ2 + bα + bβ] / (a2 x (c/a) + ab(α+β) + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= a[(α+β)2 – b(α+β)] / ac
= a[b2/a – 2c/a] – b2/a
= a[(b2 – 2c – b2)/a] / ac
= (b2 – 2c – b2) / ac = -2/a
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
= a[(α2 + β2) / αβ] + b[(α2+β2)/αβ]
= a[(α+β)2 – 2αβ] + b((α+β)2 – 2αβ) / αβ
Since,
Sum of the zeroes of polynomial= α + β = – b/a
Product of zeroes of polynomial= αβ = c/a
After putting it, we will get
= a[(-ba)2 – 3x(c/a)] + b((-b/a)2 – 2(c/a)) / (c/a)
= [(-b2a2/a2c)+(3bca2/a2)+(b/a)2 – (2bca2/a2c)] = b