1 Crore+ students have signed up on EduRev. Have you? 
If α,β and γ are the zeros of the polynomial 2x^{3} – 6x^{2} – 4x + 30, then the value of (αβ + βγ + γα) is
We have a 3 degree polynomial here of form
ax^{3} + bx^{2} + cx +d = 0
Polynomial p(x) = 2x^{3} – 6x^{2} – 4x + 30
The roots of this polynomial are α, β, γ
We have formula for products of zeroes taking 2 at a time = αβ + βγ + αγ = c/a
αβ + βγ + αγ = 4/2 = 2
The quadratic polynomial whose sum of zeroes is 3 and product of zeroes is –2 is :
Sum of zeroes = 3
Product of zeroes = 2
Let the quadratic polynomial be ax^{2} + bx + c
Sum of zeroes = 3
b/a = 3/1
So a = 1, b = 3
Product of zeroes = 2
c/a = 2/1
So a = 1, c = 2
Hence the quadratic polynomial will be x^{2}  3x  2
If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is
Given ,g(x)=x+2=0
x=2
P(x)=x^{3}2ax^{2}+16
p(2)=(2^{3}) 2 × a × (2^{2})+16
=8  8a +16
=8  8a
=>8a = 8
=>a=1
If α,β and γ are the zeros of the polynomial f(x) = x^{3} + px^{2} – pqrx + r, then =
If the parabola f(x) = ax^{2} + bx + c passes through the points (–1, 12), (0, 5) and (2, –3), the value of a + b + c is –
f(x)=ax^{2}+bx+c
It is given that f(0)=5, f(−1)=12 and f(2)=−3
f(0)=c=5 ... (i)
f(−1)=a−b+c=12
∴a−b=12−c
∵c=5
∴a−b=12−5 =7 ... (ii)
f(2)=4a+2b+c=−3
⇒4a+2b=−3−c
as c=5
⇒4a+2b=−8
⇒2a+b=−4 ... (iii)
Solving (ii) and (iii), we get
a=1
b=−6
Hence, a=1,b=−6,c=5
Thus, a+b+c =1−6+5 =0.
If a, b are the zeros of f(x) = x^{2} + px + 1 and c, d are the zeros of f(x) = x^{2} + qx + 1 the value of E = (a – c) (b – c) (a + d) (b + d) is –
The given polynomials are
f(x) = x² + px + 1
g(x) = x² + qx + 1
Since a, b are the zeroes of f(x),
a + b =  p ..... (1)
ab = 1 ..... (2)
Since c, d are the zeroes of g(x),
c + d =  q ..... (3)
cd = 1 ..... (4)
Now, E = (a  c) (b  c) (a + d) (b + d)
= {ab  (a + b)c + c²} {ab + (a + b)d + d²}
= (1 + pc + c²) (1  pd + d²) [by (1) & (2)]
= 1  pd + d² + pc  p²cd + pcd² + c²  pc²d + c²d²
= 1  pd + d² + pc  p²(cd) + p(cd)d + c²  pc(cd) + (cd)²
= 1  pd + d² + pc  p² + pd + c²  pc + 1 [by (4)]
= 2 + c² + d²  p²
= 2 + (c + d)²  2(cd)  p²
= 2 + ( q)²  2  p² [by (3) & (4)]
= q²  p²
If α,β are zeros of ax^{2 }+ bx + c then zeros of a3x^{2} + abcx + c^{3} are –
Let α,β be the zeros of the polynomial x^{2} – px + r and be the zeros of x^{2} – qx + r. Then the value
of r is –
Since a and b are roots of x^{2} – px + r = 0
We can say a+b=p....say equation 1 and ab=r
Also a/2 and 2b are roots of x^{2} – qx + r
So, a/2 +2b=q ..... say equation 2 and a/2*2b=ab=r
2(equation1)equation 2 = 2a a/2 +2b2b = 2pq
so 3a/2=2pq
a=2(2pq)/3
Substituting in equation 1
We get b=p4p/3+2q/3=(2qp)/3
So we know ab=r=2(2pq)(2qp)/9
When x^{200} + 1 is divided by x^{2} + 1, the remainder is equal to –
When x^{200 }+ 1 is divided by x^{2} + 1, the remainder is equal to
x²⁰⁰ + 1 = x²⁰⁰ + (x¹⁹⁸  x¹⁹⁸  x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴  x¹⁹⁴  x¹⁹² + x¹⁹² +x¹⁹⁰  x¹⁹⁰ +..................x⁴ + x⁴ + x²  x²  1 + 1 )+ 1
=> x²⁰⁰ + 1 = x²⁰⁰ + x¹⁹⁸  x¹⁹⁸  x¹⁹⁶ + x¹⁹⁶ + x¹⁹⁴  x¹⁹⁴  x¹⁹² + x¹⁹² +x¹⁹⁰  x¹⁹⁰ +..................x⁴ + x⁴ + x²  x²  1 + 1 + 1
=> x²⁰⁰ + 1 = x¹⁹⁸(x² + 1)  x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1)  x¹⁹²(x² +1) + x¹⁹⁰(x² +1)  x¹⁸⁸(x²+1) +..................+ x²(x² + 1)  (x² + 1) + 1 + 1
=> x²⁰⁰ + 1 = x¹⁹⁸(x² + 1)  x¹⁹⁶(x² + 1) + x¹⁹⁴(x² + 1)  x¹⁹²(x² +1) + x¹⁹⁰(x² +1)  x¹⁸⁸(x²+1) +..................+ x²(x² + 1)  1.(x² + 1) + 2
=> x²⁰⁰ + 1 = (x² + 1)(x¹⁹⁸ x¹⁹⁶ + x¹⁹⁴  x¹⁹² + x¹⁹⁰  x¹⁸⁸+..................+ x²  1) + 2
=> when x²⁰⁰ + 1 is divided by x² + 1 the reminder = 2
If a (p + q)^{2} + 2bpq + c = 0 and also a(q + r)^{2} + 2bqr + c = 0 then pr is equal to –
A*(p+q)^2 + 2bpq + c = 0 and a*(p+r)^2 +2bpr + c = 0
=> aq^2 + 2p(a+b)q + ap^2 + c = 0
and ar^2 + 2p(a+b)r + ap^2 + c = 0
=> q and r are the roots of the equation
ax^2 + 2p(a+b)x + ap^2 + c = 0
=> qr = (ap^2 + c) / a.
If a,b and c are not all equal and α and β be the zeros of the polynomial ax^{2} + bx + c, then value of (1 + α + α^{2}) (1 + β + β^{2}) is :
Two complex numbers α and β are such that α + β = 2 and α^{4} + β^{4} = 272, then the polynomial whose zeros
are α and β is –
If 2 and 3 are the zeros of f(x) = 2x^{3} + mx^{2} – 13x + n, then the values of m and n are respectively –
If α,β are the zeros of the polynomial 6x^{2} + 6px + p^{2}, then the polynomial whose zeros are (α + β)^{2} and (α  β)^{2} is
If c, d are zeros of x^{2} – 10ax – 11b and a, b are zeros of x^{2} – 10cx – 11d, then value of a + b + c + d is –
If the ratio of the roots of polynomial x^{2} + bx + c is the same as that of the ratio of the roots of x^{2} + qx + r, then –
If the roots of the polynomial ax^{2} + bx + c are of the form and then the value of (a + b + c)^{2} is–
If α, β and γ are the zeros of the polynomial x^{3} + a_{0}x^{2} + a_{1}x + a_{2}, then (1 – α^{2}) (1 – β^{2}) (1 – γ^{2}) is
If α,β,γ are the zeros of the polynomial x^{3} – 3x + 11, then the polynomial whose zeros are (α + β), (β + γ) and
(γ + α) is –
If α,β,γ are such that α + β + γ = 2, α^{2} + β^{2} + γ^{2} = 6, α^{3} + β^{3} + γ^{3} = 8, then α^{4} + β^{4} + γ^{4} is equal to–
(α+β+γ)^{2}=α^{2}+β^{2}+γ^{2}+2(αβ+βγ+αγ)
⇒4=6+2(αβ+βγ+αγ)
⇒αβ+βγ+αγ=−1
(α^{3}+β^{3}+γ^{3}−3αβγ)=(α+β+γ)(α^{2}+β^{2}+γ^{2}−αβ−βγ−αγ)
⇒8−3αβγ=2(6+1)
⇒αβγ=−2
(α^{2}+β^{2}+γ^{2})^{2}=α^{4}+β^{4}+γ^{4}+2(α^{2}β^{2}+β^{2}γ^{2}+α^{2}γ^{2})
⇒α^{4}+β^{3}+γ^{4}+2[(αβ+βγ+αγ)^{2}−2αβγ(α+β+γ)]
⇒36=α^{4}+β^{4}+γ^{4}+2[(−1)^{2}−2(−2(2)]
⇒α^{4}+β^{4}+γ^{4}=36−18
⇒18
Hence (C) is the correct answer
If α,β are the roots of ax^{2} + bx + c and α + k, β + k are the roots of px^{2} + qx + r, then k =
If α,β are the roots of the polynomial x^{2} – px + q, then the quadratic polynomial, the roots of which are (α^{2}– β^{2}) (α^{3} – β^{3}) and (α^{3}β^{2} + α^{2}β^{3}) :
If α,β are the roots of the polynomial x^{2}  px + q.
α + β = p and αβ = q
if the roots are (α^{2} β^{2})(α^{3}  β^{3}) and (α^{3}β^{2} + α^{2}β^{3})
= (α^{2} β^{2})(α^{3}  β^{3}) + (α^{3}β^{2} + α^{2}β^{3})
= (α +β)(α β)2(α^{2} + αβ+ β^{2}) + α^{2}β^{2}(α + β).
= (α +β)[ (α +β)^{2}  4αβ ] [ (α +β)^{2}  αβ ] + α^{2}β^{2}(α + β).
= p( p^{2}  4q)( p^{2}  q) + q^{2}p
= p( p4  p^{2}q 4p^{2}q + 4q^{2}) + pq^{2}
= p^{5}  5p^{3}q + 5pq^{2} .
⇒ (α^{2} β^{2})(α^{3}  β^{3})(α^{3}β^{2} + α^{2}β^{3}) = (p^{6}q^{2}5p^{4}q^{3}+4p^{2}q^{4})
∴ x^{2} (p^{5}  5p^{3}q + 5pq^{2} ) x + ( p^{6}q^{2}  5p^{4}q^{3} + 4p^{2}q^{4} ) = 0
The condition that x^{3} – ax^{2} + bx – c = 0 may have two of the roots equal to each other but of opposite signs is:
If the roots of polynomial x^{2} + bx + ac are α,β and roots of the polynomial x^{2} + ax + bc are α,γ then the values of α,β,γ respectively are –
α+β=−b
αβ=ac
Similarly,
α+γ=−a
αγ=bc
Now α is the common root.
Therefore, from the equations,
αβ=ac and αγ=bc
We get the common root α=c
Therefore
β=a and γ=b
Since α+β=−b
Hence
c+a=−b
a+b+c=0
If one zero of the polynomial ax^{2} + bx + c is positive and the other negative then (a,b,c εR, a = 0)
if α and β are two roots of equation ax^{2}+bx+c=0
product of roots α*β = c/a
now if one root is negative and other positive, then product of the roots should be negative.
i.e c/a is negative which means both a and c are of opposite signs.
If α,β are the zeros of the polynomial x^{2} – px + q, then is equal to –
If α,β are the zeros of the polynomial x^{2} – px + 36 and α^{2} + β^{2} = 9, then p =
sum of roots α+β= b/a = p
product of roots αβ = c/a= 36
α^{2}+β^{2}= 9
α^{2}+β^{2 }+2αβ  2αβ = 9
(α+β)^{2 } 2αβ = 9
p^{2}= 9 + 2*36
p^{2 }= 81
p = 9
If α,β are zeros of ax^{2} + bx + c, ac = 0, then zeros of cx^{2} + bx + a are –
α+β = b/a (eq1) αβ = c/a (eq 2)
let the second equation cx^{2} + bx + a has roots x and y
x+y = b/c (eq 3) xy = a/c (eq 4)
divide eq 1 by eq 3
α+β / x+y = (b/a) / (b/c)
= c/a
= αβ
α+β / αβ = x+y
x+y = 1/α + 1/β
A real number is said to be algebraic if it satisfies a polynomial equation with integral coefficients. Which of the following numbers is not algebraic :
a transcendental number is a complex number that is not algebraic—that is, not a root (i.e., solution) of a nonzero polynomial equation with integer or equivalently rational coefficients. The most popular transcendental numbers are π and e
The biquadraic polynomial whose zeros are 1, 2, 4/3,  1 is :
As the roots are 1,2,4/3,1
Therefore the equation will be
(x1)(x2)(x4/3)(x+1)
If we will solve this thing then we will get
3x^{4} – 10x^{3} + 5x^{2} + 10x – 8
The cubic polynomials whose zeros are 4,3/2 and – 2 is :
roots are 4,3/2 and – 2 so we can form equation as
(x4)(x3/2)(x2)=0
which will give 2x^{3} – 7x^{2} – 10x + 24
If the sum of zeros of the polynomial p(x) = kx^{3} – 5x^{2} – 11x – 3 is 2, then k is equal to :
Let α,β,γ be the zeroes of given polynomial p(x)......
then,
α+β+γ= b/a = 2 (given)
=> 2 = (5)/k
=> k =5/2
If f(x) = 4x^{3} – 6x^{2} + 5x – 1 and α, β and γ are its zeros, then αβγ =
for a cubic equation product of roots is α,β,γ = d/a
= (1)/4 = 1/4
Consider f(x) = 8x^{4} – 2x^{2} + 6x – 5 and α,β,γ,δ are it's zeros then α + β + γ + δ =
for a equation of 4 roots ax^{4}+bx^{3}+cx^{2}+dx+e=0
sum of roots is b/a
in our equation b = 0 (since there is no term of x^{3})
so sum of roots is b/a=0
If x^{2} – ax + b = 0 and x^{2} – px + q = 0 have a root in common and the second equation has equal roots, then –
x^{2} – ax + b = 0 and x^{2} – px + q = 0 are the equations
second equation has two equal roots let the roots be k
2k = p => k = p/2 and k^{2}=q
now both equations has one common root k
for first equation k+ l = a => l = a p/2 and kl = b
put k and l
p/2 (ap/2) = b
ap/2  p^{2}/4 = b
ap/2 = k^{2}+b
ap/2 = q+b
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 








