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Form a quadratic polynomial, sum of whose zeroes is 3 and product of whose zeroes is 2 solution
Sum of zeros = 3/1
b/a = 3/1
Product of zeros = 2/1
c/a = 2/1
This gives
a = 1
b = 3
c = 2,
The required quadratic equation is
ax^{2}+bx+c
So, 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
Hence, value of a=1
So option B is correct answer.
If 3 + 5 – 8 = 0, then the value of (3)^{3} + (5)^{3} – (8)^{3} is
Given :
3 + 5 + (8) = 0.
Then,
3³ + 5³ + (8)³ = ?
Using, identity...
If a+b+c=0, then a³ + b ³ + c³ = 3abc
Let a = 3, b = 5 and c = 8.
Here, a + b + c = 3 + 5 + (8) = 0.
Then, a³ + b³ + c³ = 3³ + 5³ + (8)³
= 3× 3 × 5 × 8
= 9 × 40
= 360
If one of the factors of x^{2} + x – 20 is (x + 5), then other factor is
Using midterm splitting,
x^{2}+x20=x^{2}+5x4x20=x(x+5)4(x+5)
Taking common x+5
(x+5)(x4) , so the other factor is x4
If α,β be the zeros of the quadratic polynomial 2x^{2} + 5x + 1, then value of α + β + αβ =
P(x) = 2x² + 5x + 1
Sum of roots = 5/2
Product of roots = 1/2
Therefore substituting these values,
α + β +αβ
=(α + β) + αβ
= 5/2 + 1/2
= 4/2
= 2
If α,β be the zeros of the quadratic polynomial 2 – 3x – x^{2}, then α + β =
If α and β are the zeros of the polynomial then
(x−α)(x−β) are the factors of the polynomial
Thus, (x−α)(x−β) is the polynomial.
So, the polynomial =x^{2 }− αx − βx + αβ
=x^{2 }− (α + β)x + αβ....(i)
Now,the quadratic polynomial is
2 − 3x − x^{2} = x^{2} + 3x − 2....(ii)
Now, comparing equation (i) and (ii),we get,
−(α + β) = 3
α + β = −3
Quadratic polynomial having sum of it's zeros 5 and product of it's zeros – 14 is –
The quadratic equation is of the form x^{2 } (sum of zeros) x + (product of zeros)
=x^{2 } 5x  14
If x = 2 and x = 3 are zeros of the quadratic polynomial x^{2} + ax + b, the values of a and b respectively are :
Zeros of the polynomials are the values which gives zero when their value is substituted in the polynomial
When x=2,
x^{2}+ax+b =(2)^{2}+a*2+b=0
4+2a+b=0
b=42a ….1
When x=3,
(3)^{2}+ 3a + b=0
9 + 3a + b=0
Substituting
9 + 3a  4  2a =0
5 + a =0
a = 5
b = 6
If 3 is a zero of the polynomial f(x) = x^{4} – x^{3} – 8x^{2} + kx + 12, then the value of k is –
The sum and product of zeros of the quadratic polynomial are – 5 and 3 respectively the quadratic polynomial is equal to –
On dividing x^{3} – 3x^{2} + x + 2 by polynomial g(x), the quotient and remainder were x – 2 and 4 – 2x respectively then g(x) :
We have remainder theorem as
x^{3}3x^{2}+x+2=q(x)*g(x)+r(x)
x^{3}3x^{2}+x+2=(x2)g(x)+(42x)
g(x)=
So by division method,
g(x)= x^{2 } x + 1
If the polynomial 3x^{2} – x^{3} – 3x + 5 is divided by another polynomial x – 1 – x^{2}, the remainder comes out to be 3, then quotient polynomial is –
Let's assume the quotient polynomial as ax + b.
Now, (x  1  x^2)(ax + b) + 3 = 3x^2  x^3  3x + 5
ax^3 + bx^2 + ax^2  ax^2  bx  a + 3 = x^3 + 3x^2  3x + 5
ax^3 + (b  1)x^2 + (a  b)x + (3  a) = x^3 + 3x^2  3x + 5
Equating the coefficients of the same degree terms on both sides, we get:
For x^3: a = 1 For x^2: b  1 = 3 => b = 4 For x: a  b = 3 => a  4 = 3 => a = 1 For constant: 3  a = 5 => a = 2
Therefore, the quotient polynomial is x + 4.
Hence, the quotient polynomial is x + 4.
If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is
If is the zero of the cubic polynomial f(x) = 3x^{3} – 5x^{2} – 11x – 3 the other zeros are :
If p and q are the zeroes of the polynomial x^{2} 5x  k. Such that p  q = 1, find the value of K
Let p(x) = ax^{2} + bx + c be a quadratic polynomial. It can have at most –
The graph of the quadratic polynomial ax^{2} + bx + c, a ≠ 0 is always –
So it s an equation of a parabola.
The graph will be that of a parabola.
If 2 and as the sum and product of its zeros respectively then the quadratic polynomial f(x) is –
If α and β are the zeros of the polynomial f(x) = 16x^{2} + 4x – 5 then is equal to –
If α and β are the zeros of the polynomial f(x) = 15x^{2} – 5x + 6 then is equal to –
The correct answer is a.
(1+1/α) (1+1/β)
= (α+1/α) (β+1/β)
=αβ + α + β + 1 / αβ
=( 2/5 + 1/3+ 1) / 2/5
=26/15 X 5/2 = 13/3
If the sum of the two zeros of x^{3} + px^{2} +qx + r is zero, then pq =
Sum of two solutions is zero ,so both are inverse of each other. So we have the solutions as a,a,b.
(xa)(x+a)(x+b)=(x^{2}a^{2})(x+b)=x^{3}+bx^{2}a2xa2b
pq=a^{2}b=r
Let a = 0 and p(x) be a polynomial of degree greater than 2. If p(x) leaves remainders a and – a when divided respectively by x + a and x – a, the remainder when p(x) is divided by x^{2} – a^{2 }is
Let P(x)=(x−a)(x+a)Q(x)+rx+s
When we divide by x+a, we get
P(−a)=(−a−a)(−a+a)Q(−a)+r(−a)+s=a
⇒s−ra=a ...(1)
And when we divide by x−a, we get
P(a)=(a−a)(a+a)Q(a)+ra+s=−a
⇒s+ra=−a ...(2)
Solving (1) and (2), we get
s=0,r=−1
Hence, P(x)=(x−a)(x+a)Q(x)−x
And when we divide by x2a2, it leaves a remainder −x.
If one root of the polynomial x^{2} + px + q is square of the other root, then –
Let one root be a, other root is a^{2}.
(xa)(xa^{2})=x^{2}+(a^{2}a)x+a^{3}
p^{3}=(a^{2}a)^{3}=a^{6}3a^{5}3a^{4}a^{3}
q(3p1)= a^{3}(3(a^{2}a)1)=3a^{5}+3a^{4}+a^{3}
q^{2}=a^{6}
Substituting the values
a^{6}3a^{5}3a^{4}a^{3}+3a^{5}+3a^{4}+a^{3}+a^{6}=0
If α,β are the zeros of x2 + px + 1 and γ,δ be those of x2 + qx + 1, then the value of (α–γ) (β–γ) (α+δ) (β+δ) =
Alpha(a) and beta(b) are roots of x^2 + px + 1
This implies that sum of roots= a+b = p/1=p
And the product of roots = ab = 1/1=1
Similarly ,
Gamma(c) and delta(d) are roots of x^2 + qx + 1
So c+d=q and cd =1.
The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (xp)(xq)=0 where a and b are the roots .
So x^2 (p+q)x + pq =0
Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get
Sum of the roots =p+q= b/a
And product of the roots = pq = c/a}
RHS=(ac)(bc)(a+d)(b+d)
=(c^2acbc +ab)(d^2 +bd +ad + ab)
We know ab=1
So RHS= (c^2acbc +1)(d^2 +bd +ad + 1)
= (c^2)(d^2) +(a+b)c^2(d) + c^2 (d^2)c(a+b) (a+b)^2(cd) (a+b)c + d^2 + bd + ad + 1
= 1 + ac+bc + c^2  dadb  (a^2 + b^2 + 2(1)) ac bc + bd + ad +1
Cancelling off all the common terms,
We get c^2 +d^2a^2b^2
= c^2+d^2a^2b^2
=c^2 a^2 + d^2 b^2 + 2(1) 2(1){ ab=cd =1}
=c^2 + d^2 + 2cd  a^2  b^2  2ab
LHS=(c+d)^2(a+b)^2
Therefore,
But we know p= a+b and q = c+d
LHS= q^2p^2
The quadratic polynomial whose zeros are twice the zeros of 2x^{2} – 5x + 2 = 0 is –
Let α and β be the roots of the given equation.
Then, α + β = 5/2 and αβ = 2/2 = 1
∴ 2α + 2β
∴ 2(α + β)
∴ 5,(2α)(2β) = 4
So, the requiared equation is :
x^{2}−5x+4=0
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