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Chapter 2: Polynomials Exercise – 2.1 Question: 1 Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients: (i) f(x) = x2 – 2x – 8 (ii) g(s) = 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) h(t) = t2 – 15 (v) p(x) = x2 + 2√2 x – 6 (vi) q(x) = √3 x2 + 10x + 7√3 (vii) f(x) = x2 - (√3 + 1)x + √3 (viii) g(x) = a(x2 + 1) – x(a2 + 1) Solution: (i) f(x) = x2 – 2x – 8 We have, f(x) = x2 – 2x – 8 = x2 – 4x + 2x – 8 = x (x – 4) + 2(x – 4) = (x + 2)(x – 4) Zeroes of the polynomials are – 2 and 4. Now, Hence, the relationship is verified. (ii) g(s) = 4s2 – 4s + 1 We have, g(s) = 4s2 – 4s + 1 = 4s2 – 2s – 2s + 1 = 2s(2s – 1)− 1(2s – 1) = (2s – 1)(2s – 1) Zeroes of the polynomials are 1/2 and 1/2. Hence, the relationship is verified. (iii) 6s2 − 3 − 7x = 6s2 − 7x − 3 = (3x + 11) (2x – 3) Zeros of the polynomials are 3/2 and (-1)/3 Hence, the relationship is verified. (iv) h(t) = t2 – 15 We have, Zeroes of the polynomials are - √15 and √15 Sum of the zeroes = 0 - √15 + √15 = 0 0 = 0 Hence, the relationship verified. Zeroes of the polynomials are 3√2 and –3√2 Sum of the zeroes - 6 = – 6 Hence, the relationship is verified. Zeros of the polynomials are -√3 and -7/√3 Product of the polynomials are - √3, 7/√3 7 = 7 Hence, the relationship is verified. Zeros of the polynomials are 1 and √3 Hence, the relationship is verified (viii) g(x) = a[(x2 + 1)– x(a2 + 1)]2 = ax2 + a − a2x − x = ax2 − [(a2x + 1)] + a = ax2 − a2x – x + a = ax(x − a) − 1(x – a) = (x – a)(ax – 1) Zeros of the polynomials are 1/a and 1 Sum of the zeros Product of zeros = a/a Hence, the relationship is verified. Question: 2 If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of Solution: We have, α and β are the roots of the quadratic polynomial. f(x) = x2 – 5x + 4 Sum of the roots = α + β = 5 Product of the roots = αβ = 4 So, Question: 3 If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial. p(y) = x2 – 5x + 4 Sum of the zeroes = α + β = 5 Product of the roots = αβ = 4 So, Question: 4 If α and β are the zeroes of the quadratic polynomial p (y) = 5y2 – 7y + 1, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial. p(y) = 5y2 – 7y + 1 Sum of the zeroes = α + β = 7 Product of the roots = αβ = 1 So, Question: 5 If α and β are the zeroes of the quadratic polynomial f(x) = x2 – x – 4, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial. We have, f(x) = x2 – x – 4 Sum of zeroes = α + β = 1 Product of the zeroes = αβ = - 4 So, Question: 6 If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial. We have, f(x) = x2 + x – 2 Sum of zeroes = α + β = 1 Product of the zeroes = αβ = – 2 So, Question: 7 If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k. Solution: Let, the two zeroes of the polynomial f(x) = 4x2 – 8kx – 9 be α and − α. Product of the zeroes = α × − α = – 9 Sum of the zeroes = α + (− α) = – 8k = 0 Since, α – α = 0 ⇒ 8k = 0 ⇒ k = 0 Question: 8 If the sum of the zeroes of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, then find the value of k. Solution: Let the two zeroes of the polynomial f(t) = kt2 + 2t + 3k be α and β. Sum of the zeroes = α + β = 2 Product of the zeroes = α × β = 3k Now, Question: 9 If α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2β + αβ2. Solution: Since, α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1 So, Sum of the zeroes α + β = 5/4 Product of the zeroes α × β = – 1/4 Now, α2β + αβ2 = αβ (α + β) Question: 10 If α and β are the zeroes of the quadratic polynomial f(t) = t2 – 4t + 3, find the value of α4β3 + α3β4. Solution: Since, α and β are the zeroes of the quadratic polynomial f(t) = t2 – 4t + 3 So, Sum of the zeroes = α + β = 4 Product of the zeroes = α × β = 3 Now, α4β3 + α3β4 = α3β3(α + β) = (3)3(4) = 108 Question: 11 If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2. Sum of the zeroes = α + β = -1/6 Product of the zeroes =α × β = -1/3 Now, By substitution the values of the sum of zeroes and products of the zeroes, we will get = - 25/12 Question: 12 If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of Solution: Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2. Sum of the zeroes = α + β = 6/3 Product of the zeroes = α × β = 4/3 Now, By substituting the values of sum and product of the zeroes, we will get 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 the two zeroes of the polynomial be αand β. We have, f(x) = x2 + px + 45 Now, Sum of the zeroes = α + β = – p Product of the zeroes = α × β = 45 So, Thus, in the given equation, 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 Solution: Since, α and β are the roots of the quadratic polynomial given in the question. f(x) = x2 – px + q Now, Sum of the zeroes = p = α + β Product of the zeroes = q = α × β 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: Since, 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) So, (α + 1)(β + 1) = αβ + α + β + 1 = αβ + (α + β) + 1 = (− p – c) + p + 1 = 1 – c = RHS So, 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: We have, α + β = 24 …… E-1 α – β = 8 …. E-2 By solving the above two equations accordingly, we will get 2α = 32 α = 16 Substitute the value of α, in any of the equation. Let we substitute it in E-2, 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 is-K 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 Solution: We have, f(x) = x2 – 1 Sum of the zeroes = α + β = 0 Product of the zeroes = αβ = – 1 From the question, Sum of the zeroes of the new polynomial {By substituting the value of the sum and products of the zeroes} As given in the question, Product of the zeroes 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 Solution: We have, f(x) = x2 – 3x – 2 Sum of the zeroes = α + β = 3 Product of the zeroes = αβ = – 2 From the question, Sum of the zeroes of the new polynomial So, the quadratic polynomial is, x2- (sum of the zeroes)x + (product of the zeroes) x2 - (sum of the zeroes)x + (product of the zeroes) Hence, the required quadratic polynomial is k 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: We have, f(x) = x2 + px + q Sum of the zeroes = α + β = -p Product of the zeroes = αβ = q From the question, 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 (α – β)2 = (- p)2((- p)2 - 4q) = p2 (p2–4q) So, 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 Solution: We have, 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 So, 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) f(x) = k(x2 – 6x + 11) (ii) Sum of the zeroes of new polynomial Product of the zeroes of new polynomial So, the quadratic polynomial is, x2 – (sum of the zeroes)x + (product of the zeroes) Thus, the required quadratic polynomial is f(x) = k(x2 – 23x + 13) Question: 21 If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: (i) α – β (iv) α2β + αβ2 (v) α4 + β4 Solution: f(x) = ax2 + bx + c Here, Sum of the zeroes of polynomial = α + β = -b/a Product of zeroes of polynomial = αβ = c/a Since, α + β are the roots (or) zeroes of the given polynomial, so (i) α – β The two zeroes of the polynomials are - From previous question we know that, Also, αβ = c/a Putting the values in E.1, we will get Since, Sum of the zeroes of polynomial = α + β = – b/a Product of zeroes of polynomial = αβ = c/a After substituting it in E-1, we will get (iv) α2β + αβ2 = αβ(α + β) …….. E- 1. Since, Sum of the zeroes of polynomial = α + β = – b/ a Product of zeroes of polynomial = αβ = c/a After substituting it in E-1, we will get (v) α4 + β4 = (α2 + β2)2 – 2α2β2 = ((α + β)2 – 2αβ)2 – (2αβ)2 ……. E- 1 Since, Sum of the zeroes of polynomial = α + β = – b/a Product of zeroes of polynomial = αβ = c/a After substituting it in E-1, we will get Since, Sum of the zeroes of polynomial = α + β = – b/a Product of zeroes of polynomial = αβ = c/a After substituting it, we will get Since, Sum of the zeroes of polynomial = α + β = – b/a Product of zeroes of polynomial = αβ = c/a After substituting it, we will get Since, Sum of the zeroes of polynomial= α + β = – b/a Product of zeroes of polynomial= αβ = c/a After substituting it, we will get
Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) f(x) = x2 – 2x – 8
(ii) g(s) = 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) h(t) = t2 – 15
(v) p(x) = x2 + 2√2 x – 6
(vi) q(x) = √3 x2 + 10x + 7√3
(vii) f(x) = x2 - (√3 + 1)x + √3
(viii) g(x) = a(x2 + 1) – x(a2 + 1)
We have,
f(x) = x2 – 2x – 8
= x2 – 4x + 2x – 8
= x (x – 4) + 2(x – 4)
= (x + 2)(x – 4)
Zeroes of the polynomials are – 2 and 4.
Now,
Hence, the relationship is verified.
g(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s(2s – 1)− 1(2s – 1)
= (2s – 1)(2s – 1)
Zeroes of the polynomials are 1/2 and 1/2.
(iii) 6s2 − 3 − 7x
= 6s2 − 7x − 3 = (3x + 11) (2x – 3)
Zeros of the polynomials are 3/2 and (-1)/3
Zeroes of the polynomials are - √15 and √15
Sum of the zeroes = 0 - √15 + √15 = 0
0 = 0
Hence, the relationship verified.
Zeroes of the polynomials are 3√2 and –3√2 Sum of the zeroes
- 6 = – 6
Zeros of the polynomials are -√3 and -7/√3
Product of the polynomials are - √3, 7/√3
7 = 7
Zeros of the polynomials are 1 and √3
Hence, the relationship is verified
(viii) g(x) = a[(x2 + 1)– x(a2 + 1)]2
= ax2 + a − a2x − x
= ax2 − [(a2x + 1)] + a
= ax2 − a2x – x + a
= ax(x − a) − 1(x – a) = (x – a)(ax – 1)
Zeros of the polynomials are 1/a and 1 Sum of the zeros
Product of zeros = a/a
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of
α and β are the roots of the quadratic polynomial.
f(x) = x2 – 5x + 4
Sum of the roots = α + β = 5
Product of the roots = αβ = 4
So,
Since, α and β are the zeroes of the quadratic polynomial.
p(y) = x2 – 5x + 4
Sum of the zeroes = α + β = 5
If α and β are the zeroes of the quadratic polynomial p (y) = 5y2 – 7y + 1, find the value of
p(y) = 5y2 – 7y + 1
Sum of the zeroes = α + β = 7
Product of the roots = αβ = 1
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – x – 4, find the value of
f(x) = x2 – x – 4
Sum of zeroes = α + β = 1
Product of the zeroes = αβ = - 4
If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of
f(x) = x2 + x – 2
Product of the zeroes = αβ = – 2
If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k.
Let, the two zeroes of the polynomial f(x) = 4x2 – 8kx – 9 be α and − α.
Product of the zeroes = α × − α = – 9
Sum of the zeroes = α + (− α) = – 8k = 0 Since, α – α = 0
⇒ 8k = 0 ⇒ k = 0
If the sum of the zeroes of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, then find the value of k.
Let the two zeroes of the polynomial f(t) = kt2 + 2t + 3k be α and β.
Sum of the zeroes = α + β = 2
Product of the zeroes = α × β = 3k
If α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2β + αβ2.
Since, α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1
So, Sum of the zeroes α + β = 5/4
Product of the zeroes α × β = – 1/4
α2β + αβ2 = αβ (α + β)
If α and β are the zeroes of the quadratic polynomial
f(t) = t2 – 4t + 3, find the value of α4β3 + α3β4.
Since, α and β are the zeroes of the quadratic polynomial f(t) = t2 – 4t + 3
So, Sum of the zeroes = α + β = 4
Product of the zeroes = α × β = 3
α4β3 + α3β4 = α3β3(α + β)
= (3)3(4) = 108
f(x) = 6x2 + x – 2, find the value of
Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
Sum of the zeroes = α + β = -1/6
Product of the zeroes =α × β = -1/3
By substitution the values of the sum of zeroes and products of the zeroes, we will get
= - 25/12
If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of
Sum of the zeroes = α + β = 6/3
Product of the zeroes = α × β = 4/3
By substituting the values of sum and product of the zeroes, we will get
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.
Let the two zeroes of the polynomial be αand β.
f(x) = x2 + px + 45
Sum of the zeroes = α + β = – p
Product of the zeroes = α × β = 45
Thus, in the given equation, p will be either 18 or -18.
f(x) = x2 – px + q, prove that
Since, α and β are the roots of the quadratic polynomial given in the question.
f(x) = x2 – px + q
Sum of the zeroes = p = α + β
Product of the zeroes = q = α × β
LHS = RHS
Hence, proved.
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.
Since, α and β are the zeroes of the quadratic polynomial
f(x) = x2 – p(x + 1)– c
Sum of the zeroes = α + β = p
Product of the zeroes = α × β = (- p – c)
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
= (− p – c) + p + 1
= 1 – c = RHS
So, LHS = RHS
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.
α + β = 24 …… E-1
α – β = 8 …. E-2
By solving the above two equations accordingly, we will get
2α = 32 α = 16
Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get
β = 16 – 8 β = 8
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 is-K
x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128
Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128
f(x) = x2 – 1, find a quadratic polynomial whose zeroes are
f(x) = x2 – 1
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = – 1
From the question,
Sum of the zeroes of the new polynomial
{By substituting the value of the sum and products of the zeroes}
As given in the question,
Product of the zeroes
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
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 3x – 2, find a quadratic polynomial whose zeroes are
f(x) = x2 – 3x – 2
Sum of the zeroes = α + β = 3
So, the quadratic polynomial is,
x2- (sum of the zeroes)x + (product of the zeroes)
x2 - (sum of the zeroes)x + (product of the zeroes)
Hence, the required quadratic polynomial is k
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.
f(x) = x2 + px + q
Sum of the zeroes = α + β = -p
Product of the zeroes = αβ = q
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 (α – β)2
= (- p)2((- p)2 - 4q)
= p2 (p2–4q)
= x2 – (2p2 – 4q)x + p2(p2 – 4q)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2p2 –4q) x + p2(p2 - 4q)).
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
f(x) = x2 – 2x + 3
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
So, quadratic polynomial is:
= x2 – 6x +11
Hence, the required quadratic polynomial is f(x) = k(x2 – 6x + 11)
f(x) = k(x2 – 6x + 11)
(ii) Sum of the zeroes of new polynomial
Product of the zeroes of new polynomial
Thus, the required quadratic polynomial is f(x) = k(x2 – 23x + 13)
If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate:
(i) α – β
(iv) α2β + αβ2
(v) α4 + β4
f(x) = ax2 + bx + c
Here,
Sum of the zeroes of polynomial = α + β = -b/a
Product of zeroes of polynomial = αβ = c/a
Since, α + β are the roots (or) zeroes of the given polynomial, so
The two zeroes of the polynomials are -
From previous question we know that,
Also,
αβ = c/a
Putting the values in E.1, we will get
Since,
Sum of the zeroes of polynomial = α + β = – b/a
After substituting it in E-1, we will get
= αβ(α + β) …….. E- 1.
Sum of the zeroes of polynomial = α + β = – b/ a
= (α2 + β2)2 – 2α2β2
= ((α + β)2 – 2αβ)2 – (2αβ)2 ……. E- 1
After substituting it, we will get
Sum of the zeroes of polynomial= α + β = – b/a
Product of zeroes of polynomial= αβ = c/a
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Chapter 2: Polynomials Exercise – 2.2...
Chapter 2: Polynomials Exercise – 2.3...