Class 10th Math, Pair of Linear Equations in Two Variables, Exercise-3.2 (Solution) and Exercise-3.2 (Solution)

CLASS 10th Math, Pair f Linear Equation in Two Variables   (NCERT)

 Exercise: 3.1 (Solution), Question-1 to Question-3

Exercise: 3.2 (Solution), Question-1 to Question-7

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Q1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solutions:- Let the present age of Aftab be ‘x’.

And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,

Age of Aftab = x – 7

Age of his daughter = y – 7

According to the question,

x−7 = 7(y−7)

⇒ x−7 = 7y−49

⇒ x−7y = −42         ………………………(i)

Also, three years from now or after three years,

Age of Aftab will become = x + 3.

Age of his daughter will become = y + 3

According to the situation given,

x+3 = 3(y+3)

⇒x+3 = 3y+9

⇒x−3y = 6       …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have

(x−3y)−(x−7y) = 6−(−42)

⇒−3y+7y=6+42

⇒4y=48

⇒y=12

The algebraic equation is represented by

x−7y = −42

x−3y = 6

For, x−7y =−42 or x = −42+7y

The solution table is

For,  x−3y=6   or     x=6+3y

The solution table is

The grapmathematicsraj1.blogspot.comhical representation is:

Q2. The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Solutions:- Let us assume that the cost of a bat be ‘Rs x’

And,the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

3x+6y=3900

and x+2y=1300

For, 3x+6y=3900

Or x = Or x = $\frac{3900 – 6y}{3}$

The solution table is

x	300	100	-100
y	500	600	700

For, x+2y=1300

Or x=1300-2y

The solution table is

x	300	100	-100
y	500	600	700

The graphical representation is as follows.

Q3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Solutions:- Let the cost of 1 kg of apples be ‘Rs. x’

And, cost of 1 kg of grapes be ‘Rs. y’

According to the question, the algebraic representation is

2x+y=160

And 4x+2y=300

For, 2x+y=160 or y=160−2x, the solution table is;

x	50	60	70
y	60	40	20

For 4x+2y=300 or y = $\frac{300 – 4x}{2}$ , the solution table is;

x	70	80	75
y	10	-10	0

The graphical representation is as follows;

.................................................................................................

 Exercise 3.2

Q1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46. Find the cost of one pencil and that of one pen.

Solution:-

(i) Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

x + y = 10

x – y = 4

Now, for x + y = 10 or x=10−y, the solutions are;

x	5	4	6
y	5	6	4

For x – y = 4 or x = 4 + y, the solutions are;

x	4	5	3
y	0	1	-1

The graphical representation is as follows;

From the graph, it can be seen that the given lines cross each other at point (7, 3). Therefore, there are 7 girls and 3 boys in the class.

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.

According to the question, the algebraic expression cab be represented as;

5x + 7y = 50

7x + 5y = 46

For, 5x + 7y = 50 or  x = $\frac{50-7y}{5}$, the solutions are;

x	3	10	-4
y	5	0	10

For 7x + 5y = 46 or x = $\frac{46-5y}{7}$, the solutions are;

x	8	3	-2
y	-2	5	12

Hence, the graphical representation is as follows;

From the graph, it is can be seen that the given lines cross each other at point (3, 5).

So, the cost of a eraser is 3/- and cost of a chocolate is 5/-.

Q2. On comparing the ratios , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solutions:-

(i) Given expressions;

5x−4y+8=0

7x+6y−9=0

Comparing these equations with a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;

9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a1x + b1y + c1 = 0

And a2x + b2y + c2 = 0

We get,

So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;

6x – 3y + 10 = 0

2x – y + 9 = 0

So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

Q3. On comparing the ratios a1/a2, b1/b2 & c1/c2 find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii)(3/2)x+(5/3)y = 7 ; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

(v) (4/3)x+2y = 8; 2x + 3y = 12

Solutions:-

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0

and 2x – 3y = 7 or 2x – 3y -7 = 0

(ii) Given 2x – 3y = 8 and 4x – 6y = 9

Therefore,

(iii) (3/2)x+(5/3)y = 7 and 9x – 10y = 14

Therefore,

So, the equations are intersecting  each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) Given, 5x – 3y = 11 and – 10x + 6y = –22

Therefore,

(v) (4/3)x+2y = 8 and 2x + 3y = 12

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solutions:

(i) Given, x + y = 5 and 2x + 2y = 10

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other.

Therefore, the equations have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 16

(iii) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0

From the graph, it can be seen that these lines are intersecting each other at only one point,(2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0

Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

Q5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solutions:- Let us consider.

The width of the garden is x and length is y.

Now, according to the question, we can express the given condition as;

y – x = 4

and

y + x = 36

Now, taking y – x = 4 or y = x + 4

x	0	8	12
y	4	12	16

For y + x = 36, y = 36 – x

x	0	36	36
y	36	0	20

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

Q6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

Solutions:

(i) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

Clearly, you can see another equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy below condition;

Clearly, you can see another equation satisfies the condition.

(iii) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy below condition;

Clearly, you can see another equation satisfies the condition.

Q7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:- Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = 1+y

x	0	1	2
y	1	2	3

For, 3x + 2y – 12 = 0 or x = $\frac{12-2y}{3}$

x	4	2	0
y	0	3	6

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

See next blog for more Solutions... Thank you.

Class 10th Math, Polynomials Solution

CLASS 10th Math, Polynomials  (NCERT)

 Exercise: 2.1 To Exercise: 2.4

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Exercise 2.1 (Solution)

Q1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Solutions:-

Graphical method to find zeroes:-

Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.

1. In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.
2. In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.
3. In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.
4. In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.
5. In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.
6. In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.

Exercise 2.2 (Solution)

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solutions:-

(i)x2–2x –8

⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2)

Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)

(ii)4s2–4s+1

⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )

(iii) 6x2–3–7x

⇒6x2–7x–3 = (3x+1)(2x-3)

Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)

Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )

(iv)4u2+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )

(v) t2–15

⇒ t2 = 15 or t = ±√15

Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )

(vi) 3x2–x–4

⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)

Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1

Solution:-

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–(1/4)x +(-1) = 0

4x2–x-4 = 0

Thus,4x2–x–4 is the quadratic polynomial.

(ii)√2, 1/3

Solution:-

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2 –(√2)x + (1/3) = 0

3x2-3√2x+1 = 0

Thus, 3x2-3√2x+1 is the quadratic polynomial.

(iii) 0, √5

Solution:-

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly

as:-

x2–(α+β)x +αβ = 0

x2–(0)x +√5= 0

Thus,x2+√5 is the quadratic polynomial.

(iv)1, 1

Solution:-

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–x+1 = 0

Thus , x2–x+1is the quadratic polynomial.

(v) -1/4, 1/4

Solution:-

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x +αβ = 0

x2–(-1/4)x +(1/4) = 0

4x2+x+1 = 0

Thus,4x2+x+1 is the quadratic polynomial.

(vi) 4, 1

Solution:-

Given,

Sum of zeroes = α+β =

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2–(α+β)x+αβ = 0

x2–4x+1 = 0

Thus,x2–4x+1 is the quadratic polynomial.

Exercise 2.3 (Solution)

Q1:-Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i)p(x) = x3-3x2+5x–3 , g(x) = x2–2

Solution:-Given,

Dividend = p(x) = x3-3x2+5x–3

Divisor = g(x) = x2– 2

Therefore, upon division we get,

Quotient = x–3

Remainder = 7x–9

(ii) p(x) = x4-3x2+4x+5 , g(x) = x2+1-x

Solution:-Given,

Dividend = p(x) = x4 – 3x2 + 4x +5

Divisor = g(x) = x2 +1-x

Therefore, upon division we get,

Quotient = x2 +x–3

Remainder = 8

(iii) p(x) =x4–5x+6, g(x) = 2–x2

Solution:- Given,

Dividend = p(x) =x4 – 5x + 6 = x4 +0x2–5x+6

Divisor = g(x) = 2–x2 = –x2+2

Therefore, upon division we get,

Quotient = -x2-2

Remainder = -5x + 10

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t2-3, 2t4 +3t3-2t2-9t-12

Solutions:- Given,

First polynomial = t2-3

Second polynomial = 2t4 +3t3-2t2 -9t-12

As we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t2+3t+4.

(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2

Solutions:- Given,

First polynomial = x2+3x+1

Second polynomial = 3x4+5x3-7x2+2x+2

As we can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.

(iii) x3-3x+1, x5-4x3+x2+3x+1

Solutions:- Given,

First polynomial = x3-3x+1

Second polynomial = x5-4x3+x2+3x+1

As we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1 .

Q3. Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and – √(5/3).

Solutions:- Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

√(5/3) and – √(5/3) are zeroes of polynomial f(x).

∴ (x –√(5/3)) (x+√(5/3) = x2-(5/3) = 0

(3x2−5)=0, is a factor of given polynomial f(x).

Now, when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor of f(x) and the remainder will be 0.

Therefore, 3x4 +6x3 −2x2 −10x–5 = (3x2 –5)(x2+2x+1)

Now, on further factorizing (x2+2x+1) we get,

x2+2x+1 = x2+x+x+1 = 0

x(x+1)+1(x+1) = 0

(x+1)(x+1) = 0

So, its zeroes are given by: x= −1 and x = −1.

Therefore, all four zeroes of given polynomial equation are:

√(5/3),- √(5/3) , −1 and −1.

Hence, is the answer.

Q4. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were x–2 and –2x+4, respectively. Find g(x).

Solutions:-Given,

Dividend, p(x) = x3-3x2+x+2

Quotient = x-2

Remainder = –2x+4

We have to find the value of Divisor, g(x) =?

As we know,

Dividend = Divisor×Quotient + Remainder

∴ x3-3x2+x+2 = g(x)×(x-2)+(-2x+4)

x3-3x2+x+2-(-2x+4) = g(x)×(x-2)

Therefore, g(x) × (x-2) = x3-3x2+x+2

Now, for finding g(x) we will divide x3-3x2+x+2 with (x-2)

Therefore, g(x) = (x2–x+1)

Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solutions:- According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;

Dividend = Divisor×Quotient+Remainder

∴ p(x) = g(x)×q(x)+r(x)

Where r(x) = 0 or degree of r(x)< degree of g(x).

Now let us proof the three given cases as per division algorithm by taking examples for each.

(i):deg p(x) = deg q(x)

Degree of dividend is equal to degree of quotient, only when the divisor is a constant term.

Let us take an example, 3x2+3x+3 is a polynomial to be divided by 3.

So, (3x2+3x+3)/3 = x2+x+1 = q(x)

Thus, you can see, the degree of quotient is equal to the degree of dividend.

Hence, division algorithm is satisfied here.

(ii):deg q(x) = deg r(x)

Let us take an example , p(x)=x2+x is a polynomial to be divided by g(x)=x.

So, (x2+x)/x = x+1 = q(x)

Also, remainder, r(x) = 0

Thus, you can see, the degree of quotient is equal to the degree of remainder.

Hence, division algorithm is satisfied here.

(iii):deg r(x) = 0

The degree of remainder is 0 only when the remainder left after division algorithm is constant.

Let us take an example, p(x) = x2+1 is a polynomial to be divided by g(x)=x.

So,( x2+1)/x= x=q(x)

And r(x)=1

Clearly, the degree of remainder here is 0.

Hence, division algorithm is satisfied here.

Exercise 2.4 (Solution)

Q1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x3+x2-5x+2; -1/2, 1, -2

Solutions:- Given, p(x) = 2x3+x2-5x+2

And zeroes for p(x) are = 1/2, 1, -2

∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x3-4x2+5x-2 ;2, 1, 1

Solutions:- Given, p(x) = x3-4x2+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 23-4(2)2+5(2)-2 = 0

p(1) = 13-(4×12 )+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = x3-4x2+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

Q2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Solutions:- Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x3-2x2-7x+14

Q3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.

Solutions:- We are given with the polynomial here,

p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px3+qx2+rx+s = x3-3x2+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b2

-1/1 = 1-b2

b2 = 1+1 = 2

b = √2

Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.

Q4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes.

Solutions:- Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x4-6x3-26x2+138x-35

Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get,

x2-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 –2x−35)

Now, on further factorizing (x2–2x−35) we get,

x2–(7−5)x −35 = x2– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:              

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7.

See next blog for more Solution........Thank you.