# Solutions of half-yearly maths question paper of class 10 CBSE 2020

Here you can study the** Solutions of the** **half-yearly maths question paper of 10 class CBSE 2020**, **the class 10 maths CBSE question paper** is taken from one of the reputed schools of Delhi i.e Sanjivani Senior Secondary School of Mohan Garden, New Delhi. **Solutions of this question paper** will help all the students of **class 10** in their preparation for the CBSE annual **exam**. All **questions** are solved by an expert teacher of **Maths.**

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**Sanjivani Senior Secondary School, Mohan Garden, New Delhi**

MM -80

Time-3 hr

Subject-Maths

General Instruction: Read the following instructions very carefully and strictly follow them

(i) This question paper comprises four sections – A, B, C, and D. This question paper carries 40 questions. All questions are compulsory.

(ii) Section A. question number 1 to 20 comprises 20 questions of one mark each.

Section B. question number . 21 to 26 comprises 6 questions of two marks each.

Section C. question number 27 to 34 comprises 8 questions of three marks each.

Section D.question number 35 to 40 comprises 6 questions of four marks each.

SECTION-A

question number 1 to 10 are a multiple-choice questions. Select the correct option.

Q1.The graphs of polynomial x = p(y) are given in the figure. Then no of its zeroes is

(a) 3

(b) 1

(c) 2

(d) 4

Ans.(a)

Q2.Which of the following rational number have terminating decimal

Ans.(d)

**Q3.The sum and the product of zeroes of a quadratic polynomial are 3 and -10 respectively.The quadratic polynomial is.**

**(a) x² – 3x + 10 **

**(b) x² + 3x – 10**

**(c)x² – 3x – 10**

**(d)x² + 3x + 10**

Ans. Applyig the formula for quadratic polyomial

x² – (α + β)x + αβ

Where α and β are the zereoes of the quadratic polynomial given

We are given α + β = 3, αβ = -10

x² – 3x -10

Hence and is (c) x² – 3x – 10

**Q4.Write number of solutions of following pair of linear equations**

**x + 2y -8 = 0**

**2x + 4y = 16**

(a) Unique solution

(b) No solutions

(c) Infinite many solutions

(d) None

Ans.Comparing the coefficients of given pair of linear equations with the coefficients of standard pair of linear equations a_{1}x + b_{1}y + c_{1}= 0 and a_{2}x + b_{2}y + c_{2}= 0

x + 2y -8 = 0, 2x + 4y = 16(rewritten as 2x + 4y -16 = 0)

a_{1}=1 ,b_{1}=2, c_{1}= -8and a_{2}=2 , b_{2}=4 , c_{2}= -16

Since the relationship between the coefficients of both equations are as following

Therefore the given pair has infinite solutions

**Q5.Which of the following are quadratic equation ?**

**(a) x³ + x² + 4 = 0**

**(b)2x² – 7x = 0**

Ans. The equation (a) x³ + x² + 4 = 0 is cubic, the equation (b)2x² – 7x = 0 is quadratic equation, (c) is cubic and (d) is of power 4

Hence the answer is (b)2x² – 7x = 0

Q6. If b² – 4ac ≥ 0 then what would be the roots of the quadratic equation ax² + bx + c = 0

(d) None

Ans. (a)

**Q7. Find the common difference of AP.**

**(a) 2**

**(b) -2**

**(c) -3**

**(d) None**

Ans. The given AP is as following

The common difference (d) of the given AP is

Hence the answer is (b) -2

**Q8.In figure DE ll BC, find the value of x.**

**(a) 3 cm**

**(b) 2 cm**

**(c) 4 cm**

**(d) 1 cm**

Ans.

In ΔABC we are given DE ll BC, applying the BPT theorem

x = 2

Hence the answer is (b) 2 cm

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**Q9.Sides of triangles are given below. Determine which of the sides are of right triangles.If the right triangle then finds the length of the hypotenuse.**

**7 cm, 24 cm and 25 cm**

**(a) Yes, 24 cm**

**(b) Yes, 25 cm**

**(c) No**

**(d) Yes, 7 cm**

Ans.The condition of a triangle being a right triangle is as follows

The square of longest side = sum of the squares of other sides

25² = 24² + 7²

625 = 576 + 49

625 = 625

Therefore the given triangle is right triangle and its hypotenuse is 25 cm

Hence the answer is (b) Yes, 25 cm

**Q10. Find 10 ^{th} term of AP 2, 7, 12……..**

**(a) 37**

**(b) 47**

**(c) 57**

**(d) None**

Ans.

n^{th }term of AP is = a + (n-1)d

a = 2, d = 7-2 = 5 and n = 10

a_{10} = 2 + (10-1)5

= 2 + 9×5 = 2 + 45 = 47

Therefore the answer is (b) 47

**In question number 11 to 15 , fill in blank**

**Q11. If equation kx -2y = 3 and 3x + y = 5 represent two intersection lines at unique point then value of k is ….**

Ans. The given equations are kx -2y = 3 and 3x + y = 5

Writting both equation in standard form

kx -2y – 3=0, 3x + y – 5 = 0

The condition of two lines a_{1}x + b_{1}y + c_{1}= 0 and a_{2}x + b_{2}y + c_{2}= 0 itersecting each other is as follows

Comparing the coefficients of both given lines

a_{1}=k ,b_{1}= -2, c_{1}= -3 and a_{2}=3 , b_{2}=1 , c_{2}= -5

k ≠ -6

The given line has unique solutions for all values of k except to -6

So, answer , k is not equal to -6

**Q12.If quadratic equation has equal roots 3x² – 4x + k = 0 has equal roots then value of k is………**

Ans. The given equation is 3x² – 4x + k = 0

The condition that the ax² + bx + c = 0 has equal roots is as follows

b² – 4ac = 0

Comparing the given equation with standard equation

a = 3, b = -4 and c = k

(-4)² – 4× 3 ×k = 0

12k = 16

Hence the answer is

**Q13. The HCF of two number is 27 and their LCM is 162. If one of the number is 54 . Find other number……..**

Ans. We are given HCF of two number = 27 and LCM of two number = 162

One of the number = 54, let another number = x

The relationship between two numbers and their LCM and HCF is as follows

LCM × HCF = Product of both numbers

27× 162 = 54 × x

Therefore another number is 81

**Q14. Sum of first n natural number is ……..**

Ans. The natural numbers are 1,2,3,4…

First term, a = 1, common difference, d = 1, sum of n terms is = S_{n}

Putting the value of d = 1, a = 1

**Q15. I fig. OP = 5 cm, OQ = 5√3 cm, ∠POQ = 90°, ∠RPQ = 90° and PR = 24 cm, find RQ……**

Ans. In fig. right ΔPQR,we are given PR = 24 cm , In right ΔPOQ , OQ = 5√3 cm, OP = 5 cm

Applying Pythagoras theorem in triangle ΔPQR

RQ² = PQ² + PR²

RQ² = PQ² + 24²

RQ² = PQ² + 576…….(i)

Let’s find the value of PQ from ΔPOQ

In right ΔPOQ

PQ² = OP² + OQ²

PQ² = 5² + (5√3)²

PQ² = 25 + 75

PQ² = 100

Putting the value of PQ in equation (i)

RQ² = 100 + 576

RQ² = 676

RQ = 26

The value of RQ = 26 cm

Question number 16- 20, answer the following

**Q16.The n ^{th} term of an AP is (7 – 4n), then what is its common difference.**

Ans. Let n^{th} is represented by T_{n}

We are given

T_{n}= 7 – 4n

T_{1}= 7 – 4×1 = 3

T_{2}= 7 – 4×2 = -1

……

The common difference, d = T_{2}– T_{1}= -1 -3 = -4

**Q17. ΔABC is an isosceles triangle with AC = BC, if AB² = 2AC², then find the measure of ∠C.**

Ans. We are given ΔABC is an isosceles

AC = BC

If AB² = 2AC²

AB² = AC² + AC²

Putting AC = BC

AB² = AC² + BC²

Therefore ΔABC is a right triangle in which AB is hypotenuse with sides AC and BC and ∠C

= 90°

**Q18.The decimal expansion of** **will terminates after how many places of decimal**.

Ans. We are given the fraction

Converting the denominator into the power of 10, for this multiplying the denominator and numerator by 5

Hence the decimal expansion of the given fraction will terminate after 2 places.

**Q19.Form the quadratic polynomial whose zeroes are √2 + 3 and √2 – 3.**

Ans. If zeroes of a quadratic polynomial are given as α and β then the quadratic polynomial is given as follows

x² – (α + β)x + αβ

Putting the values α = √2 + 3, β = √2 – 3

x² – (√2 + 3 + √2 – 3)x + (√2 + 3) (√2 – 3)

x² – 2√2 x + √2² – 3²

x² – 2√2 x + 2 – 9

x²- 2√2 x – 7

Hence required quadratic polynomial is x²- 2√2 x – 7

**Q20. If -5 is a root of 2x² + px – 15 = 0, find value of p.**

Ans. We are given -5 is the root of 2x² + px – 15 = 0

Therefore x = -5 will satisfy the equation 2x² + px – 15 = 0

2×(-5)² + p(-5) – 15 = 0

50 -5p -15 = 0

-5p + 35 = 0

-5p = -35

p = 7

**Section B**

**Question number 21 -26 carry 2 marks each**

**Q21. Divide the polynomial (9x² + 12x + 10) by (3x + 2).**

Ans. Dividing (9x² + 12x + 10) by (3x + 2)

Therefore the quotient is (3x + 2) and the reminder is 6

**Q22.Find HCF and LCM of 540 and 72.**

Ans. Factorizing the numbers

540 = 2²× 3³×5

72 = 3²× 2³

HCF = 2² ×3² = 36

LCM = 2³×3³× 5 = 1080

**Q23. Solve for x and y **

**x -4y = 1, 2x – 5y = 5**

Ans.We are given the equations

x -4y = 1….(i) 2x – 5y = 5…..(ii)

Multiplyig the equation (i) by 2 , we get equation (iii)

2x -8y = 2…….(iii)

Subtractig the equation (iii) form (ii), we get

3y = 3

y = 1

Putting the value y = 1 in equation (i)

x -4× 1 = 1

x = 1 + 4 = 5

Hence,the value of x =5 and y = 1

**Q24. Find roots by quadratic formula**

**4x² + 4√3 x + 3 = 0 OR 2x² + x -4 = 0**

Ans. We are given the quadratic equation

4x² + 4√3 x + 3 = 0

Applying the quadratic equation formula

Putting the value a = 4, b= 4√3 and c = 3

Both of the roots of the given equation are -√3/2, -√3/2

Similarily the equation 2x² + x -4 = 0 can be solved by using the quadratic equation formula

**Q25.In an AP find missing terms**

Ans. We are given second term 13 and fourth term 3

Applying the n^{th} term formula of AP a_{n} = a + (n-1)d

a_{2} = a + (2-1)d, a_{4} = a + (4-1)d

a_{2} = a + d, a_{4} = a + 3d

13 =a + d…..(i) 3 = a + 3d…..(ii)

Subtracting equation (ii) from equation (i)

-2d = 10

d = -5

Putting the value of d = -5 in equation (i)

a -5 = 13

a = 18

Therefore the required term of AP are

**18,** 13, **8,** 3

**Q26. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.**

Ans.

**GIVEN: Δ** PQR

S is a point on PR and T is the point on QR

∠P = ∠RTS

**CONSTRUCTION:** Joining the point T and S, we get a ΔRTS

**TO PROVE:** ΔRPQ ∼ ΔRTS

**PROOF:** In ΔPQR and ΔRTS

∠P = ∠RTS (given)

∠R = ∠R (common)

ΔRPQ ∼ ΔRTS (AA rule)

Hence proved

**Section C**

**Question number 27 to 34 carry 3 marks each**

Q27. Prove that √3 is an irrational number.

Ans.

Let √3 is a rational number

Where a and b are co-prime number (i.e the numbers who has no common factor except to 1)

b√3 = a

Squaring both sides

3b² = a²……..(i)

It is clear from here that a² is divisible by 3

So, a will also be divisible by 3

Therefore putting a = 3c in equation (i) (where c is another positive integer)

3b² = (3c)² = 9c²

b² = 3c²……(ii)

It is clear from here that b² is divisible by 3

So, b will also be divisible by 3

From equation (i) and equation (ii), we concluded to the point that 3 is a common factor between a and b, hence a and b can’t be co-prime number

Since our assumption that √3 is a rational number is wrong, therefore √3 is an irrational number.

OR

**Find HCF of 378, 180, and 420 by the prime factorization method. Is HCF × LCM of three numbers equal to the product of three numbers.**

Ans. The given numbers are 378, 180, and 420

Factorizing the numbers

378 = 2 ×3³×7

180 = 2²× 3²×5

420 = 2²× 5×3×7

HCF = 2×3 = 6

LCM = 2²×3³×7×5= 3780

HCF × LCM = 6×3780= 22680

Product of numbers = 378 × 180 ×420 =28576800

HCF × LCM ≠ Product of numbers

Therefore HCF × LCM of three numbers is not equal to the product of three numbers.

OR

**Solve graphically 2x + 3y = 2, x – 2y = 8**

Ans. We are given the equation 2x + 3y = 2, x – 2y = 8

The solutions of both equations is as following

Solutions of the equation 2x + 3y = 2

x | 1 | 4 | -2 |

y | 0 | -2 | 2 |

Solutions of the equation x – 2y = 8

x | 0 | 2 | 4 |

y | -4 | -3 | -2 |

Drawing the graph of the equation

Hence both of the lies intersect at(4,-2), so the solution of both equations is (4,-2)

**Q28. Solve for x ad y**

Ans.

We are given the equations

Let

5a + b = 2……..(iii)

6a – 3b = 1……(iv)

Multiplying the equation (iv) by 5 and equation (iii) by 6 , we get equation (v) and equation (vi)

30 a + 6b = 12….(v) and 30a – 15b = 5…..(vi)

Subtractig equation (vi) from equation (v)

21b = 7

Putting the value of b in equation (iii)

As

x – 1 = 3, y – 2 = 3

x = 4, y = 5

**Q29.Find zeroes of quadratic polynomial 6x² + 2 + 7x and verify the relationship between zeroes and coefficients.**

Ans. We are given the quadratic polynomial 6x² + 2 + 7x

Rewriting the quadratic polynomial in standard form

6x² + 7x + 2

Factorizing the polynomial by the splitting up method

6x² + 4x + 3x+ 2

2x(3x + 2) + 1(3x + 2)

(3x + 2)(2x + 1)

3x + 2 = 0, 2x + 1 = 0

Therefore zeroes of quadratic polynomial(α and β)

The relationship between zeroes and coefficients of a quadratic polynomial ax² +bx + c are as follows

comparing the equation with standard quadratic equation

a = 6, b = 7, c= 2

Putting the values of a,b,c and zeroes of the polynomial in the relationship formula of zeroes and coefiicients.

Therefore the relationship of zeroes and coefficients is verified

**OR**

**If α and β are zeroes of a quadratic polynomial 4x² + 4x + 1, then form a quadratic polynomial whose zeroes are 2α and 2β.**

Ans. We are given the quadratic polynomial 4x² + 4x + 1

The relationship between zeroes(α and β) and coefficients of a quadratic polynomial ax² +bx + c are as follows

Comparing the equation with the standard quadratic equation

a = 4, b = 4 and c = 1

Puttig the values in the relationship formula

α + β = -4/4 and αβ = 1/4

α + β = -1….(i) αβ = 1/4…..(ii)

Let’s find out the quadratic polynomial whose zeroes are 2α and 2β by the following formula

x² -(sum of zeroes)x + product of zeroes

x² -(2α+ 2β)x + 2α×2β

x² -2(α+ β)x + 4 αβ

Putting the values of α+ β and αβ

x² -2(-1) x + 4 ×1/4

x² + 2x + 1

Hence the required quadratic polynomial is x² + 2x + 1

**Q30.Find the roots of the equations**

Ans. The given equation is

Simplifying it to get the standard quadratic equation

11x² -33x – 308 = -330

11x² -33x – 308 + 330 = 0

11x² -33x + 22 = 0

Factorisig it by splitting method

11x² -22x -11x + 22 = 0

11x(x – 2) – 11(x – 2) = 0

(x – 2) (11x -11) = 0

x = 2, x = 1

Hence the solution of the given equation is x = 2, x = 1

**Q31.Which term of AP 3,15,27,39…….. is 132 more than its 54 ^{th }term.**

Ans. The given AP is 3,15,27,39……..

The first term, a of AP is = 3, common difference, d = 15-3 = 12

Let n^{th }term of given AP is 132 more than its 54^{th }term.

a_{n}= a_{54}+ 132

a + (n-1)d = a + 53d + 132

(n-1)d – 53d = 132

d(n –1 -53) = 132

12(n – 54) = 132

n – 54 = 132/12

n = 54 + 11 = 65

Hence 65^{th }term is 132 more than the 54^{th }term of the given AP

OR

In an AP, given a = 3, n = 8, S = 192, Find d

Ans. We are given a = 3, n = 8, S = 192

Applying the formula for sum of an AP

4(6 +7d) = 192

24 + 28d = 192

28d = 192 – 24 = 168

d = 4

Hence the value of d is 4

**Q32. If a line is drawn parallel to one side of a triangle to intersect other two sides in district points .Prove that the other two sides are divided in the same ratio. **

Ans.

**GIVEN:** ΔABC in which DE ll BC

**CONSTRUCTION: **Drawing EF ⊥ BD and DG ⊥ AC

**TO PROVE: **

PROOF: Triangles ΔDBE and ΔDEC made on the same base and between the same parallel lines

∴arΔDBE = ar ΔDCE……(i)

arΔADE = 1/2 AD×EF

arΔDBE = 1/2 BD×EF

arΔADE = 1/2 DG×AE

arΔDCE = 1/2 CE×DG

Dividing equation (ii) by equation (iii)

From equation (i) arΔDBE = ar ΔDCE

AD × CE = BD × AE

Hence proved

**Q33. BL and CM are median of a triangle ABC, right-angled at A.**

**Prove that 4(BL² + CM²) = 5 BC².**

Ans.

**GIVEN:** ΔABC in which

∠A = 90°

BL and CM are median of a Δ ABC

**TO PROVE: **4(BL² + CM²) = 5 BC²

**PROOF:** Applying the Pythagoras theorem in right ΔABC, ΔAMC and ΔABL

BC² = AB² + AC²…….(i)

CM² = AM² + AC²….(ii)

BL² = AB² + AL²…..(iii)

Adding equation number (ii) and (iii)

CM² + BL²= AM² + AC² + AB² + AL²

From equation number (i) AC²= BC² -AB²

CM² + BL²= AM² + BC² -AB² + AB² + AL²

CM² + BL²= AM² + BC² + AL²

AM = AB/2 (CM is the median of the ΔABC on BC)

AL = AC/2 (BL is the median of ΔABC on AC)

CM² + BL² = AB²/4 + BC² + AC² /4

Multiplying both sides by 4

4 (CM² + BL²) = AB² + 4BC² + AC²

4 (CM² + BL²) = 4BC² + AC²+ AB²

Substituting from equation (i) BC² = AB² + AC²

4 (CM² + BL²) = 4BC² + BC²

4 (CM² + BL²) = 5BC² , Hence proved

**Q34. Solve for x**

Ans. We are given the equation

(3x + 4)(x + 4) = 4(x² + 3x + 2)

3x² + 12x + 4x + 16 = 4x² + 12x + 8

4x² -3x² – 4x -8 = 0

x² -4x -8 = 0

Factorising it by quadratic formula

x = 2 ±2√3

Hence x = 2 + 2√3, 2 – 2√3 are the required solutions of the given equation

**OR**

**Yash scored 40 marks in a test getting 3 marks for each right answer and losing 1 mark for each wrong answer . Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer ,then Yash would have scored 50 marks . How many questions were there in the test.**

Ans. Let Yash answered the questions correctly = x and answered the questions incorrectly = y

According to first condition of the question

3x – y = 40……(i)

According to second condition of the question

4x -2y = 50….(ii)

Multiplying equation (i) by 2, we get equation (iii)

6x -2y = 80…..(iii)

Subtracting equation (iii) from equation (ii)

-2x = -30

x = 15

Putting x = 15 in equation (i)

3×15- y = 40

45-y = 40

y = 5

The solutions of the equation is x = 15 and y = 5

Number of questions Yash answered correctly = 15 and answered incorrectly = 5

Hence total number of questions in the test were = x + y = 15 + 5 = 20

**Section D**

**Question numbers 35 to 40 carry 4 marks each.**

**Q35. If 4 times the 4 ^{th} term of an AP is equal to 18 times the 18^{th} term ,then find 22 nd term.**

Ans. n^{th} term of an AP is given by

a_{n }= a + (n -1) d

Where a is first term, d is the common difference of an AP

We are given in the question

4 times the 4^{th} term of an AP is = 18 times the 18^{th} term

4[a + (4-1)d ]=18[a + (18 -1)d]

4a + 12d = 18a +306d

294d = -14a

21d = -a

a = -21d

Now,22_{nd} term = a + 21d

Putting the value a = -21d

22_{nd} term = -21d+ 21d = 0

Hence the 22_{nd} term of the AP is 0.

**OR**

**How many terms of AP 24, 21, 18…..must be taken so that their sum is 78.**

Ans. The given AP is 24, 21, 18….

Common difference of the given AP,d is = 21 -24 = -3 and first term, a = 24

Let the number of terms whose sum is 78 are = n

Applying the following formula for the sum(S) of an AP

We are given S = 78

(51 – 3n)n =156

-3n² + 51n = 156

3n² – 51n + 156 = 0

n² – 17n + 52= 0

Factorising it by splitting up method

n² – 13n -4n+ 52= 0

n(n -13) – 4(n – 13) = 0

(n – 13)(n -4) = 0

n = 13, n = 4

Therefore the number of terms are either 13 or 4 whose sum is 78.

**Q36.In an obtuse ΔABC, ∠B is obtuse AD is perpendicular to CB produced,then prove that AC² = AB² + BC² + 2BC.BD**

Ans.

**GIVEN:** In an obtuse ΔABC, ∠B is obtuse

AD ⊥CB

**TO PROVE:**AC² = AB² + BC² + 2BC.BD

**PROOF:** In ΔADB and ΔADC, applying the Pythagoras theorem

AC² = DC² + AD²…..(i)

AB² = AD² + BD²…..(ii)

Subtracting equation (ii) from equation (i)

AC² – AB² = DC² – BD²

In figure, we have DC = BD + BC

AC² – AB² = (BD + BC)² – BD²

AC² – AB²= BD² + BC² + 2BC.BD – BD²

AC² – AB² = BC² + 2BC.BD

AC² = AB² + BC² + 2BC.BD, Hence proved

OR

The perpendicular from A on side BC of a ΔABC intersect BC at D, such that DB = 3CD,prove that 2AB² = 2AC² + BC²

**Class 10 Maths NCERT Solutions from chapter 1-15**

**Q37.Solve for x**

Ans. The given equation is

ax + bx +x² = -ab

x² + ax + bx + ab = 0

x(x + a) + b( x +a) = 0

(x + a)(x + b) = 0

x = -a, -b

Hence the solutions of the equation is x = -a and x = -b

**Q38.A boat takes 4 hr to go 44 km downstream and it can go 20 km upward in the same time . Find speed of stream and that of the boat in still water.**

Ans.Let the speed of boat in still water = x and speed of stream = y

The speed of the boat in downward stream = speed of boat + speed of stream=x+y

The speed of the boat in downward stream = Distace covered in downwardstream/Time taken = 44/4 = 11 km/h

∴ x + y = 11……(i)

The speed of the boat in upward stream = speed of boat -speed of stream=x-y

The speed of the boat in downward stream = Distance covered in upward stream/Time taken = 20/4 = 5 km/h

∴ x – y = 5……(ii)

Adding equation (i) and (ii)

2x = 16

x = 8

Putting the value of x in equation (i)

8 + y = 11

y = 3

Hence the speed of boat in still water is 8 km/h and speed of stream is 3 km/h

**Q39. Find all zeroes of 2x ^{4}– 3x^{3}-3x^{2}+ 6x -2, if two of its zeroes are √2 and -√2.**

Ans. We are given the polynomial 2x^{4}– 3x^{3}-3x^{2}+ 6x -2

If two of zeroes are √2 and -√2 then (x -√2)(x +√2) =x² -2 will be one of its factor.

Therefore further factorizing the polynomial by the reminder’s theorem

2x^{4}– 3x^{3}-3x^{2}+ 6x -2

2x^{2}(x² -2) +4x²- 3x(x²-2)-6x-3x²+ 6x -2

2x^{2}(x² -2) – 3x(x²-2)-3x² +4x²-3x²- 2

2x^{2}(x² -2) – 3x(x²-2) + (x² – 2)

(x² -2)(2x^{2 }– 3x +1)

(x² -2)[2x^{2 }– 2x-x +1]

(x² -2)[2x(x -1) -1(x -1)]

(x² -2)(x -1)(2x -1)

(x² -2) = 0⇒x =±√2, x =1 and x = 1/2

Hence other two zeroes of the given polynomial are 1 and 1/2

**Q40. Prove that 1/√2 is an irrational number.**

Ans. Let 1/√2 is a rational number

So, 1/√2 = a/b (Where a and b are co-prime numbers)

a√2 = b

Squaring both sides

2a² = b²……(i)

b² is divisible by 2

∴ b is also divisible by 2

b is also written in the form of 2c, where c is another positive integer

b = 2c

Putting b = 2c in equation (i)

2a² =(2c)²= 4c²

a² = 2c²……(ii)

a² is divisible by 2

∴ a is also divisible by 2

Equation (i) and equation (ii) implies that 2 is a common factor between a and b,so our assumption is wrong that a and b are co-prime number

Therefore 1/√2 can’t be a rational number, so 1/√2 is an irrational number

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