Class 10 CBSE Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’

Class 10 CBSE Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’ is one of the important parts of algebra. The quadratic equations are used to evaluate different kinds of physical quantities like speed, length, height, profit, loss, and many other scientific calculations. The standard form of a quadratic equation is ax² + bx + c =0, where a ,b and c are numbers such that a≠0, x is variable with power 2. Here you can study 10 most important questions which we have selected from chapter number 4 keeping in view your CBSE board exams.You can study in future study point NCERT solutions, sample papers, assignments of maths and science, solutions of previous question papers, articles related to science and maths, most important questions with solutions, and posts related to your carrier.

Class X Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’

Q1. Find the roots of the following quadratic equations.

${\color{DarkBlue}\boldsymbol{\left ( a \right )\sqrt{2}x^{2}+7x + 5\sqrt{2}=0}}$

${\color{DarkBlue} \boldsymbol{\left ( b \right )2x^{2}-x+\frac{1}{8}=0}}$

Answer.(a) The given equation is as following

√2x ² + 7x + 5√2 = 0

The product of √2 × 5√2 =10=5×2

√2x ² + 5x +2x+ 5√2 = 0

x(√2x + 5) + √2(√2x + 5)=0

(√2x + 5)(x + √2) =0

${\color{DarkBlue} \boldsymbol{x=-\frac{5}{\sqrt{2}},-\sqrt{2}}}$

(b) The given equation is as following

${\color{DarkBlue} \boldsymbol{2x^{2}-x+\frac{1}{8}=0}}$

16x² –8x + 1 = 0

The product of 16 and 1 is 16 =4 × 4

16x² –4x –4x + 1 = 0

4x(4x –1) – 1( 4x –1) =0

(4x –1)(4x –1) = 0

${\color{DarkBlue} \boldsymbol{x=\frac{1}{4},\frac{1}{4}}}$

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Q2.A cottage industry produces certain number of toys in a day .The cost of production of each toys (in rupees) was found to be 55 minus the number of toys produced in a day.On a particular day,the total cost of production was Rs 750 .Find out the number of toys produced on that day.

Answer. Let the number of toys produced in a day = x

The cost of each toy according to the question is = 55 –x

On a day the cost of production is = Rs 750

The cost of production in a day = number of toys produced in a day × cost of a toy

According to question

x(55 –x) = 750

–x² + 55x –750 = 0

x² – 55x +750 = 0

750 = 5 × 3 × 5 ×5×2= 25 × 30

x² – 25x –30x +750 = 0

x(x– 25) – 30(x – 25) = 0

(x– 25)(x – 30) = 0

x = 25, 30

Hence number of the toys produced on that day 25 or 30.

If the number of toys produced on that day = 25 then the cost of a toy is 55 – 25 =30(rupees) and if the number of toys produced on that day=30 then the cost of a toy is 55 – 30=25(rupees) in both cases the cost production is Rs 750.

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Q3. Find two consecutive positive integers, the sum of whose squares is 365.

Answers. Let one of the positive integers is x

Next positive integer is = x+ 1

According to question the sum of squares of these number = 365

(x +1)² + x² = 365

x² + 1 + 2x + x²= 365

2x² + 2x –364 = 0

x² + x –182 = 0

182 = 2 ×13×7 =14 × 13

x² + 14x –13x –182 = 0

x(x + 14) – 13(x – 14) = 0

(x + 14)(x –13) = 0

x =–14, 13

Neglecting negative integers,therefore two consecutive positive integers are 13 and 14 .

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Q4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm , find the other two sides.

Answer. Let the base of given right Δ is x then according to question its altitude is x – 7

The hypotenuse of given Δ is =13 cm

Applying Pythagoras theorem

(x – 7)² + x² = 13²

x² + 49 – 14x + x²= 169

2x² – 14x – 120 = 0

x² – 7x – 60 = 0

60= 2 ×3×2×5 = 12×5

x² – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x +5) = 0

x = 12, – 5

Side can not be negative so neglecting negative sign

Therefore altitude of given Δ is 12 cm and base is 12 – 7 = 5 cm.

Q5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particle day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day Rs90.Find the number of articles produced and the cost of each article.

Answer. Let the number of articles produced on that day = x

Since the cost of each article was observed on that day 3 more than twice the number of articles produced

So, the cost of each article is = 2x + 3

The total cost of production as per the question=x(2x + 3)

It is given in question the total cost of production on that day= Rs90

According to question

x(2x + 3) = 90

2x² + 3x –90 = 0

2 × 90 = 2×3×3×5 ×2 =15×12

2x² + 15x –12x –90 = 0

x(2x + 15) –6(2x +15) =0

(2x + 15)(x –6) = 0

${\color{DarkBlue} \boldsymbol{x=-\frac{15}{2},6}}$

The number of articles produced can’t be negative so number of articles produced on that day equal to = 6 and the cost of each article is = Rs (2x +3 )=Rs(2×6 + 3) = Rs15

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Q6. Find the roots of the following equation by the method of complete square method .

2x² –7x + 3 = 0

Answer. The identity of complete square is (a +b)² = a² + 2ab+ b²

Let’s make the given equation 2x² –7x + 3 = 0 a complete square

⇒(√2x)² –7x + 3

⇒2ab = 7x

⇒2×√2x×b=7x

${\color{DarkBlue} \boldsymbol{b=\frac{7}{2\sqrt{2}}}}$

Adding and subtracting ${\color{DarkBlue} \boldsymbol{b^{2}=\frac{49}{8}}}$ to the given equation

${\color{DarkBlue} \boldsymbol{\left ( \sqrt{2}x \right )^{2}-7x+\left ( \frac{7}{2\sqrt{2}} \right )^{2}-\left ( \frac{7}{2\sqrt{2}} \right )^{2}+3}}$

${\color{DarkBlue} \boldsymbol{\left ( \sqrt{2}x-\frac{7}{2\sqrt{2}} \right )^{2}-\frac{49}{8}+3}}$

${\color{DarkBlue} \boldsymbol{\left ( \sqrt{2}x-\frac{7}{2\sqrt{2}} \right )^{2}-\frac{25}{8}}}$

${\color{DarkBlue} \boldsymbol{\left ( \sqrt{2}x-\frac{7}{2\sqrt{2}} \right )^{2}-\left ( \frac{5}{2\sqrt{2}} \right )^{2}}}$

${\color{DarkBlue} \boldsymbol{=\left ( \sqrt{2}x-\frac{7}{2\sqrt{2}}+\frac{5}{2\sqrt{2}} \right )\left ( \sqrt{2}x-\frac{7}{2\sqrt{2}}-\frac{5}{2\sqrt{2}} \right )}}$

${\color{DarkBlue} \boldsymbol{\left ( \sqrt{2}x-\frac{1}{\sqrt{2}} \right )\left ( \sqrt{2}x-3\sqrt{2} \right )=0}}$

${\color{DarkBlue} \boldsymbol{x=\frac{1}{2},3}}$

Q7. The sum of reciprocal of Rehman’s ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer. Let Rehman’s present age is = x

His age 3 years ago = x –3

Reciprocal of his age 3 years ago = ${\color{DarkBlue} \boldsymbol{\frac{1}{x-3}}}$

His age 5 years from now = x +5

Reciprocal of his age 5 years from now = ${\color{DarkBlue} \boldsymbol{\frac{1}{x+5}}}$

According to the question we have

${\color{DarkBlue} \boldsymbol{\frac{1}{x-3}}+\boldsymbol{\frac{1}{x+5}=\frac{1}{3}}}$

${\color{DarkBlue} \boldsymbol{\frac{x-3 +x+5}{\left ( x-3 \right )\left ( x+5 \right )}=\frac{1}{3}}}$

x² +5x –3x – 15 = 6x +6

x² – 4x –21 = 0

21 = 7 × 3

x² – 7x +3x –21 = 0

x(x –7) + 3(x –7) =0

(x –7)(x + 3) = 0

x = 7, –3

The age can’t be negative so Rehman’s present age is 7 years.

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Q8. The diagonal of a rectangular field is 60 m more than its shorter side. If the longer side is 30 m more than the shorter side. Find the sides of the field.

Answer. Let the shorter side of the rectangle ABCD is = x

The diagonal of rectangle = x + 60

Longer side of rectangle = x +30

ΔADC is a right triangle so applying Pythagoras theorem

(x +60)² = x² + (x +30)²

x² + 3600 + 120x = x² + x² +900 + 60x

x²–60x –2700= 0

a = 1, b = –60, c = –2700

Applying quadratic formula

${\color{DarkBlue} \boldsymbol{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \boldsymbol{x=\frac{60\pm \sqrt{3600-4\times 1\times -2700}}{2\times 1}}}$

${\color{DarkBlue} \boldsymbol{x=\frac{60\pm \sqrt{14400}}{2}}}$

${\color{DarkBlue} \boldsymbol{x=\frac{60\pm 120}{2}}}$

${\color{DarkBlue} \boldsymbol{x=\frac{60+ 120}{2},\frac{60-120}{2}}}$

x =90,–30

Length can’t be negative so length of shorter side of the rectangle=90 m

Length of longer side = 90 + 30 = 120 m

Q9. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number.Find the two numbers.

Answer. Let the larger number is = x

Its square is = x²

Since we are given the difference of squares of both numbers

Square of larger number – Square of smaller number =180

x² = Square of smaller number +180

Since it is given that the square of the smaller number is 8 times the larger number

So,

Square of smaller number = 8x

x²= 8x + 180=0

x² – 8x – 180= 0

180 = 2 ×3×3×5×2= 18× 10

x² –18x +10x – 180= 0

x(x –18) + 10(x –18) = 0

(x –18)(x + 10) = 0

x = 18, –10

If we suppose x =–10 since we are given that square of the smaller number is 8 times of larger number,it gives the square of smaller number =8 ×–10=–80, the square of any number can’t be negative so the required larger number is = 18

square of smaller number =8x= 8 ×18 = 144

smaller number = ${\color{DarkBlue} \boldsymbol{\sqrt{144}=12}}$

Therefore the larger number and smaller numbers are 18 and 12 respectively.

Q10.A train travels 360 km at a uniform speed.If the speed had been 5 km/h more , it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer. Let the speed of the train = x km/h

Distance covered by the train = 360 km

Time is taken by the train with x km/h = ${\color{DarkBlue} \boldsymbol{\frac{360}{x}}}$

The modified speed of the train = (x +5) km/h

Time taken by the train with (x +5) km/h = ${\color{DarkBlue} \boldsymbol{\frac{360}{x+5}}}$

According to question

${\color{DarkBlue} \boldsymbol{\frac{360}{x}-\frac{360}{x+5}=1}}$

${\color{DarkBlue} \boldsymbol{\frac{1}{x}-\frac{1}{x+5}=\frac{1}{360}}}$

${\color{DarkBlue} \boldsymbol{\frac{x+5-x}{x\left ( x+5 \right )}=\frac{1}{360}}}$

x² + 5x –1800 = 0

1800 = 2 × 3 ×3×5×2×2×5= 45× 40

x² + 45x –40x –1800 = 0

x(x + 45) – 40(x + 45) = 0

(x + 45)(x – 40) = 0

x = – 45, 40

The speed of the train can’t be negative,so the speed of the train is 40 km/h.

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Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
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Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

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Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
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