NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.3 are provided here to help students in their Class 10 exam preparations. These solutions are designed by subject experts to guide students through the process of solving quadratic equations and word problems related to quadratic equations, using different methods such as factorization, quadratic formula, and completing the square.

Exercise 4.3 focuses on solving quadratic equations that arise from real-life scenarios and word problems. It helps students apply the concept of quadratic equations to practical situations, thus deepening their understanding of the subject. The exercise also introduces students to new problem-solving techniques.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.3

Quadratic Equations

Medium

Q.

$\begin{array}{l}{Find the roots of the following quadratic equations}\\ {by factorisation:}\\ {  (i) }{\mathrm{x}}^{2}-3\mathrm{x}-10=0{               (ii) }2{\mathrm{x}}^{2}+\mathrm{x}-6=0\\ {(iii) }\sqrt{2}{\mathrm{x}}^{2}+7\mathrm{x}+5\sqrt{2}=0{{        (iv) 2x}}^{2}-\mathrm{x}+\frac{1}{8}=0\\ {{(v) 100x}}^{2}-20\mathrm{x}+1=0\end{array}$

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Quadratic Equations

Medium

Q.

$\begin{array}{l}\text{Find the roots of the following quadratic equations,}\\ \text{if they exist, by the method of completing the square:}\\ \text{ (i) }2{\mathrm{x}}^2-7\mathrm{x}+3=0\text{ (ii)\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^2+\mathrm{x}-4=0\\ \text{\hspace{0.17em}\hspace{0.17em}(iii) 4}{\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+3=0\text{ (iv) }2{\mathrm{x}}^2+\mathrm{x}+4=0\end{array}$

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Quadratic Equations

Medium

Q.

$\begin{array}{l}{Find the roots of the following quadratic equations,}\\ {by applying quadratic formula:}\\ {   (i) }2{\mathrm{x}}^{2}-7\mathrm{x}+3=0{          (ii)\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}+\mathrm{x}-4=0\\ {\hspace{0.17em}\hspace{0.17em}(iii) 4}{\mathrm{x}}^{2}+4\sqrt{3}\mathrm{x}+3=0{     (iv) }2{\mathrm{x}}^{2}+\mathrm{x}+4=0{ }\end{array}$

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Quadratic Equations

Medium

Q.

$\begin{array}{l}\text{The sum of the reciprocals of Rehman's ages, (in years) 3 years ago }\\ \text{and 5 years from now is }\frac{1}{3}.\text{ Find his present age.}\end{array}$

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Quadratic Equations

Medium

Q.

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

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Quadratic Equations

Medium

Q.

$\begin{array}{l}\text{The diagonal of a rectangular field is 60 metres more }\\ \text{than the shorter side. If the longer side is 30 metres}\\ \text{more than the shorter side, find the sides of the field.}\end{array}$

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Quadratic Equations

Medium

Q.

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.

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Quadratic Equations

Medium

Q.

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.

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Quadratic Equations

Medium

Q.

$\begin{array}{l}\begin{array}{l}\text{Two water taps together can fill a tank in \hspace{0.33em}}9\frac{3}{8}\text{\hspace{0.33em}hours. }\\ \text{The tap of larger diameter takes 10 hours less than}\end{array}\\ \begin{array}{l}\text{the smaller one to fill the tank separately. }\\ \text{Find the time in which each tap can separately fill the tank.}\end{array}\end{array}$

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Quadratic Equations

Medium

Q.

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

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Quadratic Equations

Medium

Q.

Sum of the areas of two squares is 468 m­². If the difference of their perimeters is 24 m, find the sides of the two squares.

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These solutions are aligned with the latest CBSE syllabus and offer clear, step-by-step guidance to solve problems efficiently and excel in the exams.

Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square.

(i) $2x^2 - 7x + 3 = 0$

• Solution: 1. Divide by 2: $x^2 - \frac{7}{2}x + \frac{3}{2} = 0$.

  2. Add/Subtract $(\frac{1}{2} \text{ coeff. of } x)^2 = (\frac{7}{4})^2$:

  $x^2 - \frac{7}{2}x + (\frac{7}{4})^2 = -\frac{3}{2} + \frac{49}{16}$

  3. $(x - \frac{7}{4})^2 = \frac{25}{16}$

  4. Taking square root: $x - \frac{7}{4} = \pm \frac{5}{4}$

• Answer: $x = 3$ and $x = \frac{1}{2}$.

Q2. Find the roots of the quadratic equations given in Q1 by applying the quadratic formula.

• Explanation: Use $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Example (i): $D = (-7)^2 - 4(2)(3) = 25$.

  $x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$.

• Answer: $x = 3, 1/2$.

Q3. Find the roots of the following equations:

(i) $x - \frac{1}{x} = 3, x \neq 0$

• Solution: $x^2 - 3x - 1 = 0$.

  Using Quadratic Formula: $x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2} = \frac{3 \pm \sqrt{13}}{2}$.

  (ii) $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$

• Solution: $\frac{(x-7)-(x+4)}{(x+4)(x-7)} = \frac{11}{30} \implies \frac{-11}{x^2 - 3x - 28} = \frac{11}{30}$.

  Simplifies to: $x^2 - 3x + 2 = 0 \implies (x-1)(x-2) = 0$.

• Answer: $x = 1, 2$.

Q4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

• Equation: $\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \implies x^2 - 4x - 21 = 0$.

• Solving: $(x-7)(x+3) = 0$.

• Answer: Rehman's present age is 7 years.

Q5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

• Equation: Let Maths $= x$. $(x+2)(30-x-3) = 210 \implies x^2 - 25x + 156 = 0$.

• Solving: $(x-12)(x-13) = 0$.

• Answer: If Maths $= 12$, English $= 18$. If Maths $= 13$, English $= 17$.

Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

• Equation: $(x+60)^2 = x^2 + (x+30)^2$ (By Pythagoras Theorem).

  $x^2 - 60x - 2700 = 0$.

• Solving: $(x-90)(x+30) = 0$.

• Answer: Shorter side = 90m, Longer side = 120m.

Q7. 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.

• Equation: $x^2 - y^2 = 180$ and $y^2 = 8x \implies x^2 - 8x - 180 = 0$.

• Solving: $(x-18)(x+10) = 0$.

• Answer: The numbers are (18, 12) or (18, -12).

Q8. 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.

• Equation: $\frac{360}{x} - \frac{360}{x+5} = 1 \implies x^2 + 5x - 1800 = 0$.

• Solving: $(x+45)(x-40) = 0$.

• Answer: Speed of the train = 40 km/h.

Q9. Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

• Equation: $\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} \implies 8x^2 - 230x + 750 = 0$.

• Solving: $(4x-15)(x-25) = 0 \implies x = 25$ ($3.75$ is rejected).

• Answer: Smaller tap = 25 hours, Larger tap = 15 hours.

Q10. An express train takes 1 hour less than a passenger train to travel 132 km...

• Equation: $\frac{132}{x} - \frac{132}{x+11} = 1 \implies x^2 + 11x - 1452 = 0$.

• Solving: $(x+44)(x-33) = 0$.

• Answer: Passenger train = 33 km/h, Express train = 44 km/h.

Q11. Sum of the areas of two squares is $468\text{ m}^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.

• Equation: $x^2 + y^2 = 468$ and $4x - 4y = 24 \implies x - y = 6$.

  Substituting $x = y+6$: $(y+6)^2 + y^2 = 468 \implies y^2 + 6y - 216 = 0$.

• Solving: $(y+18)(y-12) = 0$.

• Answer: Sides are 12m and 18m.

FAQs: Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.3

Q1. What is the focus of Exercise 4.3?
Answer:
Exercise 4.3 focuses on solving word problems that involve quadratic equations. Students are required to form quadratic equations from given situations and solve them using factorization, quadratic formula, or completing the square.

Q2. How do I solve quadratic equations in Exercise 4.3?
Answer:
To solve quadratic equations:

1. Form the equation from the problem’s context.

2. Use factorization if the equation is factorable, or use the quadratic formula

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$x=2a−b±b2−4ac​​ if factorization is difficult.

3. For some problems, use completing the square to solve the equation.

Q3. What methods are used in Exercise 4.3?
Answer:
The methods used to solve quadratic equations in this exercise are:

• Factorization Method: Breaking down the quadratic equation into factors.

• Quadratic Formula: Using the formula to find the roots.

• Completing the Square: A method of transforming the quadratic equation into a perfect square trinomial.

Q4. How do quadratic equations apply to real-life problems?
Answer:
Quadratic equations are useful in real-life situations such as:

• Projectile motion problems (e.g., calculating the height of an object).

• Profit and loss in business calculations.

• Geometry problems (e.g., finding the dimensions of shapes).

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide step-by-step explanations, helping students apply quadratic equations to word problems effectively. Practicing these solutions allows students to build confidence and improve their problem-solving skills, ensuring success in the Class 10 exams.