Exercise 4.3 (NCERT) – Quadratic Equations, Class 10 – All Solutions

01 Apr
Class 10, Class 10 - Chapter 4, Class 10 - Quadratic Equations, , , ,

Exercise 4.3 (NCERT) - Quadratic Equations, Class 10

Chapter 4 Exercise 4.3 Class 10

Below are the Quick links for all questions of Exercise 4.3, Quadratic Equations, Class 10.

Click a link to view the solution of corresponding question.

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NCERT

  • E.g. 7) 2x2 - 4x + 3 = 0
  • Ex: 4.3, Q1) 2x2 - 3x + 5 = 0
  • Ex: 4.3, Q3) 2x2 - 6x + 3 = 0

NCERT Exemplar

  • Ex: 4.2, Q1 (i) x2 - 3x + 4 = 0
  • Ex: 4.2, Q1 (ii) 2x2 + x - 1 = 0
  • Ex: 4.2, Q1 (iv) 3x2 - 4x + 1 = 0
  • Ex: 4.3, Q1 (i) 2x2 - 3x - 5 = 0
  • Ex: 4.3, Q1 (ii) 5x2 + 13x + 8 = 0
  • Ex: 4.3, Q1 (iii) -3x2 + 5x + 12 = 0
  • Ex: 4.3, Q1 (iv) -x2 + 7x - 10 = 0

More Questions

  • 1) 4x2 - 4 = 0
  • 2) x2 = 0
  • 3) x2 - 4 = 0
  • 4) x2 + 4 = 0
  • 5) 4x2 + 4x + 1 = 0
  • 6) 8x2 + 8x + 2 = 0
    Observe this equation has same roots as that of question 5.
  • 7) x2 + 5x + 6 = 0
  • 8) 20x2 - 15x = 0
  • 9) x2 - x + 6 = 0
  • 10) x2 - 10x + 25 = 0

Enter the coefficeints of QE

Coeff. the x2

Coeff. the x

Constant

\( \begin{aligned}\text{Coeff of x}^2\text{ cannot be 0}\end{aligned}\)

Answer

Solution

Comparing the equation with
ax2 + bx + c = 0, we get

Discriminant (D)

Nature of Roots

D > 0 ⇒ Two Real and Distinct Roots

D = 0 ⇒ Two Real and Equal Roots

D < 0 ⇒ Two Complex Roots

Roots

Below are the solutions of every question of exercise 4.3 of chapter 4, Quadratic Equations, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0

𝑎=2,𝑏=−3,𝑐=5

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−3)2−4(2)(5)=9−40=−31

Nature of Roots

D < 0 ⇒ No Real Roots

(ii) 3x2 – 4√3x + 4 = 0

Comparing the equation with ax2 + bx + c = 0 

𝑎=3,𝑏=−4√3,𝑐=4

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−43‾√)2−4(3)(4)=(−43‾√)2−4(3)(4)=(16×3)−48=48−48=0

Nature of Roots

D = 0 ⇒ Two Equal Real Roots

Roots

𝑥=−𝑏±𝐷‾‾√2𝑎=−(−43‾√)±0‾√2×3=43‾√+02×3,43‾√−02×3=23‾√3,23‾√3

(iii) 2x2 – 6x + 3 = 0

Comparing the equation with ax2 + bx + c = 0 

𝑎=2,𝑏=−6,𝑐=3

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−6)2−4(2)(3)=36−24=12

Nature of Roots

D > 0 ⇒ Two Distinct Real Roots

Roots

\( \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-6) \pm \sqrt{12}}{2 \times 2}\\ &= \frac{6 \pm 2\sqrt{3} }{4}\\ &= \frac{3 \pm \sqrt{3} }{2}\\ \end{aligned}\)

Question 2

Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x2 + kx + 3 = 0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 2, b = k, c = 3 \end{aligned}\)

We are given that the equation has two equal roots, this means

Discriminant = 0

\( \begin{aligned} \Rightarrow b^2-4ac &= 0\\ \Rightarrow k^2-4(2)(3) &=0\\ \Rightarrow k^2-24 &=0\\ \Rightarrow k^2 &=24\\ \Rightarrow k &=\pm\sqrt{24}\\ \Rightarrow k &=\pm2\sqrt{6}\\ \end{aligned}\)

(ii) kx (x – 2) + 6 = 0

𝑘𝑥(𝑥–2)+6⇒𝑘𝑥2–2𝑘𝑥+6=0=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = k, b = -2k, c = 6 \end{aligned}\)

We are given that the equation has two equal roots, this means

Discriminant = 0

⇒𝑏2−4𝑎𝑐⇒(−2𝑘)2−4(𝑘)(6)⇒4𝑘2−24𝑘⇒4𝑘(𝑘−6)=0=0=0=0
⇒  4𝑘=0  ⇒  𝑘=0  𝑜𝑟  𝑘−6=0𝑜𝑟  𝑘=6

We can clearly observe that the value of k cannot be 0.

Let us put k = 0 in the given equation.

𝑘𝑥(𝑥–2)+6⇒  0×𝑥(𝑥–2)+6⇒  0+6⇒  6=0=0=0=0

This is not possible. Hence the value of k cannot be 0.

Hence k = 6.

Important: Always verify your answer by putting the value of k in the given equation to see if we get valid quadratic equation or not and if time allows, quickly verify if discriminant is 0 or not.

Question 3

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

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For Rectangular Mango Grove

𝐿𝑒𝑡  𝑏𝑟𝑒𝑎𝑑𝑡ℎ  (𝑏)⇒𝐿𝑒𝑛𝑔𝑡ℎ  (𝑙)=𝑥  𝑚𝑒𝑡𝑒𝑟=2𝑥  𝑚𝑒𝑡𝑒𝑟

Area = 800 m2

\( \begin{aligned} \Rightarrow l \times b &= 800\\ \Rightarrow 2x \times x &= 800\\ \Rightarrow 2x^2 &= 800\\ \Rightarrow x^2 &= 400\\ \Rightarrow x &= \pm\sqrt{400}\\ \Rightarrow x &= \pm20\\\\ \Rightarrow breadth = x &= \pm20 \end{aligned}\)

Reject the -ve value because breadth cannot be -ve.

Hence 

𝑏𝑟𝑒𝑎𝑑𝑡ℎ=𝑥𝑙𝑒𝑛𝑔𝑡ℎ=2𝑥=20  𝑚𝑒𝑡𝑒𝑟=2×20=40  𝑚𝑒𝑡𝑒𝑟

Conclusion: Yes, it is possible to design the rectangular mango grove whose length is twice its breadth.

Question 4

Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

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It is recommended to solve such questions by making a table as shown below.

Let x and y are the current (today’s) ages of the two friends.

Today 4 Years Ago
Age (yrs) of Friend 1
x
x – 4
Age (yrs) of Friend 2
y
y – 4
Condition
x + y = 20
(x – 4) * (y – 4) = 48

We got two equations from the given data.

𝑥+𝑦=20  …(1)(𝑥−4)×(𝑦−4)=48  …(2)

Get the value of y in terms of x from equation 1.

(Applying substitution method)

𝑥+𝑦⇒𝑦=20=20−𝑥

Put the value of y in equation 2.

(𝑥−4)×(𝑦−4)(𝑥−4)×(20−𝑥−4)(𝑥−4)×(16−𝑥)16𝑥−64−𝑥2+4𝑥−𝑥2+20𝑥−64−48−𝑥2+20𝑥−112𝑥2−20𝑥+112=48=48=48=48=0=0=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 1, b = -20, c = 112 \end{aligned}\)

Discriminant (D)

\( \begin{aligned} D &= b^2-4ac\\ &= (-20)^2-4(1)(112)\\ &= 400-448\\ &= -48\\ \end{aligned}\)

D < 0 ⇒ No Real Roots

Conclusion: This means the situation provided in the question is not possible.

Question 5

Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

For Rectangular Park

𝐿𝑒𝑡  𝐿𝑒𝑛𝑔𝑡ℎ𝐵𝑟𝑒𝑎𝑑𝑡ℎ=𝑥=𝑦

Perimeter = 80 m

⇒2(𝑥+𝑦)⇒𝑥+𝑦⇒𝑦=80=40=40−𝑥  …(1)

Area = 400 m2

\( \begin{aligned} &\Rightarrow x \times y = 400 \end{aligned}\)
\( \begin{aligned} &\text{Substitute the value of } y \text{ from equation (1)} \end{aligned}\)
⇒𝑥×(40−𝑥)⇒40𝑥−𝑥2⇒𝑥2−40𝑥+400=400=400=0

Comparing the equation with ax2 + bx + c = 0 

\(\begin{aligned} &a = 1, b = -40, c = 400 \end{aligned}\)

Discriminant (D)

𝐷=𝑏2−4𝑎𝑐=(−40)2−4(1)(400)=1600−1600=0

D = 0 ⇒ Two Equal Real Roots

\( \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-40) \pm \sqrt{0}}{2 \times 1}\\ &= \frac{40 \pm 0}{2}\\ &= \frac{40 + 0}{2}, \frac{40 – 0}{2}\\ &= \frac{40 }{2}, \frac{40 }{2}\\ &= 20, 20 \end{aligned}\)

Hence

\( \begin{aligned} Length = x &= 20 m\\ Breadth = 40 – x &= 20 m \end{aligned}\)
\( \begin{aligned} \end{aligned}\)