Quadratic Equations | Exercise 4.1

Question 2

  1. Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 m2528\text{ m}^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • A quadratic equation in variable xx is an equation of the standard form ax2+bx+c=0ax^2 + bx + c = 0, where a,b,ca, b, c are real numbers and a≠0a \neq 0.
  • To formulate a quadratic equation from a word problem:
    • Identify the unknown quantity to be found and represent it using a variable (usually xx).
    • Express all other given quantities in terms of xx.
    • Form an equation using the given condition and simplify it into standard form.

(i) The area of a rectangular plot is 528 m2528\text{ m}^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Step 1 · Formulate the Quadratic Equation

Let the breadth of the rectangular plot be x mx\text{ m}.Length=(2x+1) m\text{Length} = (2x + 1)\text{ m}

Area of rectangle=Length×Breadth=528 m2\text{Area of rectangle} = \text{Length} \times \text{Breadth} = 528\text{ m}^2

x(2x+1)=528x(2x + 1) = 528 2x2+x=5282x^2 + x = 528 2x2+x−528=02x^2 + x - 528 = 0
Answer

(i) 2x2+x−528=02x^2 + x - 528 = 0, where xx is the breadth of the plot in metres.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Step 1 · Formulate the Quadratic Equation

Let the first positive integer be xx.

The consecutive positive integer is (x+1)(x + 1).

Given that their product is 306306

x(x+1)=306x(x + 1) = 306 x2+x=306x^2 + x = 306 x2+x−306=0x^2 + x - 306 = 0
Answer

(ii) x2+x−306=0x^2 + x - 306 = 0, where xx is the smaller positive integer.

(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.

Step 1 · Formulate the Quadratic Equation

Let Rohan's present age be x yearsx\text{ years}.

Mother’s present age=(x+26) years\text{Mother's present age} = (x + 26)\text{ years}

After 3 years3\text{ years}

  • Rohan’s age=(x+3) years\text{Rohan's age} = (x + 3)\text{ years}
  • Mother’s age=(x+26+3)=(x+29) years\text{Mother's age} = (x + 26 + 3) = (x + 29)\text{ years}

Given that the product of their ages after 3 years3\text{ years} is 360360

(x+3)(x+29)=360(x + 3)(x + 29) = 360 x2+29x+3x+87=360x^2 + 29x + 3x + 87 = 360 x2+32x+87=360x^2 + 32x + 87 = 360 x2+32x+87−360=0x2+32x−273=0\begin{aligned} x^2 + 32x + 87 - 360 &= 0 \\[0.6em] x^2 + 32x - 273 &= 0 \end{aligned}
Answer

(iii) x2+32x−273=0x^2 + 32x - 273 = 0, where xx is Rohan's present age in years.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Step 1 · Formulate the Quadratic Equation

Let the uniform speed of the train be x km/hx\text{ km/h}.

Distance=480 km\text{Distance} = 480\text{ km}

Time taken=DistanceSpeed=480x hours\text{Time taken} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{480}{x}\text{ hours}

When speed is decreased by 8 km/h8\text{ km/h}

  • New speed=(x−8) km/h\text{New speed} = (x - 8)\text{ km/h}
  • New time=480x−8 hours\text{New time} = \dfrac{480}{x - 8}\text{ hours}

According to the question

480x−8=480x+3\dfrac{480}{x - 8} = \dfrac{480}{x} + 3 480x−8−480x=3\dfrac{480}{x - 8} - \dfrac{480}{x} = 3 480x−480(x−8)x(x−8)=3\dfrac{480x - 480(x - 8)}{x(x - 8)} = 3 480x−480x+3840x2−8x=3\dfrac{480x - 480x + 3840}{x^2 - 8x} = 3 3840x2−8x=3\dfrac{3840}{x^2 - 8x} = 3 3840=3(x2−8x)3840=3x2−24x\begin{aligned} 3840 &= 3(x^2 - 8x) \\[0.6em] 3840 &= 3x^2 - 24x \end{aligned} 0=3x2−24x−38403x2−24x−3840=0\begin{aligned} 0 &= 3x^2 - 24x - 3840 \\[0.6em] 3x^2 - 24x - 3840 &= 0 \end{aligned}
Answer

(iv) 3x2−24x−3840=03x^2 - 24x - 3840 = 0 (or x2−8x−1280=0x^2 - 8x - 1280 = 0), where xx is the speed of the train in km/h\text{km/h}.

Common Mistakes
  • Sign Error in Time Difference: Writing 480x−480x−8=3\dfrac{480}{x} - \dfrac{480}{x - 8} = 3 instead of 480x−8−480x=3\dfrac{480}{x - 8} - \dfrac{480}{x} = 3. Since speed is lower at (x−8) km/h(x - 8)\text{ km/h}, that journey takes more time, so 480x−8>480x\dfrac{480}{x-8} > \dfrac{480}{x}.
  • Forgetting Future Age Addition for Both People: Adding 33 years to Rohan's age but forgetting to add 33 years to his mother's age.

More questions in Exercise 4.1

Q1
  1. Check whether the following are quadratic equations :

(i) (x+1)2=2(x−3)(x + 1)^2 = 2(x - 3)

(ii) x2−2x=(−2)(3−x)x^2 - 2x = (-2)(3 - x)

(iii) (x−2)(x+1)=(x−1)(x+3)(x - 2)(x + 1) = (x - 1)(x + 3)

(iv) (x−3)(2x+1)=x(x+5)(x - 3)(2x + 1) = x(x + 5)

(v) (2x−1)(x−3)=(x+5)(x−1)(2x - 1)(x - 3) = (x + 5)(x - 1)

(vi) x2+3x+1=(x−2)2x^2 + 3x + 1 = (x - 2)^2

(vii) (x+2)3=2x(x2−1)(x + 2)^3 = 2x(x^2 - 1)

(viii) x3−4x2−x+1=(x−2)3x^3 - 4x^2 - x + 1 = (x - 2)^3

Q2
  1. Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 m2528\text{ m}^2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

← Back to Quadratic Equations