Quadratic Equation

NCERT Exercise 4.3

Part 4

Question 5: 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.

Solution: Let us assume, marks in Mathematics `= x`

Therefore, marks in English `= 30 – x`

If she scores 2 marks more in Mathematics; then marks in mathematics `= x +2`

And if she scores 3 marks less in English, the marks in English `= 30 – x – 3 = 27 – x`

As per question;

`(x + 2)(27 – x) = 210`

Or, `27x – x^2 + 54 – 2x = 210`

Or, `25x – x^2 + 54 – 210 = 0`

Or, `25x – x^2 – 156 = 0`

Or, `x^2 – 25x + 156 = 0`

Or, `x^2 – 12x – 13x + 156 = 0`

Or, `x(x – 12) – 13(x – 12) = 0`

Or, `(x – 12)(x – 13) = 0`

Hence, `x = 12` and `x = 13`

Case 1: If `x = 13`, then marks in English `= 30 – 13 = 17`

Case 2: If `x = 12`, then marks in English `= 30 – 12 = 18`

In both the cases; after adding 2 marks to mathematics and deducting 3 marks from English; the product of resultants is 210

Question 6: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution: Let us assume, the shorter side `= x`

Then; longer side = x + 30 and diagonal `= x + 60`

Using Pythagoras Theorem, we get following equation:

`(x + 60)^2 = x^2 + (x + 30)^2`

Or, `x^2 + 120x + 3600`

= `x<sup>2</sup> + x<sup>2</sup> + 60x + 900`

Or, `-x^2 + 60x + 2700 = 0`

Or, `x^2 - 60x - 2700 = 0`

Or, `x^2 - 90x + 30x - 2700 = 0`

Or, `x(x - 90) + 30(x - 90) = 0`

Or,`(x + 30)(x - 90) = 0`

Or, `x = -30` and `x = 90`

Discarding the negative value, we have x = 90 m, longer side = 120 m and diagonal = 150 m

Question 7: The difference of squares of two numbers is 180. The square of the smaller number is 8 times the large number. Find the two numbers.

Solution: Let us assume, larger number = x

Hence, square of smaller number = 8x

As per question;

`x^2 – 8x = 180`

Or, `x^2 – 8x – 180 = 0`

Or, `x^2 – 18x + 10x – 180 = 0`

Or, `x(x – 18) + 10(x – 18) = 0`

Or, `(x + 10)(x – 18) = 0`

Hence, `x = - 10` and `x = 18`

Discarding the negative value; `x = 18`

Smaller number

`=sqrt(8xx18)=sqrt(144)=12`

Hence, the numbers are; 12 and 18