Page 14 - CHAPTER 4 (Quadratic equations)

CHAPTER 4

QUADRATIC EQUATIONS

Example 2 : The sum of squares of two given consecutive and positive

integers is 290, then find the integers.

Solution: Suppose the two consecutive odd positive integers be and

(x+2). Then,

2
x²+(x+2) = 290
⇒x²+x²+4x+4=290 ⇒x²+2x-143=0

⇒x²+13x-11x-143=0 ⇒x(x+13)- 11 (x+13)= 0

⇒(x+13)(x-11)= 0 ⇒x= -13 or x=11

But –13, is not an odd positive integer.

So, the required integers are 11 and 13.

Example 3: Seven years age Vaarun's age was five times the square

of Swaati's age. Three years hence, Swaati's age will be two fifth of

Vaarun's age. Find their present ages.

Solution: Let the present ages of Vaarun and Swaati be x years and y

years respectively.

Seven years ago,

Varun’s age = (x –7) years and Swati’s age = (y –7) years.

2
2
∴(x-7)= 5 (y-7) ⟹x-7=5 (y - 14y+49)
⇒x=5y²-70y+245+7 ⇒x=5y²-70y+252 …. (i)

Three years hence,

Vaarun's age = (x + 3) years and Swaati's age = (y +3) years.

2
∴(y+3)= (x+3)⇒5y+15=2x+6 ⇒x= 5y+9 …. (ii)
5 2

From (i) and (ii) we get 5y²-70y+252= 5y+9
2

⇒10y²-140y+504=5y+9 ⇒10y²-145y+495=0 ⇒2y²-29y+99=0

⇒2y²-18y-11y+99=0 ⇒2y (y-9)- 11 (y-9)= 0

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