Example-10.A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find its length and breadth.

Solution: the breadth of the rectangular park be xm So, its length=(x+3) m.

Therefore, the area of the rectangular park = x(x + 3) * m ^ 2 -(x^ 2 +3)

Now, base of the isosceles triangle=xm

Therefore, its area- = 1/2 * x * 12 = 6x * m ^ 2 .

According to our requirements,

x ^ 2 + 3x = 6x + 4

x ^ 2 - 3x - 4 = 0 i.e.,

Using the quadratic formula, we get

x = (3 plus/minus sqrt(25))/2 = (3 plus/minus 5)/2 = - 4or - 1

But x ne-1 (Why?). Therefore, x = 4

So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m

Verification: Area of rectangular park = 28m ^ 4 ,

area of triangular park = (28 - 4) * m ^ 2 =24m^ 2.



Example-11. Find the roots of the following quadratic equations, if they exist.

(i) x ^ 2 + 4x + 5 = 0

(ii) 2x ^ 2 - 2sqrt(2) * x + 1 = 0

Solution: (i) x ^ 2 + 4x + 5 = 0 Here, a = 1 , b = 4 c = 5 So, b ^ 2 - 4ac = 16 - 20 = - 4 < 0

Since the square of a real number cannot be negative, therefore sqrt(b ^ 2 - 4ac) will not have any real value.

So, there are no real roots for the given equation.

(i) 2x ^ 2 - 2sqrt(2) * x + 1 = 0 . Here, a = 2 b = - 2sqrt(2) , c = 1 .

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