For a rectangular field of area 3 sq. units, the length is one unit more than twice the breadth that measures ‘x’ units. The quadratic equation representing the situation is:

x²– 2x + 6 = 0

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x²– 2x + 3 = 0

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2x²+ x - 3 = 0

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2x²+ x - 6 = 0

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Solution

The correct option is C
2x²+ x - 3 = 0

Area of rectangle = $l e n g t h \times b r e a d t h$

Given, length = $(2 \times b r e a d t h$ + 1)

Let the breadth of the field be $x$ .

Length of the field= $2 x + 1$

Area of the rectangular field= $x (2 x + 1) = 3$

$2 x^{2} + x = 3$

$2 x^{2} + x - 3 = 0$

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Q. Question 5
Which of the following equations has the sum of its roots as 3?
(a)

2 x^{2} - 3 x + 6 = 0

(b)

- x^{2} + 3 x - 3 = 0

(c)

\sqrt{2} x^{2} - \frac{3}{\sqrt{2}} x + 1 = 0

(d)

3 x^{2} - 3 x + 3 = 0

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For a rectangular field of area 3 sq. units, the length is one unit more than twice the breadth that measures ‘x’ units. The quadratic equation representing the situation is:

The correct option is C 2x2 + x - 3 = 0 Area of rectangle = length×breadth Given, length =(2×breadth + 1) Let the breadth of the field be x. Length of the field= 2x+1 Area of the rectangular field= x(2x+1)=3 2x2+x=3 2x2+x−3=0