EXERCISE 4.1

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1. Check whether the following are quadratic equations.

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(i) $(x+1)^2=2(x-3)$

Solution:

Expand both sides. $$ x^2+2x+1=2x-6 $$ Bring all terms to one side. $$ x^2+2x+1-2x+6=0 $$ $$ x^2+7=0 $$ Since the highest power of $x$ is $2$, it is a quadratic equation.

Answer: Quadratic Equation.

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(ii) $x^2-2x=(-2)(3-x)$

Solution:

Expand the right side. $$ x^2-2x=-6+2x $$ Bring all terms to one side. $$ x^2-4x+6=0 $$ Since the degree is $2$, it is a quadratic equation.

Answer: Quadratic Equation.

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

Solution:

Expand both sides. $$ x^2-x-2=x^2+2x-3 $$ Cancel $x^2$. $$ -x-2=2x-3 $$ $$ 3x=1 $$ $$ 3x-1=0 $$ The highest power of $x$ is $1$.

Answer: Not a quadratic equation.

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

Solution:

Expand both sides. $$ 2x^2-5x-3=x^2+5x $$ Bring all terms to one side. $$ x^2-10x-3=0 $$ The degree is $2$.

Answer: Quadratic Equation.

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

Solution:

Expand both sides. $$ 2x^2-7x+3=x^2+4x-5 $$ Bring all terms to one side. $$ x^2-11x+8=0 $$ The degree is $2$.

Answer: Quadratic Equation.

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(vi) $x^2+3x+1=(x-2)^2$

Solution:

Expand the right side. $$ x^2+3x+1=x^2-4x+4 $$ Cancel $x^2$. $$ 3x+1=-4x+4 $$ $$ 7x-3=0 $$ The highest power of $x$ is $1$.

Answer: Not a quadratic equation.

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(vii) $(x+2)^3=2x(x^2-1)$

Solution:

Expand both sides. $$ x^3+6x^2+12x+8=2x^3-2x $$ Bring all terms to one side. $$ x^3-6x^2-14x-8=0 $$ The highest power of $x$ is $3$.

Answer: Not a quadratic equation.

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(viii) $x^3-4x^2-x+1=(x-2)^3$

Solution:

Expand the right side. $$ (x-2)^3=x^3-6x^2+12x-8 $$ Substitute. $$ x^3-4x^2-x+1=x^3-6x^2+12x-8 $$ Cancel $x^3$. $$ 2x^2-13x+9=0 $$ The degree is $2$.

Answer: Quadratic Equation.

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2. Represent the following situations in the form of quadratic equations.

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(i) The area of a rectangular plot is $528\text{ m}^2$. The length is one more than twice its breadth.

Solution:

Let the breadth be $$ x\text{ m} $$ Then the length is $$ 2x+1 $$ Area of rectangle $$ x(2x+1)=528 $$ $$ 2x^2+x-528=0 $$

Required quadratic equation:

$$ \boxed{2x^2+x-528=0} $$

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(ii) The product of two consecutive positive integers is $306$.

Solution:

Let the first integer be $$ x $$ The second integer is $$ x+1 $$ Given, $$ x(x+1)=306 $$ $$ x^2+x-306=0 $$

Required quadratic equation:

$$ \boxed{x^2+x-306=0} $$

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(iii) Rohan's mother is $26$ years older than him. The product of their ages $3$ years from now is $360$.

Solution:

Let Rohan's present age be $$ x\text{ years} $$ Mother's present age $$ x+26 $$ After $3$ years, $$ x+3,\qquad x+29 $$ Given, $$ (x+3)(x+29)=360 $$ Expand. $$ x^2+32x+87=360 $$ $$ x^2+32x-273=0 $$

Required quadratic equation:

$$ \boxed{x^2+32x-273=0} $$

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

Solution:

Let the speed of the train be $$ x\text{ km/h} $$ Time taken $$ \frac{480}{x} $$ Reduced speed $$ x-8 $$ New time $$ \frac{480}{x-8} $$ According to the question, $$ \frac{480}{x-8}=\frac{480}{x}+3 $$ Multiply by $$ x(x-8) $$ $$ 480x=480(x-8)+3x(x-8) $$ $$ 480x=480x-3840+3x^2-24x $$ $$ 3x^2-24x-3840=0 $$ Divide by $3$. $$ x^2-8x-1280=0 $$

Required quadratic equation:

$$ \boxed{x^2-8x-1280=0} $$

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