Q1. Check whether the following are quadratic equations :

1. \((x + 1)^2 = 2(x – 3)\)

   Soultion

   $$\begin{aligned}\left( x+1\right) ^{2}&=2\left( x-3\right) \\ x^{2}+1+2x&=2\left( x-3\right) \\ x^{2}+2x+1-2x+6&=0\\ x^{2}+7&=0\\ \Rightarrow x^{2}+0x+7&=0\end{aligned}$$ is a quadratic equation

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2. \(x^2 – 2x = (–2) (3 – x)\)

   Solution:

   $$\begin{aligned}x^{2}-2x&=\left( -2\right) \left( 3-x\right) \\ x^{2}-2x&=-6+2x\\ x^2-2x-2x+6&=0\\ x^{2}-4x+6&=0\end{aligned}$$ is a Quadratic equation

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3. \((x – 2)(x + 1) = (x – 1)(x + 3)\)

   Solution:

   $$\begin{aligned}\left( x-2\right) \left( x+1\right) &=\left( x-1\right) \left( x+3\right) \\ x^{2}+x-2x-2&=x^{2}+3x-x-3\\ x^{2}-x^{2}-x-2-2x+3&=0\\ -3x+1&=0\\ 3x-1&=0\end{aligned}$$ is not a quadratic equation

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4. \((x – 3)(2x +1) = x(x + 5)\)

   Solution:

   $$\begin{aligned}\left( x-3\right) \left( 2x-1\right) &=x\left( x+5\right) \\ 2x^{2}+x-6x-3&=x^{2}+5x\\ 2x^{2}-x^{2}-5x-5x-3&=0\\ x^{2}-10x-3&=0\end{aligned}$$ is a quadratic Equation

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5. \((2x – 1)(x – 3) = (x + 5)(x – 1)\)

   Solution:

   $$\begin{aligned}\left( 2x-1\right) \left( x-3\right) &=\left( x+5\right) \left( x-1\right) \\ 2x^{2}-6x-x+3&=x^{2}-x+5x-5\\ 2x^{2}-x^{2}-7x-4x+3+5&=0\\ x^{2}-11x+8&=0\end{aligned}$$ is a quadratic Equation

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6. \(x^2 + 3x + 1 = (x – 2)^2\)

   Solution:

   $$\begin{aligned}x^{2}+3x+1&=\left( x-2\right) ^{2}\\ x^{2}+3x+1&=x^{2}-4x+4\\ 3x+4x+1-4&=0\\ 7x-3&=0\end{aligned}$$ is not a quadratic Equation

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7. \((x + 2)^3 = 2x (x^2 – 1)\)

   Solution:

   $$\begin{aligned}\left( x+2\right) ^{3}&=2x\left( x^{2}-1\right) \\ x^{3}+8+3x^{2}\cdot 2+3x\cdot 4&=2x^{3}-2x\\ x^{3}-2x^{3}+8+6x^{2}+12x+4x&=0\\ -x^{3}+6x^{2}+16x+8&=0\end{aligned}$$ is not a quadratic Equation

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8. \(x^3 – 4x^2 – x + 1 = (x – 2)^3\)

   Solution:

   $$\begin{aligned}x^{3}-4x^{2}-x+1&=\left( x-2\right) ^{3}\\ x^{3}-4x^{2}-x+1&=x^{3}-8-3\cdot 2x^{2}+3\cdot x\cdot 4\\ -4x^{2}+6x^{2}-x+1-12x+8&=0\\ 2x^{2}-13x+9&=0\end{aligned}$$ is a quadratic Equation

Represent the following situations in the form of quadratic equations :

1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

   Solution:

   Area of a rectangular plot = 528 m²
   Let breadth of the plot is \(x\)
   length, which is 1 more than twice of \(x\)
   length = \(2x+1\)
   Area of Rectangular Plot= length \(\times\) breadth

   $$\begin{aligned}528&=\left( 2x+1\right) x\\ \Rightarrow \left( 2x+1\right) \left( x\right) &=528\\ 2x^{2}+x-528&=0\end{aligned}$$ To factorise by split midde term we need factors of \(ac\) such that algebraic sum of factors should be 1 $$2\times 528=33\times 32$$ Also algebraic Sum $$33-32 = 1$$ $$\begin{aligned}2x^{2}+x-528&=0\\ 2x^{2}-32x+33x-52&=0\\ 2x\left( x-16\right) +33\left( x-16\right) &=O\\ \left( x-16\right) \left( 2x+33\right) &=0\\ x-16&=0\\ x&=16\\ 2x+33&=0\\ x&=-\dfrac{33}{2}\end{aligned}$$ therefore breadth of Rectangular Park = 16 m
   and length of Park = (16*2)+1= 33 m
2. The product of two consecutive positive integers is 306. We need to find the integers.

   Solution:

   Let first number be \(x\) next consecutive number = \(x+1\)

   $$\begin{aligned}x\left( x+1\right) &=306\\ x^{2}+x-306&=0\end{aligned}$$

   To factorise by split middle term, we need to split 306 into factors so as its algebraic sum is 1

   $$306=18\times 17$$ Algebraic sum $$18-17=1$$ $$\begin{aligned}x^{2}+18x-17x-306&=0\\ x\left( x+18\right) -17\left( x+18\right) &=0\\ \left( x+18\right) \left( x-17\right) &=0\\ x-17&=0\\ x&=17\\ x+18&=0\\ x&=-18\end{aligned}$$

   Numbers are positive therefore
   first number = 17
   Second number = 18

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

   Solution:

   Let Present age of Rohan is \(x\)
   His mother, who is 26 years older, age \(x+26\)
   After 3 years, Rohans age \(=x+3\)
   and his mothers age \(=x+29\)

   Product of their age after 3 years = 360

   $$\begin{aligned}\therefore \left( x+3\right) \left( x+29\right) &=360\\ x^{2}+29x+3x+87&=360\\ x^{2}+32x+87-360&=0\\ x^{2}+32x-273&=0\end{aligned}$$

   To find factor by split middle term 273 must be factorize to give algebraic sum of 32

   Factors of 273 \(273=39\times 7\)
   Algebraic sum \(39-7=32\) $$\begin{aligned}\therefore x^{2}+32x-273&=0\\ x^{2}+39x-7x-273&=0\\ x\left( x+39\right) -7\left( x+39\right) &=0\\ \left( x+39\right) \left( x-7\right) &=0\\ x-7&=0\\ x&=7\\ x+39&=0\\ x&=-39\end{aligned}$$

   Age can not be negative, therefore Rohan's age = 7 years

4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

   Solution:

   Let speed of train is \(v\)
   Distance travelled = 480km time taken (t)

   $$t=\dfrac{480}{v}$$

   If speed of train would have 8km/h less
   then time taken

   $$\begin{aligned}t+3&=\dfrac{480}{v-8}\\\\ \dfrac{480}{v}+3&=\dfrac{480}{v-8}\\ \dfrac{480+3v}{v}&=\dfrac{480}{v-8}\\ \left( 480+3v\right) \left( v-8\right) &=480v\\ 480v-5440+3v^{2}-24v&=480v\\ -3840+3v^{2}-24v&=0\\ 3v^{2}-24v-3840&=0\\ \Rightarrow v^{2}-8v-1280&=0\end{aligned}$$

   To factorize we need factors of 1280 such that their algebraic sum = 8

   $$1280=32\times 40$$ algebraic sum $$40-32=8$$ $$\begin{aligned}v^{2}-40v+32v-1280&=0\\ v\left( v-40\right) +32\left( 1v-40\right) &=0\\ \left( v-40\right) \left( v+32\right) &=0\\ v&=40\\ v&=-32\end{aligned}$$

   Speed of train 40km/h