Question

Q.1: Represent the following situations in the form of quadratic equations:

(i) 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.

(ii) 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.What is the speed of the train?

Q.2: Find the roots of quadratic equations by factorisation:

(i) √2 x2 + 7x + 5√2=0

(ii) 100x2 – 20x + 1 = 0​

Answer 1

(1) GIVEN :-

• Area of rectangular plot is 528 m².
• The length of the plot (in metres) is one more than twice its breadth.

TO FIND :-

• The length and breadth of the rectangular plot.

SOLUTION :-

Let the breadth of the rectangular plot be "x". Hence the length of the rectangular plot is"2x + 1".

$\implies \sf \: Area \: of \: rectangular \: plot \: = 528$

$\implies \sf \:length \times breadth = 528$

$\implies \sf \:(2x + 1)x = 528$

$\implies \sf \:2x ^{2} + x = 528$

$\implies \sf \:2x ^{2} + x - 528 = 0$

By splitting the middle term,

$\implies \sf \:2x^{2} + 33x - 32x - 528 = 0$

$\implies \sf \:x(2x + 33) - 16(2x + 33) = 0$

$\implies \sf \:(2x + 33)(x - 16) = 0$

$\implies \sf \:2x + 33 = 0 \: or \: x - 16 = 0$

$\implies \sf \:2x = - 33 \: or \: x = 16$

$\implies \sf \:x = \dfrac{ - 33}{2} \: or \: x = 16$

As breadth cannot be negative so breadth is 16m.

Hence the length is 2x + 1 = 33m.

(2) GIVEN :-

• Total distance covered by the train = 480 km.
• If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance.

TO FIND :-

• The speed of the train.

SOLUTION :-

Let the speed of the train be "x".

As we know that,

$\implies \sf \: speed \: = \dfrac {distance}{time}$

$\implies \sf \: speed \times time = distance$

$\implies \sf \: x \times time = 480$

$\implies \sf \: time \: = \dfrac{480}{x}$

Now let's consider second situation,

$\implies \sf \: speed \: = \dfrac {distance}{time}$

$\implies \sf \: speed \times time = distance$

$\implies \sf \: (x - 8) \times time = 480$

$\implies \sf \: time \: = \dfrac{480}{x - 8}$

Now According to the question,

$\implies \sf \: \dfrac{480}{x - 8} = \dfrac{480}{x} + 3$

$\implies \sf \: \dfrac{480}{x - 8} = \dfrac{480 + 3x}{x}$

$\implies \sf \: 480x = (x - 8)(480 + 3x)$

$\implies \sf \: 480x = x(480 + 3x) - 8(480 + 3x)$

$\implies \sf \: 480x = 480x + 3x ^{2} - 3840 - 24x$

$\implies \sf \: 3x ^{2} - 3840 - 24x = 0$

$\implies \sf \: 3(x ^{2} - 8x) = 3840$

$\implies \sf \: x ^{2} - 8x = 1280$

$\implies \sf \: x ^{2} - 8x - 1280 = 0$

By splitting the middle term,

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

$\sf \implies x^{2}-40x+32x-1280 = 0$

$\sf \implies x(x-40)+32(x-40)=0$

$\sf \implies (x - 40)(x + 32) = 0$

$\sf \implies x - 40 = 0 \ or \ x +32 = 0$

$\sf \implies x = 40 \ or \ x= -32$

Speed cannot be negative therefore speed of the train is 40 km/hr.

Answer 2

$\large\color{purple}\underline{\underline{{ \boxed { Solution \: 01}}}}$

• Let the breadth of the rectangular plot be x
• Length of the rectangular plot is 2x + 1

$\begin{gathered} \implies \sf \: Area \: of \: rectangular \: plot \: = 528 \\ \end{gathered} \\ \\ </p><p>\begin{gathered}\implies \sf \:length \times breadth = 528 \\ \end{gathered} \\ \\ \begin{gathered}\implies \sf \:(2x + 1)x = 528 \\ \end{gathered} \\ \\ \begin{gathered}\implies \sf \:2x ^{2} + x = 528 \\ \end{gathered} \\ \\ \implies \sf \:2x ^{2} + x - 528 = 0 \\ \\ \tt \: By \: splitting \: the \: middle \: term, \\ \begin{gathered}\implies \sf \:2x^{2} + 33x - 32x - 528 = 0 \\ \end{gathered} \\ \\\begin{gathered}\implies \sf \:x(2x + 33) - 16(2x + 33) = 0 \\ \end{gathered} \\ \\ \implies \sf \:(2x + 33)(x - 16) = 0 \\ \\\begin{gathered}\implies \sf \:2x + 33 = 0 \: or \: x - 16 = 0 \\\end{gathered} \\ \\\begin{gathered}\implies \sf \:2x = - 33 \: or \: x = 16 \\\end{gathered} \\ \\ \implies \sf \:x = \dfrac{ - 33}{2} \: or \: x = 16 \\ \\ \implies \sf \:x = \dfrac{ - 33}{2} \: or \: x = 16$

As breadth cannot be negative so,

• Breadth= 16
• Length is 2x + 1 = 33m.

$\large\color{purple}\underline{\underline{{ \boxed { Solution \: 02}}}}$

• Let the speed of the train be "x".

As we know that,

$\begin{gathered}\implies \sf \: speed \: = \dfrac {distance}{time} \\\end{gathered} \\ \begin{gathered}\implies \sf \: speed \times time = distance \\\end{gathered} \\ \\ \begin{gathered}\implies \sf \: x \times time = 480 \\ \\ \end{gathered} \\ \\ \begin{gathered}\implies \sf \: time \: = \dfrac{480}{x} \\\end{gathered} \\ \\ \tt \: second \: situation, \\ \begin{gathered}\implies \sf \: speed \: = \dfrac {distance}{time} \\\end{gathered}$

$\begin{gathered}\implies \sf \: \dfrac{480}{x - 8} = \dfrac{480}{x} + 3 \\\end{gathered} \\\begin{gathered}\implies \sf \: \dfrac{480}{x - 8} = \dfrac{480 + 3x}{x} \\\end{gathered} \\\begin{gathered}\implies \sf \: 480x = (x - 8)(480 + 3x) \\\end{gathered} \\ \\ \begin{gathered}\implies \sf \: 480x = x(480 + 3x) - 8(480 + 3x) \\\end{gathered} \\ \\ \begin{gathered}\implies \sf \: 480x = 480x + 3x ^{2} - 3840 - 24x \\\end{gathered} \\ \\ \begin{gathered} \sf \: 3x ^{2} - 3840 - 24x = 0 \\\end{gathered} \\ \\ \begin{gathered}\sf \: 3(x ^{2} - 8x) = 3840 \\\end{gathered} \\ \\ \begin{gathered}\sf \: x ^{2} - 8x = 1280 \\\end{gathered} \\\begin{gathered} \sf \: x ^{2} - 8x - 1280 = 0\\\end{gathered} \\ \\ \tt \: By \: splitting \: the \: middle \: term,$

$\begin{gathered}\sf \implies {x^{2} - 8x -1280 = 0}\\ \end{gathered} \\ \\ \begin{gathered}\sf \implies x^{2}-40x+32x-1280 = 0\\\end{gathered} \\ \\ </p><p>\begin{gathered}\sf \implies x(x-40)+32(x-40)=0\\\end{gathered} \\ \\ \begin{gathered}\sf \implies (x - 40)(x + 32) = 0\\\end{gathered} \\ \\\begin{gathered}\sf \implies x - 40 = 0 \ or \ x +32 = 0\\\end{gathered} \\ \\\begin{gathered}\sf \implies x = 40 \ or \ x= -32\\\end{gathered}$

Speed of the train is 40 km/hr.

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$\sf \: @WildCat7083$