1.

Find the product by using quadratic formula​

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Answer:

QUESTION :-

\leadsto Represent the following situations in the form of quadratic EQUATIONS :

  • 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.

Given :-

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

To Find :-

  • What is the length and breadth of the plot.

Formula Used :-

\clubsuit Area Of Rectangle Formula :

\mapsto \sf\boxed{\bold{\pink{Area_{(Rectangle)} =\: Length \times Breadth}}}

Solution :-

Let,

\mapsto \bf Breadth_{(Rectangular\: Plot)} =\: x\: m

\mapsto \bf Length_{(Rectangular\: Plot)} =\: (2x + 1)\: m

According to the question by using the formula we get,

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

\implies \sf 528 =\: 2x^2 + x

\implies \sf 2x^2 + x - 528 =\: 0\: \: \bigg\lgroup \small\bold{\pink{This\: is\: the\: required\: quadratic\: equation}}\bigg\rgroup\\

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

\implies \sf 2x^2 + (33 - 32)x - 528 =\: 0

\implies \sf 2x^2 + 33x - 32x - 528 =\: 0\: \: \bigg\lgroup \small\bold{\pink{By\: doing\: middle\: term}}\bigg\rgroup

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

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

\implies \bf x - 16 =\: 0

\implies \sf\bold{\purple{x =\: 16}}

Either,

\implies \bf 2x + 33 =\: 0

\implies \sf 2x =\: - 33

\implies \sf\bold{\purple{x =\: \dfrac{- 33}{2}}}\: \: \bigg\lgroup \small\bold{\pink{Length\: and\: breadth\: can't\: be\: negative\: (- ve)}}\bigg\rgroup\\

Then, we have to take x = 16.

HENCE, the required length and breadth of the plot :

Length Of Rectangular Plot :

\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: (2x + 1)m

\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: \{2(16) + 1\}\: m

\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: (32 + 1)m

\longrightarrow \sf\bold{\red{Length_{(Rectangular\: Plot)} =\: 33\: m}}

Breadth Of Rectangular Plot :

\longrightarrow \sf Breadth_{(Rectangular\: Plot)} =\: x\: m

\longrightarrow \sf\bold{\red{Breadth_{(Rectangular\: Plot)} =\: 16\: m}}

{\small{\bold{\underline{\therefore\: The\: length\: and\: breadth\: of\: a\: rectangular\: plot\: is\: 33\: m\: and\: 16\: m\: respectively\: .}}}}



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