Question

find the product by using quadratic formula​

Answer 1

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 :

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

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

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

Answer 2

$Answer:</p><p></p><p>Question :-</p><p></p><p>\leadsto⇝ Represent the following situations in the form of quadratic equations :</p><p></p><p>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.</p><p></p><p>Given :-</p><p></p><p>The area of a rectangular plot is 528 m².The length of the plot (in metres) is one more than twice its breadth.</p><p></p><p>To Find :-</p><p></p><p>What is the length and breadth of the plot.</p><p></p><p>Formula Used :-</p><p></p><p>\clubsuit♣ Area Of Rectangle Formula :</p><p></p><p>\mapsto \sf\boxed{\bold{\pink{Area_{(Rectangle)} =\: Length \times Breadth}}}↦Area(Rectangle)=Length×Breadth</p><p></p><p>Solution :-</p><p></p><p>Let,</p><p></p><p>\mapsto \bf Breadth_{(Rectangular\: Plot)} =\: x\: m↦Breadth(RectangularPlot)=xm</p><p></p><p>\mapsto \bf Length_{(Rectangular\: Plot)} =\: (2x + 1)\: m↦Length(RectangularPlot)=(2x+1)m</p><p></p><p>According to the question by using the formula we get,</p><p></p><p>\implies \sf 528 =\: (2x + 1) \times x⟹528=(2x+1)×x</p><p></p><p>\implies \sf 528 =\: 2x^2 + x⟹528=2x2+x</p><p></p><p>\begin{gathered}\implies \sf 2x^2 + x - 528 =\: 0\: \: \bigg\lgroup \small\bold{\pink{This\: is\: the\: required\: quadratic\: equation}}\bigg\rgroup\\\end{gathered}⟹2x2+x−528=0⎩⎪⎪⎪⎧Thisistherequiredquadraticequation⎭⎪⎪⎪⎫</p><p></p><p>\implies \sf 2x^2 + x - 528 =\: 0⟹2x2+x−528=0</p><p></p><p>\implies \sf 2x^2 + (33 - 32)x - 528 =\: 0⟹2x2+(33−32)x−528=0</p><p></p><p>\implies \sf 2x^2 + 33x - 32x - 528 =\: 0\: \: \bigg\lgroup \small\bold{\pink{By\: doing\: middle\: term}}\bigg\rgroup⟹2x2+33x−32x−528=0⎩⎪⎪⎪⎧Bydoingmiddleterm⎭⎪⎪⎪⎫</p><p></p><p>\implies \sf x(2x + 33) - 16(2x + 33) =\: 0⟹x(2x+33)−16(2x+33)=0</p><p></p><p>\implies \sf (x - 16)(2x + 33) =\: 0⟹(x−16)(2x+33)=0</p><p></p><p>\implies \bf x - 16 =\: 0⟹x−16=0</p><p></p><p>\implies \sf\bold{\purple{x =\: 16}}⟹x=16</p><p></p><p>Either,</p><p></p><p>\implies \bf 2x + 33 =\: 0⟹2x+33=0</p><p></p><p>\implies \sf 2x =\: - 33⟹2x=−33</p><p></p><p>\begin{gathered}\implies \sf\bold{\purple{x =\: \dfrac{- 33}{2}}}\: \: \bigg\lgroup \small\bold{\pink{Length\: and\: breadth\: can't\: be\: negative\: (- ve)}}\bigg\rgroup\\\end{gathered}⟹x=2−33⎩⎪⎪⎪⎧Lengthandbreadthcan′tbenegative(−ve)⎭⎪⎪⎪⎫</p><p></p><p>Then, we have to take x = 16.</p><p></p><p>Hence, the required length and breadth of the plot :</p><p></p><p>❒ Length Of Rectangular Plot :</p><p></p><p>\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: (2x + 1)m⟶Length(RectangularPlot)=(2x+1)m</p><p></p><p>\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: \{2(16) + 1\}\: m⟶Length(RectangularPlot)={2(16)+1}m</p><p></p><p>\longrightarrow \sf Length_{(Rectangular\: Plot)} =\: (32 + 1)m⟶Length(RectangularPlot)=(32+1)m</p><p></p><p>\longrightarrow \sf\bold{\red{Length_{(Rectangular\: Plot)} =\: 33\: m}}⟶Length(RectangularPlot)=33m</p><p></p><p>❒ Breadth Of Rectangular Plot :</p><p></p><p>\longrightarrow \sf Breadth_{(Rectangular\: Plot)} =\: x\: m⟶Breadth(RectangularPlot)=xm</p><p></p><p>\longrightarrow \sf\bold{\red{Breadth_{(Rectangular\: Plot)} =\: 16\: m}}⟶Breadth(RectangularPlot)=16m</p><p></p><p>{\small{\bold{\underline{\therefore\: The\: length\: and\: breadth\: of\: a\: rectangular\: plot\: is\: 33\: m\: and\: 16\: m\: respectively\: .}}}}∴Thelengthandbreadthofarectangularplotis33mand16mrespectively.</p><p></p><p>$