Question

The area of a rectangular plot is 528m sq. The length of the plot in one more than twice its breadth. Formulate the quadratic equation to determine the length and breadth of the plot

Answer 1

★$\huge\underline\frak\red{To\:find}$★

$\textbf{Determine the length and breadth}$$\textbf{of the plot}$

★$\huge\underline\frak\red{Solution}$★

$\bold{Area\:of\:a\:rectangular\:plot\:=\:528m^2}$

$\bold{Let\:the\:breadth\:be\:x\:then\:its\:length}$$\bold{be\:(2x+1)}$

$\bold{According\:to\:question}$

$\bold{Area\:of\:rectangular\:plot\:=\:528}$

$\implies\sf length×breadth\;=\;528$

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

$\implies\sf 2x^2+x\;=\;528$

$\implies\sf 2x^2+x-528\;=\;0$

$\implies\sf 2x^2+33x-32x-528\;=\;0$

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

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

Either

$\implies\sf 2x+33\;=\;0$

$\implies\sf 2x\;=\;-33$

$\implies\sf x\;=\large\frac{-33}{2}$

or

$\implies\sf x-16\;=\;0$

$\implies\sf x\;=\;16$

$\large\underline\bold\blue{NOTE :-}$

$\textbf{Length and breadth never in negative}$$\textbf{or in fraction}$

$\bold{Breadth\:=x\:=16m}$

$\bold{Length\:=\:2x+1\:=\:2×16+1\:=\:33m}$

Answer 2

AnswEr :

$\bf{\purple{\large{\underline{\underline{\tt{Given\::}}}}}}$

The area of a rectangular plot is 528 m². The length of the plot is one more than twice it's breadth.

$\bf{\red{\large{\underline{\underline{\tt{To\:Find\::}}}}}}$

Formulated the quadratic equation to determine the length & breadth of the plot.

$\bf{\green{\large{\underline{\underline{\bf{Explanation\::}}}}}}$

$\bf{We\:have}\begin{cases}\sf{Area\:of\:rectangular\:plot\:=\:528m^{2} }\\ \sf{The\:length\:of\:the\:plot\:=\:(2r+1)}\\ \sf{The\:breadth\:of\:the\:plot\:=\:r}\end{cases}}$

Formula use :

$\bf{\large{\boxed{\bf{Area\:of\:rectangle=Length\times Breadth\:\:\:\:\:\:\big[sq.unit\big]}}}}}}}$

A/q

$\dashrightarrow\tt{(2r+1)(r)=528}\\\\\\\dashrightarrow\tt{2r^{2} +r=528}\\\\\\\dashrightarrow\tt{\green{2r^{2} +r-528=0}}$

$\bf{\large{\underline{\underline{\bf{Using\:Formula\:of\:quadratic\::}}}}}}}$

$\bf{\large{\boxed{\bf{x=\frac{-b\pm\sqrt{b^{2} -4ac} }{2a} }}}}}$

Firstly, we can compared with ax² + bx + c= 0 to given above quadratic equation;

• a = 2
• b = 1
• c = -528

so,

$\mapsto\sf{x=\dfrac{-1\pm\sqrt{(1)^{2}-4(2)(-528) } }{2(2)} }\\\\\\\\\mapsto\sf{x=\dfrac{-1\pm\sqrt{1-8(-528)} }{4} }\\\\\\\\\mapsto\sf{x=\dfrac{-1\pm\sqrt{1+4224} }{4} }\\\\\\\\\mapsto\sf{x=\dfrac{-1\pm\sqrt{4225} }{4} }\\\\\\\\\mapsto\sf{x=\dfrac{-1\pm65}{4} }\\\\\\\\\mapsto\sf{x=\dfrac{-1+65}{4} \:\:\:\:Or\:\:\:\:x=\dfrac{-1-65}{4} }\\\\\\\\\mapsto\sf{x=\cancel{\dfrac{64}{4}} \:\:\:\:Or\:\:\:\:x=\dfrac{-66}{4} }\\\\\\\\\mapsto\sf{\red{x=16\:m}}$

We know that negative value isn't acceptable.

Thus,

$\bf{\large{\underbrace{\bf{Breadth\:of\:the\:plot\:=\:r=16\:m}}}}}}$

$\bf{\large{\underbrace{\bf{Length\:of\:the\:plot\:=\:2(16)+1 = 32+1=33\:m}}}}}}$