find the breadth of a rectangle whose length is (3x-1) units and area is 33x²+20...

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June 2021 2 20 Report

find the breadth of a rectangle whose length is (3x-1) units and area is 33x²+20x-7

mini6911 I think area of rectangle equation is wrong it should be:
$33{x}^{2} + 10x - 7 \\$
area of rectangle = length * breadth
let length = (ax+b)
let breadth =(cx+d)
$33 {x }^{2} + 10x - 7 = (ax + b) ( cx + d)$
after multiplication of right hand side:
$33 {x }^{2} + 10x - 7 = ac {x }^{2} + (ad+cb)x +cd$
$ac = 33, ad+cd = 10, bd = -7$
$length (ax+b)= (3x - 1)$
$then a=3, b=-1$
$ac=33, a=3$
$c= 33/3 = 11$
$bd=-7, b=-1$
$d= -7/-1 = 7$
$breadth (cx+d)= (11x + 7)$

sibhiamar 33x²+20x-7 gives no root so I think the equation for the area of rectangle is wrong
the equation should be 33x²+10x-7 since it has two roots, they are (3x-1)(11x+7)
so area of rectangle = length * breath
length = 3x-1
then (3x-1)(11x+7)=breath*(3x-1)
so breath = 11x+7

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