an electric cable costs 200rs.If the cable was 5meters longer and each metre of cable cost 2rs less,the cost of the cable would have remained unchanged.Represent the above situation in the form of a quadratic equation
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cost= 200 rs
length of the cable=x meter
then cost of one meter=
length of the cable(if)= x+5
cost of one meter=
we know that
[tex] \frac{200}{x} - \frac{200}{x+5} =2 \\ \\ \frac{200x+100-200x}{ x^{2} -5x} =2 \\ \\ \frac{100}{ x^{2} -5x} =2 \\ \\ 2 x^{2} -10x=100 \\ divide/all/terms/by/2 \\ x^{2} -5x=50 \\ x^{2} -5x-50=0 \\ (x-10)(x+5)=0 \\ x=10/ or/ x=-5 \\ [/tex]
length of the cable= 10 m
length of the cable=x meter
then cost of one meter=
length of the cable(if)= x+5
cost of one meter=
we know that
[tex] \frac{200}{x} - \frac{200}{x+5} =2 \\ \\ \frac{200x+100-200x}{ x^{2} -5x} =2 \\ \\ \frac{100}{ x^{2} -5x} =2 \\ \\ 2 x^{2} -10x=100 \\ divide/all/terms/by/2 \\ x^{2} -5x=50 \\ x^{2} -5x-50=0 \\ (x-10)(x+5)=0 \\ x=10/ or/ x=-5 \\ [/tex]
length of the cable= 10 m
Answered by
0
200/x-200/x+5=2 (/ means upon)
=200x+100-200x/x sq-5x=2
100/x sq-5x=2
2xsq-10x=2
x sq-5x = 50 (divide the terms by 2 )
x sq-5x-50=0
(x-10)(x+5)=0
x= 10,x=-5
=200x+100-200x/x sq-5x=2
100/x sq-5x=2
2xsq-10x=2
x sq-5x = 50 (divide the terms by 2 )
x sq-5x-50=0
(x-10)(x+5)=0
x= 10,x=-5
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