hi there
please help..............
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(x+5) (x-4)=x^2 ..solve it x=20
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length of rectangle =x+5
breadth of rectangle=x-4
side of square =x
Area of rectangle= l×b
=(x+5)(x-4)
(x+a)(x+b)=x^2+(a+b)x+an
[(x+5)][(x)+(-4)]=x^2+[(5)+(-4)]x+5×-4
=x^2+(5-4)x-20
=x^2+x-20
=x^2-4x+5x-20
=x(x-4)+5(x-4)
x-4=0 :::;;;; x+5=0
x=0+4 x=-5
x=4
then
side of square =4cm
breadth of rectangle=x-4
side of square =x
Area of rectangle= l×b
=(x+5)(x-4)
(x+a)(x+b)=x^2+(a+b)x+an
[(x+5)][(x)+(-4)]=x^2+[(5)+(-4)]x+5×-4
=x^2+(5-4)x-20
=x^2+x-20
=x^2-4x+5x-20
=x(x-4)+5(x-4)
x-4=0 :::;;;; x+5=0
x=0+4 x=-5
x=4
then
side of square =4cm
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