Respuesta :
Hello,
Let's x the side of the cube
Volume of the cube=x^3
Volume of the rectangular block :(x+1)x²=150
So x^3+x^2-150=0
==>x^3-5x^2+6x^2-30x+30x-150=0
==>x^2(x-5)+6x(x-5)+30(x-5)=0
==>(x-5)(x^2+6x+30)=0
The trinome has no real roots.
==>x=5 (in)
Proof: (5+1)*5*5=25*6=150
Let's x the side of the cube
Volume of the cube=x^3
Volume of the rectangular block :(x+1)x²=150
So x^3+x^2-150=0
==>x^3-5x^2+6x^2-30x+30x-150=0
==>x^2(x-5)+6x(x-5)+30(x-5)=0
==>(x-5)(x^2+6x+30)=0
The trinome has no real roots.
==>x=5 (in)
Proof: (5+1)*5*5=25*6=150
Answer:
Length of edge of original block =5 inches
Step-by-step explanation:
We are given that one dimension of a cube is increased by 1 inches to form a rectangular block.
Volume of new block means rectangular block=150 cubic inches
We have to find the value of edge of the original block
Let edge length of original block=x
Length of rectangular block=x+1
Breadth of rectangular block=x
Height of rectangular block=x
Volume of rectangular block=[tex]length\times breadth\times height[/tex]
Substitute the values then we get
[tex]150=(x+1)\times x\times x[/tex]
[tex]x^2(x+1)=150[/tex]
[tex]x^3+x^2=150[/tex]
[tex]x^3-5x^2+6x^2-30x+30x-150=0[/tex]
[tex]x^2(x-5)+6x(x-5)+30(x-5)=0[/tex]
[tex](x-5)(x^2+6x+30)=0[/tex]
[tex]x-5=0 [/tex]
x=5 and [tex]x^2+6x+30=0[/tex]
For second quadratic equation
[tex]D=b^2-4ac[/tex]
[tex]D=(6)^2-4\times 1\times 30[/tex]
[tex]D=36-120=-84<0[/tex]
Therefore, the roots of second quadratic equation are imaginary .
It is not possible, because we are finding length of edge of original block. Length is always is a natural number.
So, we take x=5 only
Hence,length of an edge of original block=5 inches