SOLUTION: Find the equation whose roots are each two more than the roots of x^3-4x^2+3x+2

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Question 910555: Find the equation whose roots are each two more than the roots of x^3-4x^2+3x+2
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
first find roots:
x%5E3-4x%5E2%2B3x%2B2=0 write -4x%5E2 as -2x%5E2-2x%5E2 and 3x as 4x-x
x%5E3-2x%5E2-2x%5E2%2B4x-x%2B2=0 group
%28x%5E3-2x%5E2%29-%282x%5E2-4x%29-%28x-2%29=0
x%5E2%28x-2%29-2x%28x-2%29-%28x-2%29=0
%28x-2%29%28x%5E2-2x-1%29=0
if x-2=0 => x=2 first root
to find other two zeros solve x%5E2-2x-1=0 using quadr. formula

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+

x+=+%282+%2B-+sqrt%284%2B4+%29%29%2F2+
x+=+%282+%2B-+sqrt%282%2A4+%29%29%2F2+
x+=+%282+%2B-+2sqrt%282+%29%29%2F2+
x+=+%28cross%282%29+%2B-+cross%282%29sqrt%282+%29%29%2Fcross%282%29+
x+=+%281+%2B-+sqrt%282+%29%29+
x+=+%281+%2B-+1.41%29+

second root: x+=+1+%2B+1.41=2.41+
third root: x+=+1+-+1.41=-0.41+

now make all two more:
first root x=2 => two more will be x%5B1%5D=4
second root: x+=2.41+ => two more will be x%5B2%5D+=4.41+
third root: x+=-0.41+ => two more will be x%5B3%5D+=1.59+
Find the equation using roots:
+%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29+
+%28x-4%29%28x-4.41%29%28x-1.59%29+ multiply all
+%28x%5E2-8.41x%2B17.64%29%28x-1.59%29+
x%5E3-10x%5E2%2B31.0119x-28.0476