Solve the following?
#(x−1)(x−3)(x−5)(x+1)=−12#
↳Redirected from
"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
#x=−0.645751,0.267949,3.732051,4.645751#
#(x-1)(x-3)(x-5)(x+1)=-12#
first, we can expand each of the brackets using FOIL
#(x-1)(x-3) = x^2 - 3x - x + 3#
#(x-5)(x+1) = x^2 +x -5x - 5#
Now we can simplify these and combine them back together into the original expression.
#(x^2 - 4x + 3) xx (x^2 -4x - 5) = -12#
Now we can factorise this.
#(color(lime)(x^2 - 4x + 3)) (color(skyblue)(x^2 -4x - 5)) = (color(lime)(x^2)) (color(skyblue)(x^2)) + (color(lime)(x^2))(color(skyblue)(-4x)) + (color(lime)(x^2))(color(skyblue)(-5)) + (color(lime)(-4x))(color(skyblue)(x^2)) + (color(lime)(-4x))(color(skyblue)(-4x)) + (color(lime)(-4x))(color(skyblue)(-5)) + (color(lime)(3))(color(skyblue)(x^2)) + (color(lime)(3))(color(skyblue)(-4x)) + (color(lime)(3))(color(skyblue)(-5))#
# = (x^4) + (-4x^3) + (-5x^2) + (-4x^3) + (16x^2) + (20x) + (3x^2) + (-12x) + (-15)#
Now we can collect like terms and simplify.
# = x^4 -4x^3 - 4x^3 -5x^2 + 3x^2 + 16x^2 + 20x -12x -15#
#x^4 - 8x^3 + 14x^2 + 8x -15 = -12#
Now we can find #x#
#x^4 - 8x^3 + 14x^2 + 8x color(red)(cancel(color(black)(-15) +15)) = -12 color(red)(+15)#
#x^4 - 8x^3 + 14x^2 + 8x = 3#
The only way that I could find to solve this is through using the Quartic formula.
#color(blue)(x=−0.645751,0.267949,3.732051,4.645751)#
#+-sqrt3+2# and #+-sqrt7+2#
We have:
#(x-1)(x-3)(x-5)(x+1)=-12#
Multiply the first two and the last two parentheses together.
#=>(x^2-4x+3)(x^2-4x-5)=-12#
Hmm... We see that the first quadratic expression is 8 larger than the second.
We let #s=(x^2-4x+3)#
We now have:
#s(s-8)=-12#
#=>s^2-8s=-12# complete the square
#=>(s^2-8s+16)-16=-12#
#=>(s-4)^2=4#
#=>-2=s-4=2#
#=>2=s=6#
We can use this to solve for #x#.
When #s=2#...
#x^2-4x+3=2#
#=>x^2-4x+1=0#
#=>(x^2-4x+4)-4+1=0#
#=>(x-2)^2-3=0#
#=>(x-2)^2=3#
#=>x-2=+-sqrt(3)#
#=>x=+-sqrt(3)+2#
When #s=6#...
#x^2-4x+3=6#
#=>x^2-4x-3=0#
#=>(x^2-4x+4)-4-3=0#
#=>(x-2)^2-7=0#
#=>(x-2)^2=7#
#=>x-2=+-sqrt(7)#
#=>x=+-sqrt(7)+2#
The answers are #+-sqrt3+2# and #+-sqrt7+2#
