Asked by Tuaha
how to solve algebra equation
√(5x^2 )+7x+2-√(4x^2+7x+18) =x-4
Answers
Answered by
Steve
√(5x^2)+7x+2-√(4x^2+7x+18) = x-4
√(5x^2)-√(4x^2+7x+18) = -6x-2
now square both sides to get
(5x-2) - 2√(5x^2)(4x^2+7x+18) + (4x^2+7x+18) = (6x+2)^2
4x^2+12x+16 - 2√(20x^4+35x^3+90x^2) = 36x^2 + 24x + 4
√(20x^4+35x^3+90x^2) = -16x^2-6x+6
20x^4+35x^3+90x^2 = 256x^4+192x^3-156x^3-72x^2+36
236x^4+157x^3-246x^2-72x+36 = 0
quartics are tough, but solving graphically, I get
x = -1.23, -0.55, 0.29,0.81
Of these, discounting the spurious solutions introduced by squaring twice, I get only -0.55
better check my algebra
√(5x^2)-√(4x^2+7x+18) = -6x-2
now square both sides to get
(5x-2) - 2√(5x^2)(4x^2+7x+18) + (4x^2+7x+18) = (6x+2)^2
4x^2+12x+16 - 2√(20x^4+35x^3+90x^2) = 36x^2 + 24x + 4
√(20x^4+35x^3+90x^2) = -16x^2-6x+6
20x^4+35x^3+90x^2 = 256x^4+192x^3-156x^3-72x^2+36
236x^4+157x^3-246x^2-72x+36 = 0
quartics are tough, but solving graphically, I get
x = -1.23, -0.55, 0.29,0.81
Of these, discounting the spurious solutions introduced by squaring twice, I get only -0.55
better check my algebra
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