Question 641585

solve the equation  in real # system


real solutions of the equation are x=?

dont see why my tutor i was assigned cant do these, please help


{{{4x^3 - 3x^2 + 31x + 8 = 0}}} ---- Multiplying eq by 4 to clear the fractions


From calculations, {{{- 1/4}}} was found to be a root of the equation.

{{{x = - 1/4}}} ----- {{{x + 1/4 = 0}}} ----- 4x + 1 = 0.
This means that one real solution is {{{x = - 1/4}}}, or {{{- 0.25}}}.
 

Using long division and dividing {{{4x^3 - 3x^2 + 31x + 8}}} by 4x + 1, we get: {{{x^2 - x + 8}}}


We now have: {{{4x^3 - 3x^2 + 31x + 8 = 0}}} ----- {{{(4x + 1)(x^2 - x + 8) = 0}}}
	
							
Solving {{{x^2 - x + 8}}} using the quadratic equation formula gives us a negative discriminant ({{{b^2 - 4ac}}}), or 1 – 4 * 1 * 8, or 1 – 32, or - 31)


As a negative discriminant gives us 2 imaginary solutions, both of which DO NOT INTERSECT the x-axis, the 2 imaginary solutions are regarded as non-solutions, since we're looking for real solutions.


Therefore, the one and only REAL solution is {{{highlight_green(x = - 1/4)}}}, or {{{highlight_green(- 0.25)}}}. 


I hope this helps you understand it better!!


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