`int_a^5(x^2+4x+1)dx=132`
`[x^3/3+4x^2/2+x]_a^5=132`
`[x^3/3+2x^2+x]_a^5=132`
`(5^3/3+2(5)^2+5)-(a^3/3+2a^2+a)=132`
`(125/3+50+5)-(a^3/3+2a^2+a)=132`
`((125+150+15)/3)-((a^3+6a^2+3a)/3)=132`
`((290-a^3-6a^2-3a)/3)=132`
`290-a^3-6a^2-3a=396`
`a^3+6a^2+3a+396-290=0`
`a^3+6a^2+3a+106=0`
Since the discriminant of this cubic equation is less than zero, so the equation has one real root and two complex conjugate roots.
Now to solve this cubic equation , let us depress the cubic equation by substituting a with (y-2).
So, `(y-2)^3+6(y-2)^2+3(y-2)+106=0`
`(y^3-8-6y(y-2))+6(y^2+4-4y)+3y-6+106=0`
`y^3-8-6y^2+12y+6y^2+24-24y+3y-6+106=0`
`y^3-9y+116=0`
Now we have to solve this depressed cubic equation of the form y^3+Ay=B
`y=s-t , A=3st=-9 , B=s^3-t^3=-116`
`(A/(3t))^3-t^3=B`
`(-9/(3t))^3-t^3=-116`
`27/t^3+t^3=116`
`t^6+27=116t^3`
`t^6-116t^3+27=0`
Now to solve the above equation for t , let us reduce to the quadratic form by assuming , t^3=u.
So the equation reduces to,`u^2-116u+27=0`
`u=(116+-sqrt(116^2-4*27))/2`
`u=(116+-sqrt(13456-108))/2`
`u=(116+-sqrt(13348))/2`
`u=(116+-sqrt(4*3337))/2 = 58+-sqrt(3337)`
`:.t=root(3)(58+-sqrt(3337))`
`s=-3/t`
`s=-3/root(3)(58+-sqrt(3337))`
`y=s-t and a=y-2`
` :.a=-2+s-t`
`a=-2-3/root(3)(58+sqrt(3337))-root(3)(58+sqrt(3337))`
` or a=-2-3/root(3)(58-sqrt(3337))-root(3)(58-sqrt(3337)) `
`or a=-7.48927`
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