Solve for real y y^2+3+ 12/(y^2+3)=7

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve : y^2 +3 + 12/(y^2+3)=7

y^2 +3 + 12/(y^2+3)=7

=> (y^2 + 3)^2 + 12 = 7( y^2 + 3)

=> y^4 + 9 + 6y^2 + 12 = 7y^2 + 21

=> y^4 - y^2 = 0

=> y^2( y^2 - 1) = 0

y^2 = 0 or y = 0

y^2 - 1 = 0

=> y^2 = 1

=> y = 1 and y = -1

The required real values of y are 0 , -1 and 1

Top Answer

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll note y^2 + 3 = t

We'll re-write the equation in t:

t + 12/t = 7

We'll multiply by t both sides, after that, bringing all terms to one side:

t^2 + 12 - 7t = 0

We'll re-arrange the terms:

t^2 - 7t + 12 = 0

We'll apply quadratic formula:

t1 = [7 + sqrt(49 - 48)]/2

t1 = (7+1)/2

t1 = 4

t2 = (7-1)/2

t2 = 3

But y^2 + 3 = t

For t = 4=> y^2 + 3 = 4

y^2 = 4 - 3

y^2 = 1

y1 = 1 and y2 = -1

For t = 3 => y^2 + 3 = 3

y^2 = 0

y3 = y4 = 0

The real roots of the equation are: {-1 ; 0 ; 1}.

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