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Determine the unique pair of real numbers \((x,y)\) satisfying \((3x^2 - 12x + 16)(2y^2 + 6y + 9) = 18.\)

Enter your answer as an ordered pair in the format \((x,y)\), where \(x\) and \(y\) are replaced by appropriate numbers.

Thank you very much for your help!

Jul 12, 2020

#1
+1

The unique answer is  (2, -1.5).

One way to determine this answer:  (I'm going to put it into graphing form.)

2y2 + 6y + 9   =  18 / (3x2 - 12x + 16)

2y2 + 6y         =  18 / (3x2 - 12x + 16) - 9

2(y2 + 3y)       =  18 / (3x2 - 12x + 16) - 9

(y2 + 3y)       =  9 / (3x2 - 12x + 16) - 9/2                                       (divide by 2)

y2 + 3y + 9/4  =   9 / (3x2 - 12x + 16) - 9/2 + 9/4                             (complete the square)

(y + 3/2)2     =  9 / (3x2 - 12x + 16) - 9/4                                        (factor; simplify)

(y + 3/2)2     =  [ 36 - 9(3x2 - 12x + 16) ]  / [ 4(3x2 - 12x + 16) ]     (common denominator)

(y + 3/2)2     =  [ 36 - 273x2 + 108x - 144) ]  / [ 4(3x2 - 12x + 16) ]

(y + 3/2)2     =  [ - 27x2 + 108x - 108) ]  / [ 4(3x2 - 12x + 16) ]

(y + 3/2)2     =  [ - 27(x2 + 4x - 4) ]  / [ 4(3x2 - 12x + 16) ]

(y + 3/2)2     =  [ - 27(x - 2)2 ]  / [ 4(3x2 - 12x + 16) ]

y + 3/2        =  +/-  sqrt( [ - 27(x - 2)2 ]  / [ 4(3x2 - 12x + 16) ] )

y             =  +/-  sqrt( [ - 27(x - 2)2 ]  / [ 4(3x2 - 12x + 16) ] )  -  3/2

The graph of  y = 3x2 - 12x + 16  has only positive values for y.

The graph of  y = -27(x - 2)2  has only negative values for y except when x = 2; then its value is zero.

Therefore, the only real value for the expression under the square root sign occurs when x = 2.

And, when x = 2, the y-value in the original equation is -1.5.

I wonder who thought up this problem!

Jul 12, 2020
#2
0

EGAD!   WIsh I could give you more 'points' for THAT one !     ~ EP

ElectricPavlov  Jul 12, 2020
#3
+1

Thanks, EP -- I appreciate your comment!

geno3141  Jul 12, 2020