Question

8x^2 + y^2 - 12x - 4xy + 9 = 0 find value of 14x - 5y

Answer 1

14x - 5y = 6 if 8x²  + y²  - 12x  - 4xy  + 9 = 0

Step-by-step explanation:

8x²  + y²  - 12x  - 4xy  + 9 = 0

=> 4x² + 4x² + y² - 12x - 4xy + 9 =0

=> 4x²  + y² - 4xy  + 4x²  -12x  + 9 = 0

=> (2x - y)²  + (2x - 3)² = 0

=> 2x - 3 = 0

=> x = 3/2

2x - y = 0

=> y = 2x

=> y = 2 * 3/2

=> y = 3

14x - 5y

= 14 * (3/2)  - 5(3)

= 21 - 15

= 6

14x - 5y = 6

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1. We are aware of the identity that

${(a + b)}^{2} = {a}^{2} + 2 ab+ {b}^{2}$

2. If the sum of squares of two real numbers are zero then they are separately zero

GIVEN

$8 {x}^{2} + {y}^{2} - 12x - 4xy + 9 = 0$

TO DETERMINE

$14x - 5y$

EVALUATION

$8 {x}^{2} + {y}^{2} - 12x - 4xy + 9 = 0$

$\implies \: 4 {x}^{2} - 4xy + {y}^{2} + 4 {x}^{2} - 12x + 9 = 0$

$\implies \: \bigg[ {(2x)}^{2} - 2 \times 2x \times y + {(y)}^{2} \bigg] + \bigg[ \: {(2x)}^{2} - 2 \times 2x \times 3 + {(3)}^{2} \bigg] = 0$

$\implies \: {(2x - y)}^{2} + {(2x - 3)}^{2} = 0$

Now the sum of the squares of two real numbers are zero then they are separately zero

Hence

$2x - y = 0 \: \: \: \: and \: \: \: \: 2x - 3 = 0$

Now

$2x - 3 = 0 \: \: \: gives \:$

$2x = 3$

$\implies \: \displaystyle \: x = \frac{3}{2}$

Again

$2x - y = 0 \: \: \: gives \: \:$

$y = 2x$

$\implies \: \displaystyle \: y = 2 \times \frac{3}{2}$

$\implies \: \displaystyle \: y= 3$

RESULT

$14x - 5y$

$= (14 \times \displaystyle \: \frac{3}{2} ) - (5 \times 3)$

$= 21 - 15$

$= 6$