Question

The linear factor(s) of the equation $9x^2-24xy+16y^2-12x+16y-12=0$ is/are

• A. $3x-2y+6=0$
• B. $3x-2y+2=0$
• C. $3x-4y+2=0$
• D. $3x-4y-6=0$

Answer 1

Answer: (C), (D)

The correct options are
C $3x-4y+2=0$
D $3x-4y-6=0$

$9x^2-24xy+16y^2-12x+16y-12=0$
Writing equation in terms of $y,$ we get
$16y^2+(-24x+16)y+(9x^2-12x-12)=0$
Now, using quadratic formula, we have
$y=\frac{24x-16\pm \sqrt{(-24x+16{)}^2-4\cdot 16\cdot (9x^2-12x-12)}}{2\cdot 16}\Rightarrow y=\frac{24x-16\pm \sqrt{64(3x-2{)}^2-64(9x^2-12x-12)}}{2\cdot 16}\Rightarrow y=\frac{24x-16\pm 8\sqrt{(9x^2+4-12x)-(9x^2-12x-12)}}{2\cdot 16}\Rightarrow y=\frac{3x-2\pm 4}{4}\Rightarrow y=\frac{3x+2}{4}\text{or}y=\frac{3x-6}{4}$
​​​​​​​$\Rightarrow 3x-4y+2=0\text{or}3x-4y-6=0$



Alternate Solution:
$9x^2-24xy+16y^2-12x+16y-12=0\Rightarrow (9x^2-24xy+16y^2)-4(3x-4y)-12=0\Rightarrow (3x-4y{)}^2-4(3x-4y)-12=0$
$\Rightarrow (3x-4y+2)(3x-4y-6)=0$
​​​​​​​​​​​​​​$\Rightarrow 3x-4y+2=0\text{or}3x-4y-6=0$