Question

Which of the following value of $y$ satisfies the equation ?

$2(y^2-6y{)}^2-8(y^2-6y+3)-40=0$

• A. $y=-3\pm \sqrt{17}$
• B. $y=8\pm \sqrt{5}$
• C. $y=-8\pm \sqrt{5}$
• D. $y=3\pm \sqrt{17}$

Answer 1

Answer: (D)

$y=3\pm \sqrt{17}$

$2(y^2-6y{)}^2-8(y^2-6y+3)-40=0$
Let $y^2-6y=m$
$2m^2-8(m+3)-40=0$
$2m^2-8m-24-40=0$
$2m^2-8m-64=0$
$m^2-4m-32=0$
$m=\frac{4\pm 12}{2}$
$m=8,-4$
Now, when $m=8$
$y^2-6y=8$
$y^2-6y-8=0$
$y=\frac{6\pm \sqrt{36+32}}{2}$ = $3\pm \sqrt{17}$
Now, when $m=-4$
$y^2-6y=-4$
$y^2-6y+4=0$
$y=\frac{6\pm \sqrt{36-16}}{2}$ = $3\pm \sqrt{5}$