Question

2(y² – 6y)² – 8(y² – 6y + 3) – 40 = 0

Answer 1

Given: 2(y² – 6y)² – 8(y² – 6y + 3) – 40 = 0
Let y² – 6y = m. Then, the equation can be written as:
2m² – 8(m + 3) – 40 = 0
$\Rightarrow$ 2m² – 8m – 24 – 40 = 0
$\Rightarrow$ 2m² – 8m – 64 = 0
$\Rightarrow$ 2(m² – 4m– 32) = 0
$\Rightarrow$ m² – 4m – 32 = 0
On splitting the middle term –4m as 4m – 8m, we get:
m² + 4m – 8m – 32 = 0
$\Rightarrow$ m( m + 4) – 8( m + 4) = 0
$\Rightarrow$ ( m – 8)( m + 4) = 0
$\Rightarrow$ m – 8 = 0 or m + 4 = 0
$\Rightarrow$ m = 8 or m = –4
On substituting m = y² – 6y, we get:
y² – 6y = 8 or y² – 6y = –4
$\Rightarrow$ y² – 6y – 8 = 0…(1) or y² – 6y + 4 = 0…(2)
Now, from equation (1), we get:
y² – 6y – 8 = 0
On using a quadratic formula, we get:
$$\begin{gathered} y=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(-8\right)}}{2\left(1\right)} \\ =\frac{6\pm \sqrt{36+32}}{2} \\ =\frac{6\pm \sqrt{68}}{2} \\ =\frac{6\pm 2\sqrt{17}}{2} \\ =3\pm \sqrt{17} \\ \mathrm{Now},\mathrm{from}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}: \\ {y}^2-6y+4=0 \\ \mathrm{On}\mathrm{using}\mathrm{the}\mathrm{quadratic}\mathrm{formula},\mathrm{we}\mathrm{get}: \\ y=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)} \\ =\frac{6\pm \sqrt{36-16}}{2} \\ =\frac{6\pm \sqrt{20}}{2} \\ =\frac{6\pm 2\sqrt{5}}{2} \\ =3\pm \sqrt{5} \\ \mathrm{Thus}\mathrm{the}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{are}y=3\pm \sqrt{17},3\pm \sqrt{5}. \end{gathered}$$