Question

(y-9)(y-7)(y+3)(y+1)-384=o​

Answer 1

Solution:

Given Equation:

$\rm \longrightarrow (y - 9)(y - 7)(y + 3)(y + 1) - 384 = 0$

Can be written as:

$\rm \longrightarrow (y - 9)(y + 3)(y - 7)(y + 1) - 384 = 0$

$\rm \longrightarrow ( {y}^{2} - 6y - 27)( {y}^{2} - 6y - 7) - 384 = 0$

Now, let us assume that:

$\rm \longrightarrow x = {y}^{2} - 6y$

Therefore:

$\rm \longrightarrow (x- 27)(x- 7) - 384 = 0$

$\rm \longrightarrow {x}^{2} - 34x + 189 - 384 = 0$

$\rm \longrightarrow {x}^{2} - 34x - 195= 0$

By splitting the middle term, we get:

$\rm \longrightarrow {x}^{2} - 39x + 5x- 195= 0$

$\rm \longrightarrow x(x - 39) + 5(x- 39)= 0$

$\rm \longrightarrow (x+ 5)(x- 39)= 0$

Therefore:

$\rm \longrightarrow x = -5,39$

Now, substitute back x = y² - 6y. We get:

$\longrightarrow \begin{cases} \rm {y}^{2} - 6y = - 5 \\ \rm {y}^{2} - 6y = 39 \end{cases}$

Solving the first equation:

$\rm \longrightarrow{y}^{2} - 6y = - 5$

$\rm \longrightarrow{y}^{2} - 6y + 5 = 0$

$\rm \longrightarrow{y}^{2} - y - 5y+ 5 = 0$

$\rm \longrightarrow y(y -1) - 5(y - 1)= 0$

$\rm \longrightarrow (y - 5)(y - 1)= 0$

$\rm \longrightarrow y = 1,5 - (i)$

Now, consider the second equation:

$\rm \longrightarrow{y}^{2} - 6y = 39$

$\rm \longrightarrow{y}^{2} - 6y - 39 = 0$

$\rm \longrightarrow y = \dfrac{6 \pm \sqrt{36 + 4 \times 39} }{2}$

$\rm \longrightarrow y = \dfrac{6 \pm \sqrt{36 + 156} }{2}$

$\rm \longrightarrow y = \dfrac{6 \pm \sqrt{192} }{2}$

$\rm \longrightarrow y = \dfrac{6 \pm \sqrt{64 \times 3} }{2}$

$\rm \longrightarrow y = \dfrac{6 \pm 8\sqrt{3} }{2}$

$\rm \longrightarrow y =3 \pm4 \sqrt{3}$

$\rm \longrightarrow y = 4 + \sqrt{3} ,4 - \sqrt{3} - (ii)$

So, the values of y are:

$\rm \longrightarrow y =1,5,4 + \sqrt{3} ,4 - \sqrt{3}$

★ Which is our required answer.

Answer:

• The values of y are 1, 5, 4 + √3 and 4 - √3