Question

solve for 'x'


(x+9)(x-3)(x-7)(x+5) = 385

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Answer 1

Ans : x = 2 and - 4 and - 1+ √71 and -1 - √71

step by step explanation:

(x + 9) (x - 3) (x - 7) (x+ 5) =385

(x^2 +9x - 7x - 63)(x^2 +5x - 3x - 15 )=385

(x^2 +2x - 63 )( x^2 + 2x - 15 )= 385

Let K = x^2 +2x -15

(K - 48) (K) = 385

K^2 -48 K - 385 =0

K^2 - 55K + 7K - 385 =0

K(K - 55) +7 (K- 55) = 0

(K - 55) (K+ 7) =0

Replace k value

(x^2 +2x - 15 -55)(x^ + 2x - 15 +7) = 0

(x^2 +2x - 70)( x^2 +2x - 8) = 0

( -2)+ √ 4+280/2, (-2) -√ 4+ 280/2 or x^2+4x - 2x - 8= 0

x(x+4) -2 (x +4) =0

(x+4) (x -2) = 0

x +4 =0 or x-2=0

x= -4 or x= 2 or - 1+√71 or -1 -√71

hope this will help you mark as brainliest

Answer 2

ANSWER

x = 2 , -4 , -1 - √71 , -1 + √71.

Step By Step Explanation

We have ->

(x+9)(x-3)(x-7)(x+5) = 385

On rearranging we get

=>(x+9)(x-7)(x-3)(x+5) = 385

=>$(x^{2} - 2x - 63)(x^{2} - 2x - 15) = 385$

Now, let $(x^{2} - 2x)$ = t.

Now we get >

(t - 63)(t - 15) = 385

=>$t^{2} - 15t - 63t + 945 = 385$

=>$t^{2} - 78t + 945 = 385$

=>$t^{2} - 8t - 70t + 945 - 385 =0$

=>$t^{2} - 8t -70t + 560 = 0$

=>$t(t - 8) - 70(t - 8) = 0$

=>(t - 8) (t - 70) = 0

So,

either t = 8 ------(i)

or t = 70 ------(ii)

Now on putting $(x^{2} - 2x)$ in place of t in (i) we get =>

$x^{2} + 2x = 8$

$\implies x^{2} + 2x - 8 = 0$

$\implies x^{2} + 4x - 2x - 8 = 0$

$\implies x(x+ 4) - 2(x + 4) = 0$

$\implies (x+ 4) (x - 2) = 0$

So, either x = -4 or 2

Hence, we get two values of x as -4 and 2.

Now on putting $(x^{2} - 2x)$ in place of t in (ii) we get =>

$(x^{2} - 2x) = 70$

$(x^{2} - 2x) - 70 = 0$

$x^{2} - 2x - 70 = 0$

Here, we shall use the Quadratic formula method to solve this equation.

Now we have,

discriminant,$d = b^{2} - 4ac$

$\implies d = (-2)^{2} - 4\times 1 \times (-70)$

$\implies d = 4 + 4\times 70$

$\implies d = 4 +280$

$\implies d = 284$

Now, $x = \dfrac{-b \pm \sqrt{d} }{2a}$

$\implies x = \dfrac{-(-2) \pm \sqrt{294} }{2 \times 1}$

$\implies x = \dfrac{2 \pm 2\sqrt{71} }{2}$

So,

either

$x = \dfrac{2 + 2\sqrt{71} }{2}$

or

$x = \dfrac{2 - 2\sqrt{71} }{2}$

So, the other two values of x are $x = \dfrac{2 + 2\sqrt{71} }{2} and\: x = \dfrac{2 - 2\sqrt{71} }{2}$

So,$x = -4 , 2, x = \dfrac{2 + 2\sqrt{71} }{2} and \: x = \dfrac{2 - 2\sqrt{71} }{2}$.

or we can write it as >

x = 2 , -4 , -1 - √71 , -1 + √71 are the values.