Question

$\red{\bf\:{x}^2-(\frac{15}{x})=4\\ \text\sf{,,where\:x\:does\:not\:equal\:to\:3}\\\\\underline\green{\text\bf{,,Find\:value\:ofx(x+1)(x+2)(x+3.)}}$

ANSWER THE FOLLOWING

Answer 1

||✪✪CORRECT QUESTION ✪✪||

If X^2- (15/x) =4 , where x does not equal to 3. Find value of :- x( x+1( x+2)( x+3) ?

|| ✰✰ ANSWER ✰✰ ||

Given That :-

➼ x² - (15/x) = 4

Taking LCM,

➼ (x³ - 15)/x = 4

➼ (x³ - 15) = 4x

Adding (-12) Both Sides now, we get,

➼ (x³ - 15) - 12 = 4x - 12

➼ (x³ - 27) = 4(x - 3)

➼ (x³ - 3³) = 4(x - 3)

using (a³ - b³) = (a - b)(a² + b² + ab) in LHS Now,

➼ (x - 3)(x² + 3x + 9) = 4(x - 3)

Cancelling (x - 3) From both sides now,

➼ (x² + 3x + 9) = 4

➼ (x² + 3x) = 4 - 9

➼ (x² + 3x) = (-5) ---------------- Equation ❶

_____________________________

Now, Adding 2 Both Sides of Equation ❶ we get,

➻ (x² + 3x + 2) = (-5) + 2

➻ (x² + 3x + 2) = (-3) -------------- Equation ❷

___________________________

Now, we Have To Find The value of x(x+1)(x+2)(x+3) :-

☛ x(x+1)(x+2)(x+3)

☛ [x(x+3)] * [(x+1)(x+2)]

☛ (x² + 3x) * (x² + 3x + 2)

Putting Values of Equation ❶ & ❷ Now, we get ,

☛ (-5) * (-3)

☛ 15 (Ans.)

ஃ The Value of x(x+1)(x+2)(x+3) will be 15.

___________________________

Answer 2

We are given x² - ($\frac {15}{x}$) = 4

=> $\frac {(x³ - 15)}{x}$ = 4

=> (x³ - 15) = 4x

By adding -12 to both Sides now:-

=> (x³ - 15) - 12 = 4x - 12

=> (x³ - 27) = 4(x - 3)

=> (x³ - 3³) = 4(x - 3)

By using identity:-

(a³ - b³) = (a - b)(a² + b² + ab)

=> (x - 3)(x² + 9 + 3x) = 4(x - 3)

=> (x² + 3x + 9) = $\frac {4(x - 3)}{(x - 3)}$

=> (x² + 3x + 9) = 4

=> x² + 3x = 4 - 9

=> x² + 3x = -5 -(i)

By adding 2 to both sides:-

=> x² + 3x + 2 = -5 + 2

=> x² + 3x + 2 = -3 -(ii)

We have to find the value of x(x+1)(x+2)(x+3)

=> x(x+1)(x+2)(x+3)

=> [x(x+3)] × [(x+1)(x+2)]

=> (x² + 3x) × (x² + 3x + 2)

By putting the values of (i) and (ii):-

=> (-5) × (-3)

=> 5 × 3

=> 15

Therefore, the answer is 15.