Question

If x2-15/x=4 and x≠3 find x(x+1)(x+2)(x+3)=? a)12 b)15 c)21 d)27
Please solve this question

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Option: b is the correct answer.

                   b) 15

Step-by-step explanation:

We are given an algebraic expression as:

$x^2-\dfrac{15}{x}=4$

Now we can represent it as:

$\dfrac{x^3-15}{x}=4\\\\\\x^3-15=4x\\\\x^3-4x-15=0\\\\(x-3)(x^2+3x+5)=0\\\\Since\ x\neq 3\\\\This\ means\ that:\\\\\\x^2+3x+5=0$

Hence, on solving the quadratic inequality we get:

$x=\dfrac{-3\pm \sqrt{9-20}}{2}\\\\x=\dfrac{-3\pm \sqrt{11}i}{2}$

when,

$x=\dfrac{-3+\sqrt{11}i}{2}\\\\x+1=\dfrac{-1+\sqrt{11}i}{2}\\\\x+2=\dfrac{1+\sqrt{11}i}{2}\\\\x+3=\dfrac{3+\sqrt{11}i}{2}\\\\Hence,\\x(x+1)(x+2)(x+3)=\dfrac{(-3+\sqrt{11}i)(3+\sqrt{11}i)(-1+\sqrt{11}i)(1+\sqrt{11}i)}{16}\\\\\\x(x+1)(x+2)(x+3)=\dfrac{(-9-11)(-1-11)}{16}\\\\\\Since\ ,\ (a-b)(a+b)=a^2-b^2\\\\Hence,\\\\\\x(x+1)(x+2)(x+3)=15$

Similarly we can check for:

$x=\dfrac{-3-\sqrt{11}i}{2}$

The same value will be obtained.

                        Hence,

                x(x+1)(x+2)(x+3)=15