How many solutions the following equation has?

$3 x^{3} = (x^{2} + \sqrt{18} x + \sqrt{32}) (x^{2} - \sqrt{18} x - \sqrt{32}) - 4 x^{2}$

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Four

two

no solution

None

answer is A.

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Detailed Solution

Given equation is,

3 x^{3} = (x^{2} + \sqrt{18} x + \sqrt{32}) (x^{2} - \sqrt{18} x - \sqrt{32}) - 4 x^{2}

Solve it as,

3 x^{3} = (x^{2} + \sqrt{18} x + \sqrt{32}) (x^{2} - \sqrt{18} x - \sqrt{32}) - 4 x^{2}

3 x^{3} = (x^{4} - (\sqrt{18} x + \sqrt{32})^{2}) - 4 x^{2}

[Since

(a - b) (a + b) = (a^{2} - b^{2})]

3 x^{3} = (x^{4} - (18 x^{2} + 32 + 2 \sqrt{18} \sqrt{32} x)) - 4 x^{2}

[Since

{(a + b)}^{2} = a^{2} + b^{2} + 2 ab]

3 x^{3} = x^{4} - 18 x^{2} - 32 - 2 \sqrt{18} \sqrt{32} x - 4 x^{2}

x^{4} - 3 x^{3} - 22 x^{2} - 2 \sqrt{18} \sqrt{32} x - 32 = 0

x^{4} - 3 x^{3} - 22 x^{2} - 48 x - 32 =

0
The highest power of variable in the above equation is 4, so its degree will be 4.
Now the degree of a polynomial is equal to the number of solutions of the polynomial.
Thus, the number of solutions of the equation

3 x^{3} = (x^{2} + \sqrt{18} x + \sqrt{32}) (x^{2} - \sqrt{18} x - \sqrt{32}) - 4 x^{2}

= 4.
So, the correct option is 1.

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