If x-1-2-3 then find the value of x3-2 x2-7 x-5A- 4B- 8C- 3D- 14

The correct option is C 3 x = 1/(2 √(3))×(2 + √(3))/(2 + √(3)) = (2 + √(3))/4 3 = (2 + √(3)) ⇒ (x 2) = √(3) ∴ (x 2)^2 = 3 ⇒ x^2 4x + 4 = 3 ⇒ x^2 4 ...

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Byju's Answer

Standard VII

Mathematics

Power of a Power

If x=1/2-√3 t...

Question

If $x = \frac{1}{(2 - \sqrt{3}),}$ then find the value of $(x^{3} - 2 x^{2} - 7 x + 5)$ .

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Solution

The correct option is C
3

$x = \frac{1}{(2 - \sqrt{3})} \times \frac{(2 + \sqrt{3})}{(2 + \sqrt{3})} = \frac{(2 + \sqrt{3})}{4 - 3} = (2 + \sqrt{3}) \Rightarrow (x - 2) = \sqrt{3} ∴ (x - 2)^{2} = 3 \Rightarrow x^{2} - 4 x + 4 = 3 \Rightarrow x^{2} - 4 x + 1 = 0 ∴ x^{3} - 2 x^{2} - 7 x + 5 = x (x^{2} - 4 x + 1) + 2 (x^{2} - 4 x + 1) + 3 = x \times 0 + 2 \times 0 + 3 = 3$

Alternate approach:

Substitute $x = (2 + \sqrt{3})$ in $(x^{3} - 2 x^{2} - 7 x + 5)$ and simplify using identities $(a + b)^{3}$ and $(a + b)^{2}$

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If $x = \frac{1}{(2 - \sqrt{3}),}$ then find the value of $(x^{3} - 2 x^{2} - 7 x + 5)$ .

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If x=1(2−√3), then find the value of (x3−2x2−7x+5).

If $x = \frac{1}{(2 - \sqrt{3}),}$ then find the value of $(x^{3} - 2 x^{2} - 7 x + 5)$ .