If x 4 - 1 2 - 4 then x 3 - x 2 - 1 - -

The correct option is A 2549we have, x4+12=4 x4=4 12 x=4×(72)=14 x=14 ⇒ x^3 x^2+1 #160; =(14)^3 (14)^2+1 =2,744 196+1=2,54 ...

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Standard VIII

Mathematics

Powers with Negative Exponents

If x 4 + 1 ...

Question

If $\frac{x}{4} + \frac{1}{2} = 4$ then $x^{3} - x^{2} + 1 = ?$

2549

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Solution

The correct option is A 2549
we have, $\frac{x}{4} + \frac{1}{2} = 4$

$\frac{x}{4} = 4 - \frac{1}{2}$

$x = 4 \times (\frac{7}{2}) = 14$

$x = 14$

$\Rightarrow x^{3} - x^{2} + 1$ $= (14)^{3} - (14)^{2} + 1$

$= 2, 744 - 196 + 1 = 2, 549$ .

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Similar questions

If $x_{1}, x_{2}, x_{3}, x_{4}$ are four positive real numbers such that $x_{1} + \frac{1}{x_{2}} = 4, x_{2} + \frac{1}{x_{3}} = 1, x_{3} + \frac{1}{x_{4}} = 4$ and $x_{4} + \frac{1}{x_{1}} = 1,$ then

x^{3} - x - 4

added to

x^{4} - x^{3} - x^{2} + x + 3

to obtain

x^{4} + x^{2} - 1

?
If true then enter

1

and if false then enter

0

If $x + \frac{1}{x} = 3$ , calculate $x^{2} + \frac{1}{x^{2}}, x^{3} + \frac{1}{x^{3}} a n d x^{4} + \frac{1}{x^{4}}$ .

Q. If |x| < 1 and

y = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . . . . . . . t h e n x =

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If x4+12=4 then x3−x2+1=?

The correct option is A 2549we have, x4+12=4x4=4−12x=4×(72)=14x=14⇒x3−x2+1 =(14)3−(14)2+1=2,744−196+1=2,549.

If $\frac{x}{4} + \frac{1}{2} = 4$ then $x^{3} - x^{2} + 1 = ?$

The correct option is A 2549
we have, $\frac{x}{4} + \frac{1}{2} = 4$

$\frac{x}{4} = 4 - \frac{1}{2}$

$x = 4 \times (\frac{7}{2}) = 14$

$x = 14$

$\Rightarrow x^{3} - x^{2} + 1$ $= (14)^{3} - (14)^{2} + 1$

$= 2, 744 - 196 + 1 = 2, 549$ .