find no of solution by details

How many real roots does x^4+12x-5=0 have

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Two real roots, I think because as x==> infinity , f(x) = x^4 + 12x -5 =0 ==> infinity , again for x==> (-inf.) ,
f(x) ==> infinity...because even degree ... and f'(x) possesses a single root (i.e., single minima for f(x) )....

Saloni Gupta - 8 years, 4 months ago

That was a very good answer.

Balaji Dodda - 8 years, 4 months ago

First consider the behaviour at infinity on either side. The highest degree term is even,so this function has more of a valley behaviour. The derivative i.e. 4x^3+12=0 has only one real root. So f(x)=x^4+12*x-5 has one minima. If the value of f(x) at minima turns out to be positive. Then function has no real roots at all and if it is negative it has 2 real roots and if it is zero,then it has one root with multiplicity 2. In this case f(x) is negative at minima. So the function has 2 real roots.

Balaji Dodda - 8 years, 4 months ago

It has 2 real roots.

Balaji Dodda - 8 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$