en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((2x^2+9x-5))/((x^4+1))=0

Solution

\frac{( 2 x ^{2} + 9 x - 5 )}{( x ^{4} + 1 )} = 0

Solution

x = \frac{1}{2}, x = - 5

+1

Decimal Notation

x = 0.5, x = - 5

Solution steps

\frac{( 2 x ^{2} + 9 x - 5 )}{( x ^{4} + 1 )} = 0

Simplify

\frac{( 2 x ^{2} + 9 x - 5 )}{( x ^{4} + 1 )} : \frac{2 x ^{2} + 9 x - 5}{x ^{4} + 1}

\frac{2 x ^{2} + 9 x - 5}{x ^{4} + 1} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 x^{2} + 9 x - 5 = 0

Solve

2 x^{2} + 9 x - 5 = 0 : x = \frac{1}{2}, x = - 5

x = \frac{1}{2}, x = - 5

Popular Examples

437=((x+7))/(2*3(-4+x))

437 = \frac{( x + 7 )}{2 \cdot 3 ( - 4 + x )}

1/(2x[ 3/4+x])=-7/10

\frac{1}{2 x [ \frac{3}{4} + x ]} = - \frac{7}{10}

((n-8)!)/((n-9)!)=9

\frac{( n - 8 ) !}{( n - 9 ) !} = 9

9/(x^4)+8/(x^2)=1

\frac{9}{x ^{4}} + \frac{8}{x ^{2}} = 1

solvefor x,t= 1/((x-y))*(x+y)solve for

x, t = \frac{1}{( x - y )} \cdot (x + y)

Frequently Asked Questions (FAQ)

What is the answer to ((2x^2+9x-5))/((x^4+1))=0 ?
The answer to ((2x^2+9x-5))/((x^4+1))=0 is x= 1/2 ,x=-5
Solve for x: ((2x^2+9x-5))/((x^4+1))=0
The solution is x= 1/2 ,x=-5