What are the solutions to the equation ax^2-(4a+1)x+3a+3=0 ?

The solutions to the equation ax^2-(4a+1)x+3a+3=0 are x=3,x=(a+1)/a ;\quad a\ne 0

Find the zeros of ax^2-(4a+1)x+3a+3=0

The zeros of ax^2-(4a+1)x+3a+3=0 are x=3,x=(a+1)/a ;\quad a\ne 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

a x^{2} - (4 a + 1) x + 3 a + 3 = 0

x = 3, x = \frac{a + 1}{a}; a \neq = 0

Solution steps

a x^{2} - (4 a + 1) x + 3 a + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 a - 1 ) \pm ( - 4 a - 1 ) ^{2} - 4 a ( 3 a + 3 )}{2 a}

Simplify

(- 4 a - 1)^{2} - 4 a (3 a + 3) : 2 a - 1

x_{1, 2} = \frac{- ( - 4 a - 1 ) \pm ( 2 a - 1 )}{2 a}; a \neq = 0

Separate the solutions

x_{1} = \frac{- ( - 4 a - 1 ) + 2 a - 1}{2 a}, x_{2} = \frac{- ( - 4 a - 1 ) - ( 2 a - 1 )}{2 a}

x = \frac{- ( - 4 a - 1 ) + 2 a - 1}{2 a} : 3

x = \frac{- ( - 4 a - 1 ) - ( 2 a - 1 )}{2 a} : \frac{a + 1}{a}

The solutions to the quadratic equation are:

x = 3, x = \frac{a + 1}{a}; a \neq = 0

2 k^{2} + 5 k + 4 = 1

a, a^{2} + b^{2} = 0

5 t^{2} + t - 6 = 0

n^{2} - n - 10 = 2 n

6 u^{2} + 5 u = 1

What are the solutions to the equation ax^2-(4a+1)x+3a+3=0 ?
The solutions to the equation ax^2-(4a+1)x+3a+3=0 are x=3,x=(a+1)/a ;\quad a\ne 0
Find the zeros of ax^2-(4a+1)x+3a+3=0
The zeros of ax^2-(4a+1)x+3a+3=0 are x=3,x=(a+1)/a ;\quad a\ne 0