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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 2, x = - \frac{1}{a - 1}; a \neq = 1

Solution steps

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

Expand

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

a x^{2} - x^{2} - 2 a x + 3 x - 2 = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = - \frac{1}{a - 1}; a \neq = 1

x + x^{2} - 5 = 0

4 x^{2} - 101 x + 25 = 0

2 + 6 t - 4.9 t^{2} = 0

(s + 3) (s + 5) = 0

(5 x + 2) (5 x - 2) = (A + B) (A - B)

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