Question

The discriminant of the equation
(x+1)^3=4-X+X^3
​

Answer 1

Answer:

=x^3+3(x^2)(1)+3(x)(1^2)+(1^3)=4-x+x^3

=x^3+3x^2+3x+1=4-x+x^3

3x^2+4x-3

comparing ax^2+bx+c

a=3,b=4,c=-3

the discriminant

b^2-4ac

D=4^2-4(3)(-3)

D=16+36

D=52

Answer 2

The discriminant of the equation is 52

Given :

The equation (x + 1)³ = 4 - x + x³

To find :

The discriminant of the equation

Concept :

General form of a quadratic equation is

ax² + bx + c = 0

The Discriminant of the quadratic equation is denoted by D and defined as

D = b² - 4ac

Solution :

Step 1 of 3 :

Write down the given quadratic equation

Here the given quadratic equation is

(x + 1)³ = 4 - x + x³

Step 2 of 3 :

Simplify the given equation

$\displaystyle \sf {(x + 1)}^{3} = 4 - x + {x}^{3}$

$\displaystyle \sf{ \implies }{x}^{3} + 3 {x}^{2} + 3x + 1= 4 - x + {x}^{3}$

$\displaystyle \sf{ \implies } 3 {x}^{2} + 3x + 1= 4 - x$

$\displaystyle \sf{ \implies } 3 {x}^{2} + 4x - 3 = 0$

Step 3 of 3 :

Find the discriminant

The given equation is simplified to

3x² + 4x - 3 = 0

Comparing with the general equation ax² + bx + c = 0 we get

a = 3 , b = 4 , c = - 3

Hence discriminant of the quadratic equation

= b² - 4ac

$\sf = {( 4)}^{2} - 4 \times 3 \times ( - 3)$

$\sf = 16 + 36$

$\sf = 52$

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