Question

(x+1)2-x2=0 has number of real roots equal to​

Answer 1

Answer:

The equation has one real root.

And the root is -1/2.

Step-by-step explanation:

(x + 1)^2 - x^2 = 0

x^2 + 2x + 1 -x^2 = 0

2x + 1 = 0

x = -1/2

Thus, the given equation has only one real root.

:)

Answer 2

(x + 1)² - x² = 0 has one real root

Given :

The equation (x + 1)² - x² = 0

To find :

The number of real roots

Solution :

Step 1 of 3 :

Write down the given equation

The given equation is

(x + 1)² - x² = 0

Step 2 of 3 :

Simplify the given equation

$\displaystyle \sf{ {(x + 1)}^{2} - {x}^{2} = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + 2x + 1 - {x}^{2} = 0 }$

$\displaystyle \sf{ \implies 2x + 1 = 0 }$

Step 3 of 3 :

Find number of real roots

$\displaystyle \sf{ {(x + 1)}^{2} - {x}^{2} = 0 }$

$\displaystyle \sf{ \implies 2x + 1 = 0 }$

$\displaystyle \sf{ \implies 2x = - 1 }$

$\displaystyle \sf{ \implies x = - \frac{1}{2} }$

Hence (x + 1)² - x² = 0 has one real root

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