Question

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

Answer 1

$\underline{\textbf{Given:}}$

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

$\underline{\textbf{To find:}}$

$\textsf{The real roots of the given equation}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

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

$\textsf{This can be written as,}$

$\mathsf{x^2+1+2x-x^2=0}$

$\mathsf{2x+1=0}$

$\mathsf{2x=-1}$

$\implies\boxed{\mathsf{x=\dfrac{-1}{2}}}$

$\therefore\underline{\textsf{The given equation has only one real root}}$

$\underline{\textbf{Find more:}}$

Find the roots of the following quadratic (if they exist) by the method of completing the square.

√3x²+10x+7√3=0  

https://brainly.in/question/15926794

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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