Question

${x}^{2} + 2ax + {a}^{2} - {b}^{2} = 0$
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Answer 1

Answer:

x = -(a+b) Or x = b-a

Step-by-step explanation:

x²+2ax+a²-b² = 0

=> x²+2ax+(a+b)(a-b)=0

=> x²+(a+b)x+(a-b)x+(a+b)(a-b)=0

=> x(x+a+b)+(a-b)(x+a+b)=0

=> (x+a+b)(x+a-b)=0

=> x+a+b=0 Or x+a-b=0

=> x = -(a+b) Or x = b-a

Therefore,

x = -(a+b) Or x = b-a

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Answer 2

Given

quadratic equation is

x²+2ax+a²-b²

To find

The roots of the equation

Explanation:

$= > {x}^{2} + 2ax + {a}^{2} - {b}^{2} = 0$

$\underline{\sf (a^2-b^2)=(a+b)(a-b)}$

$= > {x}^{2} + 2ax + (a + b)(a - b) = 0$

Splitting the middle term

$= > {x}^{2} + (a + b)x + (a - b)x + (a + b)(a - b) = 0$

$= > x(x + a + b) + (a - b)(x + a + b) = 0$

$= > (x + a - b)(x + a + b) = 0$

$= > x + a -b = 0 \\ \\ = > \boxed{ \sf x = (b - a)}$

$= > x + a + b = 0 \\ \\ = > \boxed{ \sf \: x = - (a + b)}$

Some more information:

It the roots are not real

then the roots can be found using the formula given below

$\sf x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\sf sum \: of \: roots =\dfrac{-coefficeint \: of \: x}{coefficient\: of \: x^2}$

$\sf product \: of \: roos=\dfrac{constant \: term}{coefficient\: of \: x^2}$