Question

$x {}^{2} - 4ax + 4a {}^{2} - b { }^{2} = 0$
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Answer 1

Step-by-step explanation:

Answer:

\begin{gathered}Roots \:of\\\: given\: quadratic\: equation\: are\: ,\\x = 2a+b \: Or \: x = 2a-b\end{gathered}

Rootsof

givenquadraticequationare,

x=2a+bOrx=2a−b

Step-by-step explanation:

Given quadratic equation:

x²-4ax+4a²-b²=0

Finding roots of the quadratic equation By completing the square method:

Rearranging the equation, we get

=> x²-4ax+4a² = b²

x^{2}-2\times x \times 2a +(2a)^{2}=b^{2}x

2

−2×x×2a+(2a)

2

=b

2

\implies (x-2a)^{2}=b^{2}⟹(x−2a)

2

=b

2

\implies x-2a = ±\sqrt{b^{2}}⟹x−2a=±

b

2

\implies x-2a=±b⟹x−2a=±b

\implies x = 2a±b⟹x=2a±b

\implies x = 2a+b \: Or \: x = 2a-b⟹x=2a+bOrx=2a−b

Therefore,

x = 2a+b \: Or \: x = 2a-bx=2a+bOrx=2a−b

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