Question

using quadratic formula solve the equation :- 2x square-2 root over 2 X +1 = 0

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Answer 1

Answer:

8×-8 + 8 × 1 =0

Step-by-step explanation:

hope it helps you

Answer 2

Answer :

$\large \implies \dfrac{\sqrt{2}}{2}$

Solution :

Question :

using quadratic formula solve the equation :-

• 2x² - 2√2x -1 = 0

$\LARGE\color{blue} \mathfrak{Quadratic\: Formula }:$

$\:\:\: \:\red\bigstar\boxed{\tt x = \dfrac{-b± \sqrt{b^2-4ac}}{2a}}$

Here,

• a = 2
• b = -2√2
• c = 1

$\\ \large\underline{ \mathfrak{Substituting \: the \: values}}$

$\leadsto x = \dfrac{-(-2\sqrt{2}) ± \sqrt{(-2\sqrt{2})^2-4\times 2\times 1}}{2\times 2} \\ \\ \implies x = \dfrac{2\sqrt{2} ± \sqrt{ 8 - 8}}{4} \\ \\ \implies x = \dfrac{2\sqrt{2} ± 0 }{4} \\ \\ \implies x = \dfrac{\cancel{2}\sqrt{2} }{^2\cancel{4}} \\\\\implies \underline{\boxed{\pink{x = \dfrac{\sqrt{2}}{2}}}}$

Hence,

Value of x is 2√2.

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☞ Verification :

Quadratic equation :

• 2x² - 2√2x -1 = 0

Putting value of x

$\sf 2(\dfrac{\sqrt{2}}{2})^2 - 2\sqrt{2}(\dfrac{\sqrt{2}}{2}) - 1 = 0 \\ \\ \implies 2(\frac{\cancel{2}}{^2\cancel{4}}) - \cancel{2}\sqrt{2}(\dfrac{\sqrt{2}}{\cancel{2}}) + 1 = 0 \\ \\ \implies \dfrac{2}{2} - \sqrt{2} \times \sqrt{2} + 1 \\ \\ \implies 1 -2 + 1 = 0 \\ \\ \implies 2-2 = 0 \\ \\ \implies 0 = 0$

L.H.S = R.H.S

Hence verified !

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