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Solve this quadratic equation by factorising first

{2x}^(2) - 8 = 0


2 Answers

6 votes


\large\underline{\mathbb{SOLUTION:}}

Given:


\boxed{\bm{{2x}^(2) - 8 = 0 }}

Now, factorizing the given equation will give:


\longrightarrow \tt{ 2(x^(2) -4)=0 }

Now, divide both sides by 2:


\longrightarrow \tt{ \frac{ 2( {x}^(2) - 4) }{2} = (0)/(2) }


\longrightarrow{ \tt{ {x}^(2) - 4 = 0}}


\longrightarrow{ \tt{ {x}^(2) - {2}^(2) = 0}}

Since,


\boxed{\tt{ {a}^(2) - {b}^(2) = (a + b)(a - b)}}


\longrightarrow\tt{{x}^(2) - {2}^(2) = 0 }


\longrightarrow\tt{(x + 2)(x - 2) = 0 }


\tt{x + 2 = 0 } \qquad \qquad \tt{x - 2 = 0}


\boxed{ \tt{ \red{x = - 2 }}}\qquad \qquad \boxed{ \tt{ \red{x = 2 }}}


\large\underline{\mathbb{ANSWER:}}

Therefore, the final answer is
\rm{ x=-2 \qquad or \qquad x=2}

User Stanowczo
by
8.6k points
2 votes

Answer:

x = - 2, x = 2

Explanation:

2x² - 8 = 0 ( divide through by 2 )

x² - 4 = 0 ← x² - 4 is a difference of squares and factors in general as

a² - b² = (a + b)(a - b) , then

x² - 4 = 0

x² - 2² = 0

(x + 2)(x - 2) = 0 ← in factored form

equate each factor to zero and solve for x

x + 2 = 0 ( subtract 2 from both sides )

x = - 2

x - 2 = 0 ( add 2 to both sides )

x = 2

solutions are x = - 2 , x = 2

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