Question

Solve this quadratic equation by factorising first

`{2x}^(2) - 8 = 0`
​

Answer 1

`\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}`

Answer 2

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