Question

Find the roots of 2(x²-4x) + 2 = 0 and choose the correct option

A) 3,-3
B) V3, -v3
C) 2+ √3, 2-√3
D) 5, -1​

Answer 1

Question:-

Find the roots of 2(x²-4x) + 2 = 0

Solution:-

$\begin{array}{l l l} \rightarrow\bf{2(x {}^{2} - 4x) + 2} \\ \\ \rightarrow \: \bf{2x {}^{2} - 8x + 2 = 0} \\ \\ \rightarrow \: \bf{2(x {}^{2} - 4x + 1) = 0 } \\ \\ \underline \bold \red{✰ \: Divide \: both \: sides \: of \: the \: equation \: by \: the \: same \: term} \\ \\ \rightarrow \: \bf{x- 4x + 1 = 0} \\ \\ \underline \bold \purple{✰ \: Using \: the \: quadratic \: formula} \\ \\ \: \bf \large \orange{x = \frac{ - b \pm \: \sqrt{b² -4ac} }{2a}} \\ \\ \rightarrow \: \bf{x- 4x + 1 = 0} \\ \: \\ \end{array}$

Here,

• a = 1
• b = -4
• c = 1

$\begin{array}{ l l l}\large\bf { \mapsto \: x = \frac{ - ( - 4) \: \pm \: \sqrt{( - 4)² -4 \times 1 \times 1} }{2 \times 1}} \\ \\ \large \mapsto \bf{x = \frac{ 4 \: \pm \: \sqrt{ 16 - 4} }{2}} \\ \\ \large \mapsto \bf{x = \frac{ 4 \: \pm \: \sqrt{ 12} }{2}} \\ \\ \large\mapsto \bf{x = \frac{ 4 \: \pm \: \sqrt{ 2 \times 2 \times 3} }{2}} \\ \\ \large\mapsto \bf{x = \frac{ 4 \: \pm \: 2 \sqrt{3} }{2}} \\ \\ \underline \bold \green{✰ \: Separate \: the \: equations} \\ \\ \large\mapsto \bf{x = \frac{ 4 \: + \: 2 \sqrt{3} }{2}} \: \: \large \mapsto\bf{x = \frac{ 4 \: - 2 \sqrt{3} }{2}} \\ \\ \large \mapsto \bf{x = \frac{ \cancel2(2 + \sqrt{3}) }{ \cancel 2} } \: \large\mapsto \bf{x = \frac{ \cancel2(2 - \sqrt{3}) }{ \cancel 2} } \\ \\ \underline\bold \pink{✰ After \: Rearranging} \\ \\ \: \bf{ \mapsto \: x = 2 + \sqrt{3}} \\ \bf{\mapsto \: x = 2 - \sqrt{3}} \end{array}$

Hence the roots of the equation are 2 + √3 and 2 - √3

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Answer 2

Answer:

Question:-

Find the roots of 2(x²-4x) + 2 = 0

Solution:-

→ 2 ( x²- 4x ) + 2 = 0

→ 2x² - 8x + 2 = 0

→ 2 ( x² - 4x + 1 ) = 0

✰ Divide both sides of the equation by the same term i.e 2

→ x² − 4x + 1 = 0

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✰ Using the quadratic formula

x = { (-b) ± √D } / 2a

D = b² - 4ac

Here :

• a = 1
• b = (-4)
• c = 1

→ x² − 4x + 1 = 0

D = (-4)² - 4(1)(1)

• 16 - 4
• 12

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• ➡ x = { (-b) ± √D } / 2a

• { -(-4) ± √12 } / 2(1)

• ( 4 ± 2√3 ) / 2

• 2 ( 2 ± √3 ) / 2

• 2 ± √3

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✰ Separate the equations

x = 2 + √3

x = 2 - √3

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Hence the roots of the equation are :

2 + √3 and 2 - √3

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Hope it helps you :)