Solve- -2x2-2x-1-2x-3

The correct option is C 1,4 Given equation is: √(2x^2 2x+1)= 2x 3 Squaring on both sides, ∴ 2x^2 2x+1=4x^2 12x+9 ⇒ x^2 5x+4=0 ⇒ ...

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Standard X

Mathematics

Solving Using Quadratic Formula When D>0

Solve: √2x2...

Question

Solve: $\sqrt{2 x^{2} - 2 x + 1} = 2 x - 3$

2, 3

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1, 3

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1, 4

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- 3, 2

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Solution

The correct option is C $1, 4$
Given equation is:
$\sqrt{2 x^{2} - 2 x + 1} = 2 x - 3$
Squaring on both sides,
$∴ 2 x^{2} - 2 x + 1 = 4 x^{2} - 12 x + 9$
$\Rightarrow$ $x^{2} - 5 x + 4 = 0$
$\Rightarrow$ $(x - 1) (x - 4) = 0$
Then $x - 1 = 0$ or $x - 4 = 0$
$\Rightarrow$ $x = 1$ or $4$

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Solve: √2x2−2x+1=2x−3

The correct option is C 1,4Given equation is:√2x2−2x+1=2x−3Squaring on both sides,∴2x2−2x+1=4x2−12x+9⇒ x2−5x+4=0⇒ (x−1)(x−4)=0Then x−1=0 or x−4=0⇒ x=1 or 4

Solve: $\sqrt{2 x^{2} - 2 x + 1} = 2 x - 3$

The correct option is C $1, 4$
Given equation is:
$\sqrt{2 x^{2} - 2 x + 1} = 2 x - 3$
Squaring on both sides,
$∴ 2 x^{2} - 2 x + 1 = 4 x^{2} - 12 x + 9$
$\Rightarrow$ $x^{2} - 5 x + 4 = 0$
$\Rightarrow$ $(x - 1) (x - 4) = 0$
Then $x - 1 = 0$ or $x - 4 = 0$
$\Rightarrow$ $x = 1$ or $4$