The equation whose roots are twice the roots of $x^{2} - 3 x + 3 = 0$ is

x^{2} - 6 x + 12 = 0

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x^{2} - 3 x + 6 = 0

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2 x^{2} - 3 x + 3 = 0

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4 x^{2} - 6 x + 3 = 0

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Solution

The correct option is A $x^{2} - 6 x + 12 = 0$
$x^{2} - 3 x = 3 = 0$ ........ (1)
Here, $a + β = 3$ and $a β = 3$ . Therefore,
$2 (a + β) = 6$
$2 \times 2 a β = 4 a β = 4 \times 3 = 12$
The equation whose roots are double of (1),
$x^{2} -$ (sum of the roots)x + product of the roots $= 0$
will be $x^{2} - 6 x + 12 = 0$

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The equation whose roots are twice the roots of x2−3x+3=0 is

The correct option is A x2−6x+12=0x2−3x=3=0 ........ (1)Here, a+β=3 and aβ=3. Therefore,2(a+β)=62×2aβ=4aβ=4×3=12The equation whose roots are double of (1),x2−(sum of the roots)x + product of the roots =0will be x2−6x+12=0

The equation whose roots are twice the roots of $x^{2} - 3 x + 3 = 0$ is