Question

решите уравнение
√17+2х-3х^2=х+1​​

Answer 1

Ответ:

$\sqrt{17 + 2x - {3x}^{2} } = x + 1$

${ \sqrt{17 + 2x - {3x}^{2} } }^{2} = {(x + 1)}^{2}$

$17 + 2x - {3x}^{2} = {x}^{2} + 2x + 1$

$17 - {3x}^{2} = {x}^{2} + 1$

$- {3x}^{2} - {x}^{2} = 1 - 17$

$- {4x}^{2} = - 16| \div - 4$

${x}^{2} = 4$

${x}^{2} - 4 = 4 - 4$

${x}^{2} - 4 = 0$

${1x}^{2} - 4 = 0$

${1x}^{2} + 0x - 4 = 0$

${1x}^{2} + 0x + ( - 4) = 0$

$a = 1.b = 0.c = - 4$

$x = \frac{ - 0± \sqrt{ {0}^{2} - 4 \times 1 \times ( - 4)} }{2 \times 1}$

$x = \frac{± \sqrt{0 + 16} }{2}$

$x = \frac{± \sqrt{16} }{2}$

$x = \frac{± \sqrt{ {4}^{2} } }{2}$

$x = \frac{±4}{2}$

$x = \frac{4}{2} \\ x = \frac{ - 4}{2}$

Сокращаем обе дроби на общий делитель 2

$x = 2 \\ x = - 2$

${x}_{1}=-2, {x}_{2}=2$