Question

Для каждого значения a решите уравнение: x² + (1 - 3a)x + 2a² - 2=0

Answer 1

x² + (1 - 3a)x + 2a² - 2=0

определим кол-во корне с помощью D

D=$(1-3a)^{2}$-4×1$(2a^{2} -2)$

D=$a^{2}$-6a+9

$\left \{ {{a^{2} -6a+9>0} \atop {a^{2} -6a+9=0}} \right \left \{ {{a^{2} -6a+9$  только это все на 3 уравния в 1 скобке.

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Answer 2

To solve the quadratic equation $x^2 + (1 - 3a)x + 2a^2 - 2 = 0$, we can use the quadratic formula:

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Applying this formula to our equation $x^2 + (1 - 3a)x + 2a^2 - 2 = 0$, we have:

$a = 1 - 3a$ $b = 2a^2 - 2$ $c = 0$

Now, let's plug these values into the quadratic formula and solve for $x$:

$x = \frac{- (1 - 3a) \pm \sqrt{(1 - 3a)^2 - 4 \cdot 2a^2}}{2}$

Simplifying further:

$x = \frac{-1 + 3a \pm \sqrt{(1 - 6a + 9a^2) - 8a^2}}{2}$ $x = \frac{-1 + 3a \pm \sqrt{1 - 6a + 9a^2 - 8a^2}}{2}$ $x = \frac{-1 + 3a \pm \sqrt{1 + a^2}}{2}$

So, the solutions for $x$ in terms of $a$ are:

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

Please note that for each value of $a$, you'll have two solutions for $x$, one with the positive sign and one with the negative sign.