Question

(4x-3)^2+(3x-1)(3x+1)=0

Answer 1

Ответ:

Корней нет

Объяснение:

(4х-3)^2+(3х-1)(3х+1)=0

16х^2-24х+9+9х^2-1=0

25х^2-24х+8=0

D = (-24)^2-4*25*8 = 576-800=-224

-224<0

Корней нет

Answer 2

To solve the quadratic equation $(4x-3)^2 + (3x-1)(3x+1) = 0$, we'll first simplify and expand the expressions, and then proceed with solving for $x$.

Let's break down the equation step by step:

1. Expand $(4x-3)^2$: $(4x-3)^2 = 16x^2 - 24x + 9$

2. Expand $(3x-1)(3x+1)$: $(3x-1)(3x+1) = 9x^2 - x^2 - 3x + 3x - 1 = 8x^2 - 1$

Now the equation becomes: $16x^2 - 24x + 9 + 8x^2 - 1 = 0$

Combine like terms: $24x^2 - 24x + 8 = 0$

Divide the entire equation by $8$ to simplify: $3x^2 - 3x + 1 = 0$

Now we have a quadratic equation in standard form $ax^2 + bx + c = 0$: $a = 3, \quad b = -3, \quad c = 1$

We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plugging in the values: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$ $x = \frac{3 \pm \sqrt{9 - 12}}{6}$ $x = \frac{3 \pm \sqrt{-3}}{6}$

Since the discriminant ($b^2 - 4ac$) is negative, the quadratic equation has complex solutions: $x = \frac{3 \pm i\sqrt{3}}{6}$

So, the solutions for the equation $(4x-3)^2 + (3x-1)(3x+1) = 0$ are: $x = \frac{3 + i\sqrt{3}}{6}$ $x = \frac{3 - i\sqrt{3}}{6}$