Question

B) 3x2 – 147 = 0;1) x² – 16x + 63 = 0.​

Answer 1

Ответ:

$3 {x}^{2} - 147 = 0 \\ {x}^{2} = 49 \\ x = + - 7 \\ \\ {x }^{2} - 16x + 63 = 0 \\ x1 + x2 = 16 \\ x1x2 = 63 \\ \\ x1 = 7 \\ x2 = 9$

Answer 2

To solve the equations B) 3x^2 - 147 = 0 and 1) x² - 16x + 63 = 0, we can use the quadratic formula:

For any quadratic equation in the form of ax^2 + bx + c = 0, the quadratic formula is:

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

Let's apply this formula to both equations:

B) 3x^2 - 147 = 0:

In this equation, a = 3, b = 0, and c = -147. Plug these values into the quadratic formula:

$x = \frac{-0 \pm \sqrt{0^2 - 4(3)(-147)}}{2(3)}$

Simplify:

$x = \frac{\pm \sqrt{4(3)(147)}}{6}$

$x = \frac{\pm 2\sqrt{3 \cdot 7 \cdot 7 \cdot 3}}{6}$

$x = \frac{\pm 2 \cdot 7 \cdot \sqrt{3}}{6}$

$x = \frac{\pm 14\sqrt{3}}{6}$

Now, simplify the fraction:

$x = \frac{7\sqrt{3}}{3}$ (one solution)

$x = \frac{-7\sqrt{3}}{3}$ (another solution)

So, for equation B), the solutions are:

$x = \frac{7\sqrt{3}}{3}$ and $x = \frac{-7\sqrt{3}}{3}$.

1. x² - 16x + 63 = 0:

In this equation, a = 1, b = -16, and c = 63. Plug these values into the quadratic formula:

$x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(63)}}{2(1)}$

Simplify:

$x = \frac{16 \pm \sqrt{256 - 252}}{2}$

$x = \frac{16 \pm \sqrt{4}}{2}$

$x = \frac{16 \pm 2}{2}$

Now, simplify the fraction:

For the plus sign:

$x = \frac{16 + 2}{2} = \frac{18}{2} = 9$

For the minus sign:

$x = \frac{16 - 2}{2} = \frac{14}{2} = 7$

So, for equation 1), the solutions are:

$x = 9$ and $x = 7$.