Question

6.16. 1) (2x - 1)^2 = 16;3) 25 - (5x + 1)^2= 0​

Answer 1

1) (a - b)² = a² - 2ab + b²

(2х - 1)² = 16

(2х)² - 2 · 2х · (-1) + (-1)² = 16

4х² + 4х + 1 - 16 = 0

4х² + 4х - 15 = 0      

D = b² - 4ac = 4² - 4 · 4 · (-15) = 16 + 240 = 256

√D = √256 = 16

х₁ = (-4-16)/(2·4) = (-20)/8 = -2,5

х₂ = (-4+16)/(2·4) = 12/8 = 1,5

Ответ: (-2,5; 1,5).    

3) (a + b)² = a² + 2ab + b²    

25 - (5х + 1)² = 0

25 - ((5х)² + 2 · 5х · 1 + 1²) = 0

25 - (25х² + 10х + 1) = 0

25 - 25х² - 10х - 1 = 0  (умножим обе части уравнения на (-1))

25х² + 10х + 1 - 25 = 0

25х² + 10х - 24 = 0

D = b² - 4ac = 10² - 4 · 25 · (-24) = 100 + 2400 = 2500

√D = √2500 = 50

х₁ = (-10-50)/(2·25) = (-60)/50 = -1,2

х₂ = (-10+50)/(2·25) = 40/50 = 0,8

Ответ: (-1,2; 0,8).

                   

Answer 2

It seems like you've provided a couple of quadratic equations. Let's go through each of them and solve for the values of 'x'.

1. $(2x - 1)^2 = 16$: Start by expanding the left side of the equation: $4x^2 - 4x + 1 = 16$ Now, subtract 16 from both sides: $4x^2 - 4x - 15 = 0$ Divide the entire equation by 4 to simplify: $x^2 - x - \frac{15}{4} = 0$ Now, you can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In this case, $a = 1$, $b = -1$, and $c = -\frac{15}{4}$. Plugging these values in: $x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot \left(-\frac{15}{4}\right)}}{2 \cdot 1}$ $x = \frac{1 \pm \sqrt{1 + 15}}{2}$ $x = \frac{1 \pm \sqrt{16}}{2}$ $x = \frac{1 \pm 4}{2}$

So, you have two solutions: $x = \frac{5}{2}$ or $x = -\frac{3}{2}$

2. $25 - (5x + 1)^2 = 0$: Start by expanding the squared term: $25 - (25x^2 + 10x + 1) = 0$ Simplify the equation: $25 - 25x^2 - 10x - 1 = 0$ $-25x^2 - 10x + 24 = 0$ Divide the equation by -1 to make the coefficient of $x^2$ positive: $25x^2 + 10x - 24 = 0$ This quadratic equation doesn't factor easily, so you can again use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ In this case, $a = 25$, $b = 10$, and $c = -24$. Plugging these values in: $x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 25 \cdot (-24)}}{2 \cdot 25}$ $x = \frac{-10 \pm \sqrt{100 + 2400}}{50}$ $x = \frac{-10 \pm \sqrt{2500}}{50}$ $x = \frac{-10 \pm 50}{50}$

So, you have two solutions: $x = \frac{4}{5}$ or $x = -1$

These are the solutions for the given quadratic equations.