Question

(x+4)^2=4x^2+5(x-4)(4x-3)=-3

Answer 1

(х+4)²=4х²+5
х²+8х+16=4х²+5
х²+8х+16-4х²-5=0
-3х²+8х+11=0|:(-1)
3х²-8х-11=0
х1,2=8±√-8²-4*3*(-11)/2*3=
=8±√196/6=8±14/6
х1=8+14/6=22/6=11/3
х2=8-14/6=-6/6=-1
Ответ:-1;11/3

(х-4)(4х-3)=-3
4х²-3х-16х+12=-3
4х²-19х+12+3=0
4х²-19х+15=0
х1,2=19±√-19²-4*4*15/2*4=
=19±11/8
х1=19+11/8=30/8=15/4
х2=19-11/8=8/8=1
Ответ:1;15/4

Answer 2

Let's solve the given equation step by step. We have:

1. (x + 4)^2 = 4x^2 + 5(x - 4)(4x - 3) - 3

First, we'll expand the right-hand side of the equation:

2. (x + 4)^2 = 4x^2 + 5(4x^2 - 3x - 16x + 12) - 3

Next, simplify the expression:

3. (x + 4)^2 = 4x^2 + 5(4x^2 - 19x + 12) - 3

Now, distribute the 5 through the parentheses:

4. (x + 4)^2 = 4x^2 + 20x^2 - 95x + 60 - 3

Combine like terms:

5. (x + 4)^2 = 24x^2 - 95x + 57

To proceed, expand the left-hand side of the equation:

6. x^2 + 8x + 16 = 24x^2 - 95x + 57

Now, move all terms to one side to set the equation to zero:

7. 0 = 24x^2 - 95x + 57 - x^2 - 8x - 16

Combine like terms:

8. 0 = 23x^2 - 103x + 41

Now, the equation is in quadratic form (ax^2 + bx + c = 0), where a = 23, b = -103, and c = 41. To solve for x, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values:

x = (103 ± √((-103)^2 - 4 * 23 * 41)) / (2 * 23)

x = (103 ± √(10609 - 3812)) / 46

x = (103 ± √(6797)) / 46

x = (103 ± 82.454) / 46

Now, we have two solutions for x:

1. x = (103 + 82.454) / 46 ≈ 4.572
2. x = (103 - 82.454) / 46 ≈ 0.283

So the solutions for the given equation are approximately x ≈ 4.572 and x ≈ 0.283.