Question

9х2-7(х+4)(4-х)-(1-4х)2=15

Answer 1

9x^2-7(x+4)*(4-x)-(1-4x)^2=15
16x^2-112-1+8x-16x^2-15=0
16x^2-113+8x-16x^2-15=0
-113+8x-15=0
-128+8x=0
x=128/8
x=16

Answer 2

Let's solve the equation step by step:

9x2 - 7(x + 4)(4 - x) - (1 - 4x)2 = 15

First, let's simplify the equation:

9x2 - 7(x2 - 4x + 4(4 - x)) - (1 - 4x)2 = 15

Next, we expand the terms inside the parentheses:

9x2 - 7(x2 - 4x + 16 - 4x) - (1 - 4x)2 = 15

Simplifying further:

9x2 - 7(x2 - 8x + 16) - (1 - 4x)2 = 15

Expanding the squared terms:

9x2 - 7x2 + 56x - 112 - (1 - 8x + 16x2) = 15

Combining like terms:

9x2 - 7x2 + 56x - 112 - 1 + 8x - 16x2 = 15

Simplifying:

-7x2 - 16x2 + 9x2 + 56x + 8x - 112 - 1 - 15 = 0

Combining like terms again:

-14x2 + 64x - 128 - 16 = 0

-14x2 + 64x - 144 = 0

Now, we need to solve this quadratic equation. We can either factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -14, b = 64, and c = -144.

x = (-64 ± √(64^2 - 4(-14)(-144))) / (2(-14))

Simplifying inside the square root:

x = (-64 ± √(4096 - 8064)) / (-28)

x = (-64 ± √(-3968)) / (-28)

Since the square root of a negative number results in a complex number, this equation has no real solutions.