Question

(4x+3)^2-(3x-1)^2=0​

Answer 1

1) $1 )(4x+3)^2-(3x-1)^2=0 \\\:16x^2+24x+9-9x^2+6x-1\\7x^2+30x+8=0x_{1,\:2}=\frac{-30\pm \sqrt{30^2-4\cdot \:7\cdot \:8}}{2\cdot \:7}x= -2/7 x=-4$

Answer 2

Использовав разность квадратов, получим

(4х+3-3х+1)*(4х+3+3х-1)=0

(х+4)(7х+2)=0, приравнивая к нулю скобки, получим х=-4, х=-2/7

Ответ  -4; -2/7

Answer 3

To solve the equation $(4x+3)^2 - (3x-1)^2 = 0$, we can use the difference of squares formula, which states that $a^2 - b^2 = (a + b)(a - b)$. By applying this formula, we can simplify the equation:

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

Now, we can simplify the expressions inside the parentheses:

$7x + 2)(x + 4) = 0$

To find the solutions for $x$, we set each factor equal to zero and solve for $x$:

1. $7x + 2 = 0$ $7x = -2$ $x = -2/7$

2. $x + 4 = 0$ $x = -4$

So the solutions to the equation are $x = -2/7$ and $x = -4$.