Question

CAN YOU SOLVE?!?!?!?!!

Answer 1

Is this the answer for the top part The equation you provided is a quadratic equation in the form of ax²+bx+c=0, where a = 1, b = b/a, and c = -c/a + (b/2a)². To solve for x, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a). Substituting the values of a, b, and c in the formula, we get:

x = (-b/a ± √((b/a)² - 4(1)(-c/a + (b/2a)²))) / (2(1))

Simplifying the expression inside the square root:

x = (-b/a ± √(b²/a² + 4c/a - 4b²/4a²)) / 2

x = (-b/a ± √(b²/a² + 4c/a - b²/a²)) / 2

x = (-b/a ± √(4c/a)) / 2

x = (-b/2a ± √c/a)

Therefore, the solutions to the quadratic equation are:

x = (-b/2a + √c/a) and x = (-b/2a - √c/a)

Answer 2

`x=(-b \pm √(b^2-4ac))/(2a)`

Explanation:

Equation 1

Given equation:

`x^2+(b)/(a)x+\left((b)/(2a)\right)^2=-(c)/(a)+\left((b)/(2a)\right)^2`

`\textsf{Subtract $\left((b)/(2a)\right)^2$ from both sides of the equation:}`

`x^2+(b)/(a)x=-(c)/(a)`

`\textsf{Add $(c)/(a)$ to both sides of the equation:}`

`x^2+(b)/(a)x+(c)/(a)=0`

Multiply all terms by
`a`:

`ax^2+bx+c=0`

`\hrulefill`

Equation 2

Given equation:

`\left(x+(b)/(2a)\right)^2=-(c)/(a)+(b^2)/(4a^2)`

Expand the left side of the equation and apply the power of a quotient exponent rule to b²/4a²:

`\left(x+(b)/(2a)\right)\left(x+(b)/(2a)\right)=-(c)/(a)+\left((b)/(2a)\right)^2`

`x^2+(2b)/(2a)x+\left((b)/(2a)\right)^2=-(c)/(a)+\left((b)/(2a)\right)^2`

`\textsf{Subtract $\left((b)/(2a)\right)^2$ from both sides of the equation:}`

`x^2+(b)/(a)x=-(c)/(a)`

`\textsf{Add $(c)/(a)$ to both sides of the equation:}`

`x^2+(b)/(a)x-(c)/(a)=0`

Multiply all terms by
`a`:

`ax^2+bx+c=0`

`\hrulefill`

Solution

The equation ax² + bx + c = 0 represents a quadratic equation in the form of a polynomial of degree two, where a, b, and c are coefficients, and x is the variable.

To solve a quadratic in the form ax² + bx + c = 0, we can use the quadratic formula.

`\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=(-b \pm √(b^2-4ac))/(2a)\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}`

Therefore, the solution to both equations is:

`x=(-b \pm √(b^2-4ac))/(2a)`