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The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by:

• \(x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are coefficients of the equation. This formula helps to find the roots of the equation, which means solving for the values of \(x\) that make the equation true.
To use the quadratic formula, you need to identify the coefficients \(a\), \(b\), and \(c\). Then, substitute these values into the formula and simplify step by step:

• Calculate the discriminant \(\Delta = b^2 - 4ac\).
• Evaluate the square root of the discriminant.
• Use both the plus and minus signs in the formula to get the two possible solutions.

This method can provide both real and complex solutions, depending on the value of the discriminant.

Complex numbers are an extension of real numbers and include a real part and an imaginary part. They are written as \(a + bi\), where \(a\) is the real part, and \(bi\) is the imaginary part, with \(i\) being the imaginary unit defined as \(i^2 = -1\).

• Real numbers: A subset of complex numbers where the imaginary part is zero (\(b = 0\)).
• Imaginary numbers: A subset of complex numbers where the real part is zero (\(a = 0\)).

When solving polynomial equations, it's crucial to recognize that solutions might not always be real numbers. In cases where the discriminant in a quadratic equation is negative, the results are complex numbers.
For example, in the equation \(x^2 = -7\), there are no real solutions because no real number squared gives a negative result. Instead, the solutions involve complex numbers:

• \(x = \pm i\sqrt{7}\)

These solutions account for the negative sign under the square root.

The substitution method is used to simplify complex equations by replacing a variable with another expression. In polynomial equations, it can convert higher-degree equations into more manageable quadratic equations.
For instance, given the equation \(x^4 + 6x^2 - 7 = 0\), we notice that it can be simplified:

• Let \(y = x^2\) to reduce the fourth-degree equation to a second-degree one: \(y^2 + 6y - 7 = 0\).

This adjustment makes it easier to apply the quadratic formula. After solving for \(y\), we reverse the substitution to find the original variable \(x\):

• For example, if \(y = 1\), then \(x^2 = 1\), which means \(x = \pm 1\).
• And if \(y = -7\), then \(x^2 = -7\), leading to \(x = \pm i\sqrt{7}\).

Using substitution simplifies the process and helps in tackling more complicated polynomial equations.

Imaginary solutions arise when solving equations with a negative discriminant. They are part of the set of complex numbers and reflect solutions that are not real.
When using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), if \(b^2 - 4ac < 0\), the square root of a negative number results in imaginary solutions. For example, with \(x^2 = -7\), we solve it as:

• \(x = \pm i\sqrt{7}\)

Imaginary solutions can be verified similarly to real solutions. Substitute them back into the original equation to check if they satisfy it:

• For \(x = i\sqrt{7}\): \br>
  • \((i\sqrt{7})^4 + 6(i\sqrt{7})^2 - 7\)
  • Simplify step-by-step:
    • \(49(-1) + 6(-7) - 7 = 0\)
    • And it confirms the solution.

This process helps ensure that even imaginary solutions are valid.