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A quadratic equation is a polynomial equation of the second degree. This means it has the general form ewline ewline $$ax^2 + bx + c = 0$$ewline ewline where:

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term.

ewline ewline It's essential to recognize the standard form because it helps in identifying coefficients for further calculations, such as finding the discriminant.

The discriminant is a value that can be calculated from the coefficients of a quadratic equation. It is given by the formula: ewline ewline $$\triangle = b^2 - 4ac$$ ewline ewline The discriminant helps you determine the nature of the roots of the quadratic equation. By plugging in the identified coefficients (\(a = 1\), \(b = 8\), \(c = 12\)) into the formula, you get: ewline ewline $$\triangle = 8​^2 - 4(1)(12) = 64 - 48 = 16.$$ewline ewline A positive discriminant (\(\triangle > 0\)) indicates the equation has two distinct real roots.

Factoring trinomials involves turning the quadratic equation into a product of two binomials. For the equation ewline ewline $$x^2 + 8x + 12 = 0$$ewline ewline you check if it can be written as: ewline ewline $$(x + m)(x + n) = 0$$ewline ewline You'll find such \(m\) and \(n\) that:

• m + n = b = 8
• m * n = c = 12

ewline ewline In this case, \(m = 6\) and \(n = 2\) work because \(6 + 2 = 8\) and \(6 * 2 = 12\). So, the quadratic trinomial \(x^2 + 8x + 12\) can be factored as \((x + 6)(x + 2)\), showing it is not prime.

Real roots of a quadratic equation are the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). Using the quadratic formula \((x = \frac{-b \frac{plus or minus}{-b \frac{minus}{minus}ewline ewline Using the quadratic formula $$x= \frac{- b \frac{plus or minus} \frac{\frac{plus \frac{plus}-4ac}{ -2a}$$ ,ewline ewline these roots can be found: ewline ewline In the given example: $$ Δ = 8^2-4(1)(12) = 16$$,ewline ewline leading to: $$ (\frac{-8\frac{+4}{+4} \frac{ \frac{roots+6}{2} = -2$$,ewline ewline Since Δ is positive, the roots are distinct and real, confirmed by solving \)(x+6)(x+{0}$ instead.ewline ewline Therefore, the quadratic equation has two distinct real roots values.