Solving using Quadratic Formula

Solving using Quadratic Formula: The quadratic formula is a formula used to solve quadratic equations of the form of $$ax^2 + bx + c = 0$$ where a, b, and c...

Solving using Quadratic Formula:

The quadratic formula is a formula used to solve quadratic equations of the form of $ax^2 + bx + c = 0$ where a, b, and c are real numbers.

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where a, b, and c are the coefficients of the quadratic equation.

Steps to solve a quadratic equation using the quadratic formula:

Identify the coefficients a, b, and c of the quadratic equation $ax^2 + bx + c = 0$
Substitute the values of a, b, and c into the quadratic formula.
Calculate the discriminant, which is the value of $$b^2 - 4ac$.
Use the discriminant to determine the nature of the roots (real or complex).
Apply the different results from step 4 to determine the solutions to the quadratic equation.

Examples:

Solution: $x^2 - 4x + 3 = 0$

$a = 1, b = -4, c = 3$

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)}$

$x = 2 \pm \sqrt{16 - 12}$

$x = 2 \pm 2$

The solutions are x = 3 and x = 1.

Solution: $x^2 + 5x - 6 = 0$

$a = 1, b = 5, c = -6$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-(-6)}}{2(1)}$

$x = 3 \pm \sqrt{25 + 24}$

$x = 3 \pm \sqrt{49}$

$x = 3 \pm 7$

The solutions are x = 4 and x = -1

Solving using Quadratic Formula:

The quadratic formula is a formula used to solve quadratic equations of the form of $ax^2 + bx + c = 0$ where a, b, and c are real numbers.

The quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where a, b, and c are the coefficients of the quadratic equation.

Steps to solve a quadratic equation using the quadratic formula:

Identify the coefficients a, b, and c of the quadratic equation $ax^2 + bx + c = 0$
Substitute the values of a, b, and c into the quadratic formula.
Calculate the discriminant, which is the value of $$b^2 - 4ac$.
Use the discriminant to determine the nature of the roots (real or complex).
Apply the different results from step 4 to determine the solutions to the quadratic equation.

Examples:

Solution: $x^2 - 4x + 3 = 0$

$a = 1, b = -4, c = 3$

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)}$

$x = 2 \pm \sqrt{16 - 12}$

$x = 2 \pm 2$

The solutions are x = 3 and x = 1.

Solution: $x^2 + 5x - 6 = 0$

$a = 1, b = 5, c = -6$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-(-6)}}{2(1)}$

$x = 3 \pm \sqrt{25 + 24}$

$x = 3 \pm \sqrt{49}$

$x = 3 \pm 7$

The solutions are x = 4 and x = -1

Solving using Quadratic Formula

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