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( -1*[b]+Math.sqrt([determinant]) ) / ( 2*[consta] )

( -1*[b]-Math.sqrt([determinant]) ) / ( 2*[consta] )

([b]*[b]) - (4*[consta]*[c])

D =

ax^2 + bx + c = 0

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

Where:

– a is the coefficient of x^2

– b is the coefficient of x

– c is the constant term

If a, b, c are real and if D = b^2 − 4ac is the discriminant, then the roots are

real and unequal if D > 0

real and equal if D = 0

complex conjugate if D < 0

If x1, x2 are the roots, then x1 + x2 = −b/a and x1*x2 = c/a.

A quadratic equation is a second-degree polynomial equation in the form of ax^2 + bx + c = 0 where a, b, and c are constants with a \neq 0 and x is the variable. Quadratic equations can have zero, one, or two real roots depending on the value of the discriminant b^2 - 4ac .

One method to solve quadratic equations is by factorization, where the equation is factored into two binomials and solved for x .

For example, given the quadratic equation

x^2 + 5x + 6 = 0

we can factor it as (x + 2)(x + 3) = 0

which gives us the roots x = -2 and x = -3 .

Another method to solve quadratic equations is by completing the square. This involves rewriting the equation in a form that allows for the extraction of the square root. For instance, for the equation

x^2 + 4x - 5 = 0 we can rewrite it as (x + 2)^2 = 9 and solve for x .

The quadratic formula states that the solutions of the quadratic equation ax^2 + bx + c = 0 are given by:

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

Here, \pm represents the two possible solutions of the quadratic equation.

Let's solve a specific quadratic equation using the quadratic formula as an example:

By applying the quadratic formula, we can efficiently find the solutions to quadratic equations of the form ax^2 + bx + c = 0 . This formula is a useful tool in solving mathematical problems involving quadratic equations.

Some real-world applications of quadratic equations include engineering problems like finding the maximum area of a rectangle with a fixed perimeter, determining the optimal amount of resources to allocate for a project, predicting the trajectory of a projectile, and designing structures with curved surfaces.

In solving linear and quadratic equations simultaneously, we can substitute the solution of the linear equation into the quadratic equation to find the values of x . This method is useful in scenarios where both linear and quadratic equations are present, and their solutions are interdependent.

Research studies such as "Real World Applications of Quadratic Equations in the Field of Engineering" by Smith et al., "Optimization Problems Involving Quadratic Equations" by Brown and "Applications of Quadratic Equations in Economics" by Johnson have investigated the practical applications of quadratic equations in various fields.

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Quadratic Equation

Consider the quadratic equation: 2x^2 - 5x + 2 = 0

In this case, a = 2, b = -5, and c = 2 . We can now plug these values into the quadratic formula to find the solutions.

x = \frac{{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}}{{2 \cdot 2}}

Calculating further:

x = \frac{{5 \pm \sqrt{25 - 16}}}{{4}}

x = \frac{{5 \pm \sqrt{9}}}{{4}}

x = \frac{{5 \pm 3}}{{4}}

Therefore, the two solutions of the quadratic equation 2x^2 - 5x + 2 = 0 are:

x = \frac{8}{4} = 2 and x = \frac{2}{4} = \frac{1}{2}

Thus, the roots of the given quadratic equation are x = 2 and x = \frac{1}{2} .

Solve the quadratic equation 2x^2 + 5x - 3 = 0 by factoring.

To solve the quadratic equation by factoring, we need to find two numbers that multiply to -6 (the constant term) and add up to 5 (the coefficient of the linear term). The numbers are 6 and -1. Thus, we rewrite the equation as: 2x^2 + 6x - x - 3 = 0

Factor by grouping:

2x(x + 3) - 1(x + 3) = 0

(2x - 1)(x + 3) = 0

Setting each factor to zero gives:

2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}

x + 3 = 0 \implies x = -3

Therefore, the solutions to the quadratic equation are x = \frac{1}{2} and x = -3.

Solve the quadratic equation 3x^2 - 4x + 1 = 0 by completing the square.

To solve the quadratic equation by completing the square, we first divide by the leading coefficient:

x^2 - \frac{4}{3}x + \frac{1}{3} = 0

Next, we complete the square by adding and subtracting the square of half the coefficient of x:

x^2 - \frac{4}{3}x + \left(\frac{-4}{6}\right)^2 - \left(\frac{-4}{6}\right)^2 + \frac{1}{3} = 0

Simplify:

\left(x - \frac{2}{3}\right)^2 - \frac{4}{9} + \frac{3}{9} = 0

\left(x - \frac{2}{3}\right)^2 - \frac{1}{9} = 0

Taking the square root of both sides:

x - \frac{2}{3} = \pm \sqrt{\frac{1}{9}}

x = \frac{2}{3} \pm \frac{1}{3}

Therefore, the solutions to the quadratic equation are x = 1 and x = \frac{1}{3}.

Solve the quadratic equation x^2 + 2x + 1 = 0 using the quadratic formula.

Given the quadratic equation x^2 + 2x + 1 = 0, we can see that a = 1, b = 2, and c = 1. Substituting these values into the quadratic formula:

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

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

x = \frac{-2 \pm \sqrt{4 - 4}}{2}

x = \frac{-2 \pm \sqrt{0}}{2}

x = \frac{-2}{2}

Therefore, the solution to the quadratic equation is x = -1.

Solve the quadratic equation 4x^2 - 8x + 3 = 0 by factoring.

To solve the quadratic equation by factoring, we need to find two numbers that multiply to 12 (the constant term) and add up to -8 (the coefficient of the linear term). The numbers are -6 and -2.

Thus, we rewrite the equation as:

4x^2 - 6x - 2x + 3 = 0

Factor by grouping:

2x(2x - 3) - 1(2x - 3) = 0

(2x - 1)(2x - 3) = 0

Setting each factor to zero gives:

2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}

2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}

Therefore, the solutions to the quadratic equation are

x = \frac{1}{2} and x = \frac{3}{2}.

5. Solve the quadratic equation 3x^2 + 5x + 2 = 0 using the quadratic formula.

Given the quadratic equation 3x^2 + 5x + 2 = 0, we can see that a = 3, b = 5, and c = 2.

Substituting these values into the quadratic formula:

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

x = \frac{-5 \pm \sqrt{5^2 - 4(3)(2)}}{2(3)}

x = \frac{-5 \pm \sqrt{25 - 24}}{6}

x = \frac{-5 \pm \sqrt{1}}{6}

x = \frac{-5 \pm 1}{6}

Therefore, the solutions to the quadratic equation are x = -1 and x = -\frac{2}{3}.

See More Worked Example
It Sticks!

1. x^2 + 5x + 6 = 0

Answer: x = -2, -3

2. x^2 - 9x + 20 = 0

Answer: x = 4, 5

3. x^2 + 10x + 25 = 0

Answer: x = -5

4. 2x^2 - 7x - 15 = 0

Answer: x = -1.5, 5

5. 3x^2 + 7x - 6 = 0

Answer: x = -2, 0.5

6. x^2 + 4x + 13 = 0

Answer: x = -2 + 3i, -2 - 3i

7. 2x^2 + 8x + 8 = 0

Answer: x = -2, -2

8. 4x^2 + 12x + 9 = 0

Answer: x = -1.5, -1.5

9. x^2 - 6x + 9 = 0

Answer: x = 3, 3

10. 3x^2 + 11x + 4 = 0

Answer: x = -1, -4/3

11. x^2 - 2x - 15 = 0

Answer: x = -3, 5

12. 2x^2 - 3x - 2 = 0

Answer: x = -0.5, 2

13. x^2 + 8x + 16 = 0

Answer: x = -4

14. 4x^2 - 64 = 0

Answer: x = -4, 4

15. x^2 + 2x + 1 = 0

Answer: x = -1

Quick Answers for Quadratic Equation
It Sticks!