Roots of a Quadratic Equation

Completing the square method, Factorisation method, Quadratic Formula method are other different method for finding the roots or zeros of quadratic equations.

The quadratic formula is of form

x = - b±(b ² - 4ac)½/(2a)

a = 1 b = –3 c = - 4

x = - (-3) ± ((-3)² - 4(1)(-4))½/(2×1)

x = 3 ± (9 +16)½/2

x = 3 ± (25)½/2

x = (3 ± 5)/2

x = (3+5)/2 or (3 - 5)/2

x = 8/2 or -2/2

x = 4 or - 1

The use of the discriminant.

Factorisation is also key as there are 2 numbers that multiply to -4 and that add to -3.

There are several ways to solve the roots of a quadratic equation. Some of them include: factorisation method, completing the square, quadratic formular and graphical method. Here is an example:

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

Using the quadratic formula

1. Identify the coefficients: a = 2, b = -5, c = 3.
2. Apply the formula: x = (-b ± √(b^2 - 4ac)) / 2a.
3. Calculate: x = (-(-5) ± √((-5)^2 - 4 * 2 * 3)) / (2 * 2).
4. Simplify: x = (5 ± √(1)) / 4.
5. Solve for two possible solutions:

x1 = (5 + 1) / 4 = 3/2

x2 = (5 - 1) / 4 = 1

Using the Factorisation method:

1. Observe that the equation doesn't factor easily into two linear expressions.
2. Consider trial and error to find two numbers that multiply to 3 (ac) and add up to -5 (b). In this case, the numbers are -1 and -3.
3. Rewrite the equation: 2x^2 - x - 4x + 3 = 0.
4. Group the terms: (2x^2 - x) + (-4x + 3) = 0.
5. Factor out common factors: x(2x - 1) - 2(2x - 1) = 0.
6. Combine like terms: (2x - 1)(x - 2) = 0.
7. Set each factor equal to zero and solve for x:

2x - 1 = 0 => x = 1/2

x - 2 = 0 => x = 2

There are also easy approachable factorisation methods to solve the same equation.

Use the quadratic formula to get answer by direct substitution

The roots of a quadratic equation ax2 + bx + c = 0 can be found using the quadratic formula that says x = (-b ± √ (b2 - 4ac)) /2a.

What's the question

A quadratic equation is an equation of the form:

ax² + bx + c = 0

where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero.

1. At x=−1x = -1x=−1:
2. (−1)2−3(−1)−4=1+3−4=0(-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0(−1)2−3(−1)−4=1+3−4=0 ✅
3. At x=4x = 4x=4:
4. 42−3(4)−4=16−12−4=04^2 - 3(4) - 4 = 16 - 12 - 4 = 042−3(4)−4=16−12−4=0 ✅

These roots, x=−1x = -1x=−1 and x=4x = 4x=4, are the solutions of the quadratic equation.

x=4 and x=-1 as the roots for x2 - 3x - 4 = 0 as this can be factorised into (x-4)(x+1)

It seems like you're discussing the roots of a quadratic equation, which are the values of x that satisfy the equation when plugged in.

The roots can indeed be found using various methods, including the quadratic formula or calculators designed for this purpose.

The derivative of a quadratic equation, however, typically refers to the derivative of the equation itself concerning x, not its roots.

The derivative of a quadratic equation \(ax^2 + bx + c = 0\) with respect to x is \(2ax + b\).

You can find the roots of a quadratic equation with many different methods:

• Quadratic equation; namely x = [-b+-sqrt(b^2-4ac)] / 2a
• Factorising; e.g. x^2+7x+12=0 therefore (x+4)(x+3) = 0, so either x+4=0 or x+3=0, therefore x=-4 or -3
• Completing the square; e.g. x^2+2x-5 = 0, therefore (x+1)^2-5-1 = 0; (x+1)^2-6 = 0; (x+1)^2=6; (x+1) = +-sqrt(6); x = -1 +- sqrt(6)
• Graphically; e.g. if ax^2 + bx + c = 0, you can graph f(x) = ax^2 + bx + c, and find solutions for x where f(x) = 0.

The derivative of a quadratic equation (not the quadratic formula) results in a linear equation. For instance, for the quadratic equation ��2+��+�=0

ax2

+bx+c=0, its derivative is 2��+�

2ax+b.

The roots of a quadratic equation are the values of �

x that make the equation equal to zero. In your example, for the quadratic equation �2−3�−4=0

x2

−3x−4=0, the roots are �=−1

x=−1 and �=4

x=4

The quadratic formula x=2a

−b±b2

−4ac

If you have further questions or specific points you'd like clarification on, feel free to ask!

To derive the quadratic formula for solving equations of the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, we can use the method of completing the square:

1. Start with the standard quadratic equation: ax^2+bx+c=0
2. Divide all terms by a (assuming a≠0) to normalize the coefficient of x^2: x^2+(b/a)x+c/a=0
3. Subtract c/a​ from both sides to isolate the quadratic and linear terms: x^2+(b/a)x=−(c/a)
4. Complete the square by adding (b/2a)^2 to both sides: x^2+(b/a)x+(b/2a)^2=−(c/a)+(b/2a)^2
5. Rewrite the left side as a perfect square and simplify the right side: (x+b/2a)^2=(b^2−4ac)/4a^2
6. Take the square root of both sides: x+b/2a=±sqrt(b^2−4ac)/2a​
7. Solve for x by isolating it: x=−b/2a ± sqrt(b^2−4ac)/2a​​
8. Combine the terms over a common denominator: x=(−b ± b^2−4ac)/2a

This is the quadratic formula, which provides the solutions to any quadratic equation where a≠0.

The quadratic formula is used to find the roots (solutions) of a quadratic equation of the form:

ax2+bx+c=0

The solutions are given by:

x=−b±b2−4ac / 2ax ​​

For example, consider the quadratic equation:

x2−3x−4=0

Using factorization, we find the roots:

(x+1)(x−4)=0

Thus, the solutions are x = -1 and x = 4 because substituting them into the equation satisfies it:

• At x=−1

(−1)2−3(−1)−4=1+3−4=0

• At x=4x

(4)2−3(4)−4=16−12−4=0

These are the roots of the quadratic equation.

Derivative of a Quadratic Equation

The derivative of a quadratic function f(x)=ax2+bx+c is found using differentiation:

f′(x)=2ax+b

This derivative represents the slope of the quadratic function at any given value of x. It does not give the roots but instead tells us how the function changes.

To find the critical point (where the slope is zero), we solve:

2ax+b=0

which gives:

x=−b / 2a​

This xxx-value represents the vertex of the parabola.

Conclusion

The quadratic formula is used to find the roots of a quadratic equation, while differentiation helps determine the rate of change and critical points of a quadratic function.