Roots of a Quadratic Equation

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 are what it makes the y value equal to 0, one way to find the roots is by factorising the equation (in the form ax^2 + bx + c), into a form (x -k)(x-j) where k and j add up to make b, and multiply together to make c. The other way is using the quadratic formula.

There's a few ways you can approach this Richard. Here are some for you to help find the roots of a quadratic equation:

Using the quadratic formula (as you mentioned):

x = (-b +- sqrt(b^2 - 4ac) ) / 2a

Where a, b and c are coefficients of the quadratic equation ax^2 + bx + c = 0.

So you can sub in the coefficients into the above formula and get two roots for x.

Using a graph:

You could plot the equation using a graphical calculator or software etc and see where it intersects the x-axis, those points of intersection would be the roots for the x coordinate.

You could complete the square of the equation:

If you rearrange to put the quadratic equation in vertex form to complete the squar - (a(x-i)^2 + j) then you can find the roots by making the expression in square brackets equal to zero and solving for x.

You could factor the quadratic equation if it is factorable:

If you can factor the quadratic equation into the form (x+-a)(x+-b) = 0 for example then you solve for x by set each expression in those brackets to zero and rearranging.

I hope this helps solve your problem Richard. If you have any questions about the above methods please do let me know and I will elaborate.

Quadratic equation always have two roots.

Wr can find the roots in anyone of the following ways.

1. Factorizing.

2. Quadratic formula

3. Completing the Square.

Roots can be real, equal and imaginary depends upon equation.

For more details contact me on my cell no. 07849 122342

The other possible method is to use complete the square.

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!

You can use quadratic formula or by factorisation.

You can simply factorise

Yes you are correct!

The quadratic formula is used to find the roots of a quadratic equation, which is generally written as Ax^2 + Bx + C = 0. The formula is given by:

x = (-B ± sqrt(B^2 - 4AC)) / (2A)

For the quadratic equation x^2 - 3x - 4 = 0, the coefficients are A = 1, B = -3, and C = -4.

Plugging these values into the quadratic formula:

x = (3 ± sqrt((-3)^2 - 4(1)(-4))) / (2(1))

x = (3 ± sqrt(9 + 16)) / 2

x = (3 ± sqrt(25)) / 2

x = (3 ± 5) / 2

Now, solve for x in both cases:

1. When using +5:
2. x = (3 + 5) / 2 = 8 / 2 = 4
3. When using -5:
4. x = (3 - 5) / 2 = -2 / 2 = -1

So, the roots of the quadratic equation x^2 - 3x - 4 = 0 are x = 4 and x = -1. Your substitution into the original quadratic equation to verify these roots is also correct:

At x = -1, (-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0

At x = 4, (4)^2 - 3(4) - 4 = 16 - 12 - 4 = 0

complete the square