Roots of a Quadratic 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.

What's the question

The quadratic formulae is another quick & easy way to find out the solutions of the roots in case trial & error or algebraic factorization methods are not easily ticking!

Roots of the Quadratic Equation = x1/x2 = {-b +/- square root ( b squared - 4ac)}/2a where a = coefficient of x2, b = coefficient of y2 & c are the constant terms on the given quadratic equation.

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 \(2ax + b\).

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

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!

The above equation can easily be solve by factorization

by looking at the pair factors of -4 that add up to -3

These include -4 and +1

So, the above quadratic equation becomes (x+1)(x-4)=0

either x+1=0 or x-4=0

when x+1=0 , x=-1

When X-4=0, then x=4

hence x=-1 or 4.

Two distinct real roots.

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

X^2 - 3x - 4 = 0

(x-4)(x+1)= 0

x=4 or x=-1

My explanation is shown below: