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

Robert You got the correct roots but you have to work more on the concept of derivative. The derivative of the formula is 2x-3 and will be used to determine the rate of change or slope of the function, not the values of the formula that satisfy the equation. The method to use is to find the roots of the derivative

To solve x2 - 3x - 4, we need to identify a b and c, which are constants. 

A quadratic expression is written in the format ax2 + bx +c, where a b and c are constants (numbers)

In x2 -3x-4, a=1, b=-3 and c = -4 

Next, we find ac and b

Ac = 1 x -4 = -4 

b = -3

Next, we find two numbers that multiply to give ac and add to make b

So 1 x -4 = -4 and 1 + -4 = -3 

So the numbers are 1 and -4

We then put these in the brackets opposite x to factorise 

X2 = X x X so we have X in the brackets

(x+1)(x-4)

To get the roots, we need to solve the equations by making it =0 

(x+1)(x-4) =0 

x=-1 and x =4, because that makes the brackets = 0 and anything x 0 = 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.

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.

Another way to find the derivative of a quadratic equation is

Find the factors of the constant term (usually the number at the end of the equation). In this example its -4

The factors are

1 and 4

2 and 2

From these pairs you need to pick the pair that multiplies to get -4 and then adds to get the coefficient of x (in this example its -3)

-4 and 1 are going to be the correct pair as

-4 x 1 = -4 and -4 + 1 = -3

Then write these numbers in two brackets

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

To Solve take each bracket individually as an equation equal to 0

x-4=0 x+1=0

Solve each equation to find x

x=4 x=-1

Hi Robert,

I can see you have explained the roots of a quadratic equation beautifully here with an example. Kudos on that.

I feel that you can also use the generalised formula for finding the roots of a quadratic equation which goes as follows:

If your quadratic equation is of the form:

ax2+bx+c=0

Your roots will be X=(-b+/-sqrt(b^2-4ac))/2a

When using this for your problem, we get a= 1, b= -3, c= -4

Plugging these values in our quadratic formula we get: (3+/-5)/2

Which will give us the same roots as you have stated X=-1 and X=4.

Hope this helped your problem.

The roots of the equation x2−3x−4=0x^2 - 3x - 4 = 0x2−3x−4=0 are x=−1x = -1x=−1 and x=4x = 4x=4.

The derivative of the function f(x)=x2−3x−4f(x) = x^2 - 3x - 4f(x)=x2−3x−4 is f′(x)=2x−3f'(x) = 2x - 3f′(x)=2x−3.

The absolute values of the roots would be 111 and 444.

X^2 - 3x - 4 = 0

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

x=4 or x=-1

My explanation is shown below: