Roots of a Quadratic Equation

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.

The quadratic equation is given as

x= (-b plus or minus square root b^2-4ac) / 2a

a is the coefficient of x^2

b is the coefficient of x

and c is the number without x.

You insert the values for a, b and c into the quadratic equation and you find the value for x which is the root.

If (b^2 - 4ac) is greater than 0, there are two distinct roots.

If (b^2 - 4ac) is equal to 0, there is only 1 root.

If (b^2 - 4ac) is less than 0, there are no real roots (only complex roots).

Ax2+Bx+c=0 x1+x2=-(B/A),, X1.X2=C/A

Start with the quadratic equation: x2−3x−4=0

Identify the coefficients:

• A=1
• B=−3
• C=−4

Use the quadratic formula and insert the values.

-Will give you -1 and 4 as the x values.

Insert -1 and 4 into the original formula x2−3x−4.

Inserting -1 and 4 gives you 0 and hence x = 4 and x = -1 are the correct roots.

x=2(1)−(−3)±(−3)2−4(1)(−4)

​​ x=3±9+162x = \frac{3 \pm \sqrt{9 + 16}}{2}x=23±9+16

​​ x=3±252x = \frac{3 \pm \sqrt{25}}{2}x=23±25

​​ x=3±52x = \frac{3 \pm 5}{2}x=23±5​

Thus, the two roots are:

x=3+52=4andx=3−52=−1x = \frac{3 + 5}{2} = 4 \quad \text{and} \quad x = \frac{3 - 5}{2} = -1x=23+5​=4andx=23−5​=−1

As shown, these roots are x=4x = 4x=4 and x=−1x = -1x=−1, which both satisfy the original equation.

Verification:

• For x=−1x = -1x=−1: (−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
• For x=4x = 4x=4: 42−3(4)−4=16−12−4=04^2 - 3(4) - 4 = 16 - 12 - 4 = 042−3(4)−4=16−12−4=0

So, x=4x = 4x=4 and x=−1x = -1x=−1 are indeed the correct roots.

Derivatives of a Quadratic Equation:

If you're interested in the derivative of a quadratic equation, you're finding the rate of change of the quadratic function. The derivative of a quadratic equation f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c is:

f′(x)=2ax+bf'(x) = 2ax + bf′(x)=2ax+b

This gives the slope of the tangent line to the quadratic curve at any point xxx. However, this is different from solving the quadratic equation for its roots.

4o

Roots of a quadratic equation are simply those values of the variable which satisfy the equation.

When those values are substituted in the equation,they make the equation equal to zero and hence satisfy the equation.

Yes I verified your values of x=4 and x=-1. These are called solutions of the equation rather than absolute values. They are the values of x for which the expression = zero. You can use quadratic equation here but you should also be able to factorise to get (x-4)(x+1).

The formula for solving a quadratic equation is:

x = (-b +/-sq root (b2 -4ac))/2a

If we compare the equation, x2 - 3x - 4 = 0  to ax2 + bx + c

we see that a = 1, b= -3, c=-4

So, substituting into the quadratic equation formula, we get:

-b=3, b2= -3 x -3 = 9, 4ac= 4x1 x -4 = -16, 2a = 2

x = (-b +/-sq root (b2 -4ac))/2a

x = (3 +/- sq root (9+16))/2

x =(3 +/- sq root 25)/2

x =(3 +/- 5)/2

So the solution to the equation is either x = (3 +5)/2 = 8/2 = 4 

or x = (3-5)/2 = -2/2 = -1

To find the derivative of a quadratic equation you differentiate the equation and then that is the value. If the question means how do you solve the quadratic equation then you could factorise or use the quadratic formula.

Completing the square method is another, iterative method too is also another and factorisation method.

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.

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!

What's the question

I don't think you are refering to the derivative, which is the resultant equation when differentiating an equation.

I think you are talking about the roots of the equation, which are -1 and 4, as you have correctly stated.

The standard method utilised to solve a quadratic equation is by considering the coefficients of the equation in this way:

ax^2 + bx + c = 0. "a" is the coefficient of the squared term, b of the non-squared x term and c of the term without the variable x. The formula is: x1,2 = (-b (+/-) sqrt(b^2-4ac))/2a

For instance: x^2-6x+5 = 0 can be solved in this way:

x1,2 (roots!) = (6 (+/-) sqrt(36-20))/2 = (6 (+/-) 4)/2 = 3 (+/-) 2 --> x1 = 5, x2 = 1

The second method can only be applied for equations of the form: x^2 +sx+ p = 0. This method consists in factorising the initial equation into two factors. To do that we have to consider the coefficient s as the sum of the two factors and p as their product. The two factors are called f1 and f2: x^2+sx+p = (x+f1) * (x+f2)

The values that cancel out one or the other parenthesis are the roots of the initial quadratic equation.

Same example: x^2-6x+5 = 0

Apply second method: choose f1 = -5 and f2 = -1.

-5 + (-1) = -6 = s, OK

-5 * (-1) = 5 = p OK

Rewrite the initial equation: x^2-6x+5 = (x-5) * (x-1) = 0. Hence, the solutions are: x1 = 5, x2 = 1, as before.