Factorise the equation.

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

Hence x=4 and x=-1

Another solution to finding the roots of the quadratic above is to factorise the equation and then solve.

So if we are looking at the above where x2-3x-4.

What 2 numbers multiply together to equal -4 but also add together to equal -3.

So let’s look at the multiplication first:

For -4 we could have the following.

1 x -4

-1 x 4

-2 x 2

2 x -2

Now let’s add each of these.

1 + (-4) =-3

-1 + 4 =3

-2 + 2 =0

2 + -2 =0

which of the above equals -3

1 and -4 these must be our factors

so we have (x + 1)(x - 4)

what would x have to equal to make each bracket total 0

so what + 1 = 0 … -1 great

and same for the other = 4

so our values for x are -1 and 4

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.

The roots of a quadratic equation can be found in many ways:

1. Attempting to factor the quadratic.
2. Using the quadratic formula.
3. Taking the derivative and setting it to zero to find the vertex/stationary point.

Dear Robert. The information you have provided in the text is incorrect. The roots of a quadratic equation are the values of 'x' that satisfy the equation, not the derivatives. But the are correct.

The roots of a quadratic equation (ax² + bx + c = 0) are the values of 'x' that make the equation equal to zero.

In the example x² - 3x - 4 = 0, the roots are x = -1 and x = 4, as you correctly verified. Therefore the derivation of the derivative is not correct concept. The derivative of a quadratic function gives the slope of the tangent line, not the roots.The methods to use to solve the quadratic equation are:Factoring: If the equation can be factored, it's often the quickest method.

Quadratic formula This formula can be used to find the roots of any quadratic equation. It is given by:

x = (-b ± √(b² - 4ac)) / 2a

where 'a', 'b', and 'c' are the coefficients of the quadratic equation (ax² + bx + c = 0)

The roots of a quadratic equation are the values of  that satisfy the equation, meaning they make the expression equal to zero. A general quadratic equation has the form:

The roots (also called solutions or zeros) can be found using the quadratic formula:

This formula gives the two values of  where the graph of the quadratic equation (a parabola) intersects the x-axis.

In the example mentioned in the image, the equation given is:

To solve this, we apply the quadratic formula:

• , ,

So, the two solutions are:

These are the correct roots of the equation. The term "derivative" is misused in the original post. The derivative of a function refers to its rate of change or slope, not the roots. What the post refers to as "derivatives" are simply the roots or solutions.

In summary:

• Roots are the values that make the quadratic expression zero.
• They are found using factorisation, completing the square, or the quadratic formula.
• The term derivative should not be used when referring to the roots.

The roots of a quadratic equation are the values of  that satisfy the equation, meaning they make the expression equal to zero. A general quadratic equation has the form:

The roots (also called solutions or zeros) can be found using the quadratic formula:

This formula gives the two values of  where the graph of the quadratic equation (a parabola) intersects the x-axis.

In the example mentioned in the image, the equation given is:

To solve this, we apply the quadratic formula:

• , ,

So, the two solutions are:

These are the correct roots of the equation. The term "derivative" is misused in the original post. The derivative of a function refers to its rate of change or slope, not the roots. What the post refers to as "derivatives" are simply the roots or solutions.

In summary:

• Roots are the values that make the quadratic expression zero.
• They are found using factorisation, completing the square, or the quadratic formula.
• The term derivative should not be used when referring to the roots.

Method 1:

Method 2:

Two values of x in quadratic equation are called roots of quadratic equation. We can solve quadratic equation by four methods, factorising, by completing square, Using quadratic formula and by graphical method. Roots of quadratic equation always satisfies equation

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.

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 there’s some confusion in the terminology. Let me clarify the concepts for you:

Roots of a Quadratic Equation

The roots (or solutions) of a quadratic equation are the values of that satisfy the equation . These values make the equation equal to zero. The quadratic formula used to find these roots is:

Example:

For the quadratic equation:

Using the quadratic formula:

• , , and .

So, the roots are:

•

•

Verification:

• At :

• At :

Both values satisfy the equation, so they are the roots.

Derivative of a Quadratic Function

The derivative of a quadratic function, such as , is related to calculus and represents the slope of the function at any given point. The derivative is given by:

For :

This derivative tells you how the function is changing at any value of . It is not the same as the roots of the equation. The derivative equals zero at the vertex of the parabola, which indicates where the slope is zero (the turning point).

Key Differences:

• Roots of the quadratic equation are the solutions where .

• Derivative of the quadratic function gives the slope of the function and is zero at the vertex.

Conclusion:

The term “derivative of a quadratic equation†seems to be mixed up with “roots of a quadratic equationâ€. The roots are not derivatives. Instead, the quadratic formula finds the roots, while differentiation gives the derivative.

If you meant solving quadratic equations, the quadratic formula or factorisation can help. If you meant calculus, the derivative shows the rate of change of the function.

This is a case where a minor error can result in a completely different solution, however, Robert has gotten away lucky this time!

The main misunderstanding here is the following: Derivative & Roots.

Derivative refers to a formula (commonly dy/dx or y') made from an original equation (commonly y) which can be used to find the slope of a tangent at a given point.

Roots (also called solutions) to the equation are (usually) x values which, when inputted, output y=0.

For the given formula: x2 - 3x -4 = 0, we know that the roots (x values) are likely to be our end goal, as the output is set to 0. Note that the format of a quadratic is: y = ax2 +bx + c, thus in this question a=1,b=-3,c=-4; the plotted graph can be seen below.

There are, at GCSE, 4 main methods for finding these x values: Graphing, Factoring, Quadratic and Completing the square; we will only cover Factoring and Quadratic for this example.

Factoring

Factoring refers to finding 2 values, which can add together to make b (-3) and multiply together to make c (-4) (note that in this example a=1, which greatly simplifies the process).

1 ) I've found that the best way of approaching this is by first listing the factors of |c| (absolute value of c) = 4.

This gives:

2,2

1,4

2 ) Now we have to consider that c is negative, so we can conclude that one of factors must also be negative, e.g. -1,4, 2,-2 etc. Next, we need to identify the values which can be added together to make -3.

-2 + 2 = 0

2 + -2 = 0

1 + -4 = -3

-1 + 4 = 3

3 ) We can clearly see that using 1 and -4, a value of -3 is acquired, satisfying our requirements. All that is left is to input them into brackets of the format (x + factor) and multiply them together:

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

For the result to be 0, either x+1 = 0 or x-4 = 0. Both of these are 'events' which can occur so we must consider each of them.

if x+1=0, x=-1

if x-4=0, x=4

This gives our solutions to be x=4, x=-1 and can be confirmed by looking at the graph below.

Quadratic Formula

This is an alternative approach in which you simply input some values into a formula to get the roots, but what is the equation and what are the values?

Quadratic formula: x = ( -b ± sqrt ( b2 - 4ac ) ) / 2a

The values are taken from the quadratic formula, following the same format as mentioned in the Factoring solution.

quadratic equation format: ax2 + bx + c = 0(or y); our equation is x2 -3x -4 = 0, giving us a =1, b=-3, c=-4.

Our quadratic formula now becomes: x = ( ( -(-3) ± sqrt ( (-3)2 - 4(1)(-4) ) ) / 2(1) = ( ( 3 ± sqrt ( 9 + 16 ) ) / 2 = (3 ± 5)/2

The result has this plus-minus sign in it, which just means that the calculation can be taken as a plus or as a minus; in the same way as the Factoring solution, we must consider both.

(3 + 5) / 2 = 4

(3 - 5) / 2 = -1

Which, as expected, has given us the same solutions, x = 4, x = -1.

Factorising

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

thefefore x = -1 x= 4

completing the square

(x-3/2)^2 - (3/2)^2 - 4 = 0

(x-3/2)^2 - 9/4 - 4= 0

(x-3/2)^ = 25/4

x-3/2 = 5/2 x-3/2=-5/2

x= 4. x=-1

Graphing

plug x values between -3 and 3 into the equation to find y. Plot all x and values on a graph. Find the 2 points where the graph crosses the x axis

correct answer, the question solved and the answer are properly addressed

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.