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Roots of a...
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Robert Richard
The derivative of the quadratic formula is both values of x, which are obtained by addressing the quadratic equation. These derivatives of a quadratic equation are also called absolute nos of the formula. For example, the roots of the formula x2 - 3x - 4 = 0 are x = -1 and x = four because each satisfies the formula. that is,
At x = -1, (-1 )2 - 3( -1) - 4 = 1 + 3 - 4 = 0
At x = 4, (4 )2 - 3( 4) - 4 = 16 - 12 - 4 = 0
There are different methods for finding the derivative of a quadratic equation. The use of the quadratic formula calculator is one of them.
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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.
Use the quadratic formula to get answer by direct substitution
Ax2+Bx+c=0 x1+x2=-(B/A),, X1.X2=C/A
There are 4 different ways of finding the roots of the quadratic equation.
The first is using a factoring method
This method involves expressing the quadratic equation in a factored form as the product of two binomials. For x2−3x−4=0x2−3x−4=0, you look for two numbers that multiply to give the constant term (-4) and add to give the coefficient of xx (-3). The numbers -4 and 1 satisfy this condition, so you can write:
x2−3x−4=(x−4)(x+1)=0
x2−3x−4=(x−4)(x+1)=0
Setting each factor equal to zero gives the roots:
x−4=0⇒x=4
x−4=0⇒x=4
x+1=0⇒x=−1
x+1=0⇒x=−1
The second method is completing the square
This method involves manipulating the equation to form a perfect square trinomial, which can then be solved by taking the square root of both sides. For the given equation:
The third method is using the quadratic formula
The quadratic formula is a universal method that can solve any quadratic equation. The roots of the equation ax2+bx+c=0ax2+bx+c=0 can be found using:
x=−b±b2−4ac2a
x=2a−b±b2−4ac
For x2−3x−4=0x2−3x−4=0, a=1a=1, b=−3b=−3, and c=−4c=−4, so:
x=−(−3)±(−3)2−4(1)(−4)2(1)=3±9+162=3±252=3±52
x=2(1)−(−3)±(−3)2−4(1)(−4)
=23±9+16=23±25
=23±5
Thus, x=4x=4 or x=−1x=−1.
The last method is using a graphical method
The roots of the quadratic equation are the x-coordinates of the points where the graph of the equation intersects the x-axis. You can graph the quadratic function y=x2−3x−4y=x2−3x−4 and find the points where y=0y=0. This method visually represents the solutions but requires graphing tools or software.
Very good Robert, I agree with your statement.
To solve a quadratic equation, two methods are commonly used:
Factorising is a quick method to use where the roots are whole numbers and the factors of c in the general equation for a quadratic formula (ax^2 + bx + c = 0) are easily identified.
Otherwise, the quadratic formula is another alternative method and necessary when the roots are not whole numbers!
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x = (-b +/- square root(b2 - 4ac)) /2a
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There are 3 methods to solve a quadratic equation, these are
Completing square method
Factorisation
Using the quadratic formula
Here let's use the factorisation method,
The equation is x2-3x-4 = 0
So you have to find two numbers their sum is -3( the coefficient of X) and their product is -4( That is the power of the leading coefficient, in this case the 1 multiplied by the constant, in this case -4, therefore 1*-4= -4)
The two numbers are -4 and 1, because -4*1= -4 and -4+1= -3
Therefore, the equation becomes (x-4 ) (x+1)=0
so, x-4 =0 and x+1=0
x= 4 and x=-1
therefore the solution is x= 4 and x=-1
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At x
=
−
1
x=−1:
(
−
1
)
2
−
3
(
−
1
)
−
4
=
1
+
3
−
4
=
0
(−1)2
−3(−1)−4=1+3−4=0
At x
=
4
x=4:
(
4
)
2
−
3
(
4
)
−
4
=
16
−
12
−
4
=
0
(4)2
−3(4)−4=16−12−4=
I'm available for 1:1 private online tuition!
Click here to view my profile and arrange a free introduction.The quadratic formula is:
x = [-b (+/-) sqrt(b2 - 4ac)] / 2a
Original form of a quadratic equation is:
ax2 +bx + c = 0
Therefore:
a = 1, b = -3 and c = -4
x = {-(-3) + sqrt[(-3)2 - 4(1)(-4)]} / 2(1) = [3 + sqrt(25)] / 2 = 4
and
x = {-(-3) - sqrt[(-3)2 - 4(1)(-4)]} / 2(1) = [3 - sqrt(25)] / 2 = -1
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!
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What's the question
I'm available for 1:1 private online tuition!
Click here to view my profile and arrange a free introduction.A quadratic equation is an equation with an x^2. For example x^2 + 7x + 12 = 0
The root of a quadratic equation is the solution of the equation - what values of x in the equation above make the answer 0? In this case, we can factorise the equation to get (x+3)(x+4) = 0, so roots would be x = -3 and x = -4. Substituting these values back into the equation would give 0.
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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.
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