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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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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.
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).
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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.
I'm available for 1:1 private online tuition!
Click here to view my profile and arrange a free introduction.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.
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.
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At x
=
−
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x=−1:
(
−
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2
−
3
(
−
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−
4
=
1
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(−1)2
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At x
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I'm available for 1:1 private online tuition!
Click here to view my profile and arrange a free introduction.A quadratic is a function that may contain 2, 1 or no real roots, depending on the nature of the quadratic function.
Step 1: the first step to solving a quadratic and finding its roots is finding out how many roots there are, using the discriminant.
All quadratic equations take the form ax^2 + bx + c = 0 and the discriminant is given by the value or (b^2 - 4ac). When the discriminant is positive, there are 2 real roots. When is it equal to zero there is one real root. When it is negative, there are no real roots.
Step 2: Once we know how many roots there are, we can go about finding them. Roots can be found easily in one of two methods:
Method one: factorisation
use the ‘c’ term of the quadratic and identify pairs of factors which multiply to give the value of c. Then, identify which of these pairs will add/subtract to give the value of ‘b’ from the quadratic. Place the two correct factors into two sets of brackets (x-1)(x+3) etc.
Method two: quadratic formula
the two roots of a quadratic are also given by
x= (-b±sqrt(b^2 - 4ac))/2a
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Completing the square method is another, iterative method too is also another and factorisation method.
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Click here to view my profile and arrange a free introduction.Start with the quadratic equation: x2−3x−4=0
Identify the coefficients:
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.
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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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What's the question
I'm available for 1:1 private online tuition!
Click here to view my profile and arrange a free introduction.Use the quadratic formula to get answer by direct substitution
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
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).
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