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How to solve the quadratic equation where the coeficients are a = 1, b = 5 and c = 6?
Quadratic Equation Calculator
Simple and best practice solution for f(x)=2x+5 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. If it's not what You are looking for type in the equation solver your own equation and let us solve it. Jun 25, 2008 f(x) = 2x - 5. So: f(-1) = 2(-1) - 5. F(-1) = -2 - 5. That's my lesson when I was in 7th grade. Now I'm in 9th grade,but I'm going to high school ASAP (grade 10th). Aug 02, 2017 determine whether the given quadratic function has a maximum value or a minimum value,and then find the value.
Please, to solve a equation like ax² + bx + c = 0, enter or replace the coefficients a, b and c. Where, a is mandatory and nonzero. Ex.: To find the roots of the equation x² + 5x +6 = 0, enter a = 1, b = 5 and c = 6.
a = b = c =
In this case we have a = 1, b = 5 and c = 6. So, the equation x² + 5x + 6 = 0 has 2 real roots when solved: x₁ = -2 and x₂ = -3
What is quadractic equation (In elementary algebra).
A quadratic equation is any equation having the form ax² + bx + c = 0 where x represents an unknown, and a, b, and c are just known numbers; they are called the 'numerical coefficients'. 0 is not allowed for the value of a because if a = 0, then the equation will be linear, not quadratic. The coeficient 'a' is the quadratic coefficient, 'b' the linear coefficient and 'c' the constant or free term.
How to Solve Quadratic Equations With the Quadratic Formula (Baskara)
A way to solve quadractic equations is making use of this formula
The part (b² - 4ac) is called the 'discriminant', because it can 'discriminate' between the possible types of answer. If it is positive, you will get two real solutions, if it is zero you get just ONE solution, and if it is negative you get complex solutions. The 'discriminant' is represented by D or the Greek letter Delta (Δ):
Δ = b² - 4acFor the Quadratic Formula to work, you must arrange the equation in the form 'ax² + bx + c = 0' known as 'Standard Form'. Fastest web browser for mac os x. Examples on how to find the coefficients:
- 1) x² + 2x - 3 = 0, a = 1, b = 2 and c = 1;
- 2) -x² + 2x + 4 = 0, a = -1, b = 2 and c = -4;
- 3) x² - x + 2-√8 = 0, a = 1, b = -1 and c = 2-√8;
- 4) x² + π = 0, a = 1, b = 0 and c = π;
- 5) x² - x = 0, a = 1, b = -1 and c = 0;
Example 1:
Lets's solve the equation x² - 5x + 6 = 0:- let's apply the quadratic formula to calculate the roots.
In this case we have a = 1, b = -5 and c = 6. - Replacing the values
- x = (5 ± √(-5)² - 4.1.6)/2.1
- x = (5 ± √25 - 24)/2
- x₁ = (5 + 1)/2 = 6/2 = 3 and
- x₂ = (5 - 1)/2 = 4/2 = 2
Example 2:
Lets's solve the equation x² + 2x - 3 = 0:- In this case we have a = 1, b = 2 and c = 1.
- Replacing the values
- x = (-2 ± √2² - 4.1.1)/2.1
- x = (-2 ± √4 - 4)/2
- x = (-2 ± 0)/2
- x = (-2)/2 = -1
- x = -2/2 = -1.
- so, in this particular case, x₁ = x₂ = 1
Quadratic Equations - Samples
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-x-5
The first term is, x2 its coefficient is 1.
The middle term is, -x its coefficient is -1.
The last term, 'the constant', is -5
Step-1 : Multiply the coefficient of the first term by the constant 1 • -5 = -5
Step-2 : Find two factors of -5 whose sum equals the coefficient of the middle term, which is -1.
The first term is, x2 its coefficient is 1.
The middle term is, -x its coefficient is -1.
The last term, 'the constant', is -5
Step-1 : Multiply the coefficient of the first term by the constant 1 • -5 = -5
Step-2 : Find two factors of -5 whose sum equals the coefficient of the middle term, which is -1.
-5 | + | 1 | = | -4 |
-1 | + | 5 | = | 4 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 1 :
Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2-x-5
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting 'y' because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 0.5000
Plugging into the parabola formula 0.5000 for x we can calculate the y -coordinate :
y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 - 5.0
or y = -5.250 Flash player download for mac os x 10.5 8.
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting 'y' because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 0.5000
Plugging into the parabola formula 0.5000 for x we can calculate the y -coordinate :
y = 1.0 * 0.50 * 0.50 - 1.0 * 0.50 - 5.0
or y = -5.250 Flash player download for mac os x 10.5 8.
Parabola, Graphing Vertex and X-Intercepts :
Solve For F X 2x If X Is 5 6
https://energyvirtual223.weebly.com/mac-os-x-check-video-file-for-drm-protection.html. Root plot for : y = x2-x-5
Axis of Symmetry (dashed) {x}={ 0.50}
Vertex at {x,y} = { 0.50,-5.25}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.79, 0.00}
Root 2 at {x,y} = { 2.79, 0.00}
Axis of Symmetry (dashed) {x}={ 0.50}
Vertex at {x,y} = { 0.50,-5.25}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.79, 0.00}
Root 2 at {x,y} = { 2.79, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving x2-x-5 = 0 by Completing The Square .
Add 5 to both side of the equation :
x2-x = 5
Now the clever bit: Take the coefficient of x , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4
Add 1/4 to both sides of the equation :
On the right hand side we have :
5 + 1/4 or, (5/1)+(1/4)
The common denominator of the two fractions is 4 Adding (20/4)+(1/4) gives 21/4
So adding to both sides we finally get :
x2-x+(1/4) = 21/4
Adding 1/4 has completed the left hand side into a perfect square :
x2-x+(1/4) =
(x-(1/2)) • (x-(1/2)) =
(x-(1/2))2
Things which are equal to the same thing are also equal to one another. Since
x2-x+(1/4) = 21/4 and
x2-x+(1/4) = (x-(1/2))2
then, according to the law of transitivity,
(x-(1/2))2 = 21/4
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(1/2))2 is
(x-(1/2))2/2 =
(x-(1/2))1 =
x-(1/2)
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-(1/2) = √ 21/4
Add 1/2 to both sides to obtain:
x = 1/2 + √ 21/4
Since a square root has two values, one positive and the other negative
x2 - x - 5 = 0
has two solutions:
x = 1/2 + √ 21/4
or
x = 1/2 - √ 21/4
Note that √ 21/4 can be written as
√ 21 / √ 4 which is √ 21 / 2
Add 5 to both side of the equation :
x2-x = 5
Now the clever bit: Take the coefficient of x , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4
Add 1/4 to both sides of the equation :
On the right hand side we have :
5 + 1/4 or, (5/1)+(1/4)
The common denominator of the two fractions is 4 Adding (20/4)+(1/4) gives 21/4
So adding to both sides we finally get :
x2-x+(1/4) = 21/4
Adding 1/4 has completed the left hand side into a perfect square :
x2-x+(1/4) =
(x-(1/2)) • (x-(1/2)) =
(x-(1/2))2
Things which are equal to the same thing are also equal to one another. Since
x2-x+(1/4) = 21/4 and
x2-x+(1/4) = (x-(1/2))2
then, according to the law of transitivity,
(x-(1/2))2 = 21/4
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-(1/2))2 is
(x-(1/2))2/2 =
(x-(1/2))1 =
x-(1/2)
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-(1/2) = √ 21/4
Add 1/2 to both sides to obtain:
x = 1/2 + √ 21/4
Since a square root has two values, one positive and the other negative
x2 - x - 5 = 0
has two solutions:
x = 1/2 + √ 21/4
or
x = 1/2 - √ 21/4
Note that √ 21/4 can be written as
√ 21 / √ 4 which is √ 21 / 2
Solve Quadratic Equation using the Quadratic Formula
Graph F X 2x
Mac os x snow leopard free download for pc. 2.3 Solving x2-x-5 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ———————--
2A
In our case, A = 1
B = -1
C = -5
Accordingly, B2 - 4AC =
1 - (-20) =
21
Applying the quadratic formula :
1 ± √ 21
x = ————--
2
√ 21 , rounded to 4 decimal digits, is 4.5826
So now we are looking at:
x = ( 1 ± 4.583 ) / 2
Two real solutions:
x =(1+√21)/2= 2.791
or:
x =(1-√21)/2=-1.791
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ———————--
2A
In our case, A = 1
B = -1
C = -5
Accordingly, B2 - 4AC =
1 - (-20) =
21
Applying the quadratic formula :
1 ± √ 21
x = ————--
2
√ 21 , rounded to 4 decimal digits, is 4.5826
So now we are looking at:
x = ( 1 ± 4.583 ) / 2
Two real solutions:
x =(1+√21)/2= 2.791
or:
x =(1-√21)/2=-1.791
Two solutions were found :
- x =(1-√21)/2=-1.791
- x =(1+√21)/2= 2.791