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# Roots of a Quadratic Equation

1 year ago

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R

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.

M
Mawiya Jalil

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.

N
Naheed Fatima Razi Ahmed

R

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

H
Henry Oliver-Edwards

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:

1. Move the constant term to the other side: x2−3x=4x2−3x=4.
2. Divide the coefficient of xx by 2, square it, and add it to both sides: x2−3x+(−32)2=4+(−32)2x2−3x+(2−3​)2=4+(2−3​)2.
3. This gives x2−3x+94=4+94x2−3x+49​=4+49​, or x2−3x+94=254x2−3x+49​=425​.
4. Now, you have (x−32)2=254(x−23​)2=425​.
5. Taking the square root of both sides gives x−32=±52x−23​=±25​.
6. Solving for xx gives x=32±52x=23​±25​, resulting in x=4x=4 or x=−1x=−1.

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.

K
Khalid

I'm not sure what your question is but I'm guessing you want to know roughly what the other methods are for solving quadratic equations, other than the quadratic formula.

1. Factorising
2. Completing the square
3. Factor theorem (for A level students, usually those over 16 years of age)
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Following these rules and ensuring the quadratic always equate to zero before you apply the formula you will find all roots.

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Delia Munteanu
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There are 3 methods to solve a quadratic equation, these are

Completing square method

Factorisation

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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x = (-b ± √(b^2 - 4ac)) / (2a)

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I

There are several ways to solve the roots of a quadratic equation. Some of them include: factorisation method, completing the square, quadratic formular and graphical method. Here is an example:

Solve the quadratic equation 2x^2 - 5x + 3 = 0.

1. Identify the coefficients: a = 2, b = -5, c = 3.
2. Apply the formula: x = (-b ± √(b^2 - 4ac)) / 2a.
3. Calculate: x = (-(-5) ± √((-5)^2 - 4 * 2 * 3)) / (2 * 2).
4. Simplify: x = (5 ± √(1)) / 4.
5. Solve for two possible solutions:

x1 = (5 + 1) / 4 = 3/2

x2 = (5 - 1) / 4 = 1

Using the Factorisation method:

1. Observe that the equation doesn't factor easily into two linear expressions.
2. Consider trial and error to find two numbers that multiply to 3 (ac) and add up to -5 (b). In this case, the numbers are -1 and -3.
3. Rewrite the equation: 2x^2 - x - 4x + 3 = 0.
4. Group the terms: (2x^2 - x) + (-4x + 3) = 0.
5. Factor out common factors: x(2x - 1) - 2(2x - 1) = 0.
6. Combine like terms: (2x - 1)(x - 2) = 0.
7. Set each factor equal to zero and solve for x:

2x - 1 = 0 => x = 1/2

x - 2 = 0 => x = 2

There are also easy approachable factorisation methods to solve the same equation.

J
Jack Williams

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.

D
Dale D'Cruz

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

b±b2

−4ac

If you have further questions or specific points you'd like clarification on, feel free to ask!

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H
Hamza Aqil Mahmood

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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