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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.

47 Answers

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

Use the quadratic formula to get answer by direct substitution

R

Rajender Reddy Humnabad

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:

- Move the constant term to the other side: x2−3x=4x2−3x=4.
- 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.
- This gives x2−3x+94=4+94x2−3x+49=4+49, or x2−3x+94=254x2−3x+49=425.
- Now, you have (x−32)2=254(x−23)2=425.
- Taking the square root of both sides gives x−32=±52x−23=±25.
- 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.

- Factorising
- Completing the square
- Factor theorem (for A level students, usually those over 16 years of age)

Margaret O
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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

Fekadu G
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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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Syed T
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x = (-b ± √(b^2 - 4ac)) / (2a)

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I

IFEANYI MADUEKE

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.

**Using the quadratic formula**

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

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

x2 = (5 - 1) / 4 = 1

Using the Factorisation method:

- Observe that the equation doesn't factor easily into two linear expressions.
- 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.
- Rewrite the equation: 2x^2 - x - 4x + 3 = 0.
- Group the terms: (2x^2 - x) + (-4x + 3) = 0.
- Factor out common factors: x(2x - 1) - 2(2x - 1) = 0.
- Combine like terms: (2x - 1)(x - 2) = 0.
- 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

*ax*2

+*bx*+*c*=0, its derivative is 2��+�

2*ax*+*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

*x*2

−3*x*−4=0, the roots are �=−1

*x*=−1 and �=4

*x*=4

The quadratic formula *x*=2*a*

−*b*±*b*2

−4*ac*

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.

Yuri M
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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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