Maths

>

KS3

Geometry and Measures

Question

Roots of a Quadratic Equation

1 year ago

·

90 Replies

·

6182 views

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.

90 Answers

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.

A
Anthony Kaminskyj

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).

Yuri M Profile Picture
Yuri M Verified Sherpa Tutor ✓

Maths and Physics Tutor for Primary, KS3, GCSE and A-Level Students

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.
A
Anthony John

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.

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.

Hafsa F Profile Picture
Hafsa F Verified Sherpa Tutor ✓

KCL medical student with top grades and years of tutoring experience

12 reviews

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.
A
Amogh

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

Dr John Profile Picture
Dr John Verified Sherpa Tutor ✓

10 years SEN tutoring and lecturing experience in Maths, Sciences

Completing the square method is another, iterative method too is also another and factorisation method.

I'm available for 1:1 private online tuition!

Click here to view my profile and arrange a free introduction.
M
Minhaj Ali

Start with the quadratic equation: x2−3x−4=0

Identify the coefficients:

  • A=1
  • B=−3
  • C=−4

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.

Fekadu G Profile Picture
Fekadu G Verified Sherpa Tutor ✓

Today's decision will determine your future destination.

2 reviews

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

I'm available for 1:1 private online tuition!

Click here to view my profile and arrange a free introduction.
D
Delia Munteanu
David M Profile Picture
David M Verified Sherpa Tutor ✓

Full Time Professional Maths Tutor

14 reviews

What's the question

I'm available for 1:1 private online tuition!

Click here to view my profile and arrange a free introduction.
N
Naheed Fatima Razi Ahmed

Use the quadratic formula to get answer by direct substitution

J
Jim Ross

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

H
Halima Rahim

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).

Think you can help?

More Maths KS3 Questions
Sherpa Badge

Need a KS3 Maths tutor?

Get started with a free online introductions with an experienced and qualified online tutor on Sherpa.

Find a KS3 Maths Tutor