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Last updated on July 5th, 2025

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Standard Form of Quadratic Equation

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The quadratic equation is a second-degree polynomial, meaning that it is the highest degree exponent of two (x2). The word quadratic comes from quad, and it means square, as here the highest degree is two. ax2 + bx + c = 0 is the standard form of a quadratic equation.

Standard Form of Quadratic Equation for UK Students
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What is the Standard Form of Quadratic Equation?

The standard form of a quadratic equation is ax2 + bx + c = 0, where 
x is the variable, a ≠ 0, 
a, b, and c are real numbers. 
Here, a is the coefficient of x2
b is the coefficient of x
And c is the constant

 

 

There are three forms to write the quadratic equation, including:

 

  • Standard form: ax2 + bx + c = 0
  • Vertex form: a(x - h)2 + k = 0
  • Intercept form: a(x - p)(x - q) = 0
     
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What are the Characteristics of Standard Form of Quadratic Equation?

A second-degree polynomial equation is the quadratic equation, and its standard form is ax2 + bx + c = 0. The characteristics of the standard form of quadratic equations are:

 

 

  •  Degree of equation: The degree of the equation is determined using the highest value of the exponent. In a quadratic equation, the highest power of x is 2. So the quadratic equation is a second-degree polynomial. 

 

  • Shape of the graph: The graph of a quadratic equation forms a parabola; the shape of the parabola depends on the value of a. If a > 0, then the parabola opens upwards, and if a < 0, then the parabola opens downwards.

 

  • Vertex of parabola: The vertex of the parabola represents the minimum or maximum point of the graph, depending on the sign of a. It is located at (-b/2a, f (-b/2a)), where f(x) is the quadratic expression. 

 

  • Axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex, dividing the parabola into halves. The vertical line is given by x = -b/2a.
     
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How to Write a Quadratic Function in Standard Form?

The standard form of a quadratic function is written as:
f(x) = ax2 + bx + c = 0
Here, a, b, and c are the constant coefficients, and x is the variable. It is also known as the second-degree equation. In a quadratic function, the value of a ≠ 0, because if the value of a is 0, then the function will not be quadratic, as the highest degree in a quadratic is 2. 
 

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How to Convert Standard Form of Quadratic Equation into Vertex Form?

The standard form of a quadratic equation is ax2 + bx + c = 0, and the vertex form is a(x - h)2 + k = 0, where (h, k) are the vertices of the quadratic function. To convert the equation from standard form to vertex form, we compare these two equations:
ax2 + bx + c = a(x - h)2 + k
Substituting the value of (x - h)2 in the equation, (x - h)2 = x2 - 2xh + h2
ax2 + bx + c = a (x2 - 2xh + hj) + k
ax2 + bx + c = ax2 -2axh + ah2 + k
Comparing the coefficients of x on both sides:
bx = -2axh
h = -bx/2ax
h = -b/2a, let’s consider this equation as (1)

Comparing the constants on both sides
c = ah2 + k

Substituting the value of h from (1)
c = a(-b/2a)² + k
c = a(b2/4a2) + k
c = (b2/4a) + k
c - (b2/4a) = k
k = c - (b2/4a)

Therefore, we can use the formulas h = -b/2a and k = c - (b2/4a) to convert a standard form of a quadratic equation into vertex form. 

For example, convert 3x2 + 6x - 5 = 0 to vertex form
Here, a = 3
b = 6
c = -5

Given, equation is a(x - h)2 + k = 0
Finding the value of h and k:
h = -b/2a
h = -(6/2 × 3)
= -6/6 
= -1

k = c - (b2/4a)
= -5 - (62/4 × 3)
= -5 - (36/12)
= -5 - 3 
= -8

Substituting the value of h and k in: a(x - h)2 + k = 0
 3(x - (-1))2 - 8 = 0
3(x + 1)2 -8 
 

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How to Convert Vertex Form to Standard Form?

To convert vertex form to standard form, we simplify (x - h)2 = (x - h)(x - h). Let’s see how to convert with an example, converting 2(x + 3)² - 5

 

 

Here, a = 2
h = -3
k = -5
2(x + 3)² - 5 = 0
2(x + 3)(x + 3) - 5 = 0
2 (x2 + 6x + 9) -5 = 0
2x2 + 12x + 18 - 5 =0
2x2 + 12x + 13 = 0

 


 
How to Convert Standard Form of Quadratic Equation into Intercept Form? 

 


The quadratic equation in intercept form is a(x - p)(x - q) = 0, where (p, 0) and (q, 0) are the x-intercepts. To convert a standard form to an intercept form, we first find the roots of the quadratic equation, as p and q are the roots of the quadratic equation. Let’s learn it with an example,


For example, converting the quadratic equation x2 - 7x + 12 = 0 into intercept form
We first find the root of the quadratic equation.
x2 - 7x + 12 = 0
Here, a = 1
b = -7
c = 12
To find the value of x we use quadratic equation:
x = (-b ± (√b2 - 4ac)2a
x = (-(-7) ± (√(-7)2 - (4 × 1 × 12)/2
x = (7 ± (√49 - 48))/2
x = (7 ± √1)/ 2
x = (7 ± 1)/2
So, x = (7 + 1)/2 ⇒ 8/2 = 4
x = (7 -1)/2 ⇒ 6/2 = 3

As x = 4 and x = 3
Therefore, p = 4 and q = 3
The intercept form of the quadratic equation is:
a(x - p)(x - q) = 0
Substituting the value of p and q:
1(x - 4)(x - 3) = 0

 

 

How to Convert Intercept Form to Standard Form?

 

To convert a quadratic equation in intercept form to standard form, we simply use the intercept form. In other words, by simplifying (x - p)(x - q) = 0. 
For example, convert (2x + 3)(x -4) = 0 into standard form
(2x + 3)(x - 4) = 2x2 - 8x + 3x - 12 = 2x2 - 5x - 12 

 

 

How to Represent Quadratic Functions in Standard Form in Graph?


The standard form of a quadratic function is f(x) = ax2 + bx + c, where a ≠ 0. The curve in the graph of a quadratic function is a parabola. 
The width and slope of the parabola depend on the value of a. The vertex of the parabola is where the axis of symmetry crosses, and as it has the axis of symmetry, all parabolas are symmetric. The parabola opens upwards, and it opens downwards if a is negative. 
 

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Real-world Applications of Standard Form of Quadratic Equation

In real-life, we use the standard form of a quadratic equation, where the relationships involve squared terms. The few applications of the standard form of quadratic equations

 

 

  • To calculate the maximum height, range, and landing point of the object in projectile motion such as basketball shot and thrown rescue rope we use the quadratic equations. For example, in sports quadratic equations analyze the trajectory of a football to see the player's performance. 

 

  • Companies use the quadratic equation in profit modeling. For example, to predict the optimal number of products to produce and the best price to maximize revenue, quadratic equations are required.

 

  • In building bridges and arches, quadratic equations are used to distribute the weight and prevent collapse. For example, to ensure stability and prevent structural failure.
     
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Common Mistakes and How to Avoid Them in Standard Form of the Quadratic Equation

Students often find it hard to convert quadratic equations from one form to another. Here are some common mistakes and the ways to avoid them. Students can master the quartic equations by understanding these mistakes. 
 

Mistake 1

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Not rearranging the equation
 

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Sometimes students start solving the equation without arranging it to the standard form, for example, 5x2 + 5x - 5 = 0. Before solving the equation, it is important to arrange the equation in the standard form. The standard form of a quadratic equation is ax2 + bx + c = 0, so the correct form is 5x2 + 5x + 5 = 0.
 

Mistake 2

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Misidentifying the values of a, b, and c
 

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Mixing up coefficients is also an error that students often repeat, so always list out the values of a, b, and c before solving. 

Mistake 3

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Thinking all the parabolas open upwards 
 

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The graph of quadratic equations forms a parabola, and students assume that all parabolas open upwards, which is wrong. The opening of the parabola is based on the value of a, if a > 0, it opens upwards, and if a < 0, it opens downwards.

Mistake 4

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Incorrect expansion of (x - h)2 

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When converting the quadratic equation in vertex form to standard form, students make mistakes, especially when expanding. For example, (x - h)2 = x2 - h2 instead of x2 - 2hx + h2. This is incorrect. So, always remember (x - h)2 = x2 - 2xh + h2. 
 

Mistake 5

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Not expanding the binomials when converting the intercept to standard form 
 

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When converting the quadratic equations from intercept form to standard form, students make errors by not expanding the binomials, that is, a(x - p)(x - q). So always expand the binomial, (x - p)(x - q), then multiply the result by a and then simplify it to ax2 + bx + c = 0.
 

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Solved Examples of Standard Form of Quadratic Equation

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Problem 1

Convert x2 + 6x + 5 = 0 to vertex form

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 (x + 3)2 - 4 = 0
 

Explanation

To convert a quadratic equation from standard to vertex form, we find the value of h and k using the formulas: 
h = -b/2a
k = (4ac - b2)/4a
Here, a = 1
b = 6
c = 5

h = -b/2a
= -6/2×1 = -3

k = (4ac - b2)/(4a)
= ((4 × 1 × 5) - 62)/4 × 1
= 20 - 36/4 
= -16/4  
= -4
Here, h = -3 and k = -4
The standard form of vertex form is a(x - h)2 + k = 0
Substituting the value of h and k, 
1(x - -3)2 + -4 = (x + 3)2 -4 = 0
x2 + 6x + 5 in vertex form is: (x + 3)2 -4 = 0
 

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Problem 2

Convert 2(x + 1)2 - 5 = 0 to standard form

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2(x - 2)(x + 1/2) = 0
 

Explanation

To convert the standard form to intercept form, we first find the root of the quadratic equation, 2x2 - 3x - 2
Here, a = 2
b = -3
c = -2

x = (-b ± √b2 - 4ac)/2a
(-(-3) ± √(-3)2 - 4 × 2 × -2)/2×2
= (3 ± √9 - -16)/4
= (3 ± √25)/4
= (3 ± 5)/4
So, x = (3 + 5)/4 = 8/4 = 2
x = (3 - 5)/4 = -2/4 = -1/2
So, p = 2 and q = -1/2
So, the intercept form is:
a(x - p)(x - q) = 0
2(x - 2)(x - -1/2) = 0
2(x - 2)(x + 1/2) = 0
 

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Problem 3

Convert 2x2 - 3x - 2 to intercept form

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In intercept form 2x2 - 3x - 2 can be written as 2(x - 2)(x + 1/2) = 0
 

Explanation

To convert a quadratic equation in vertex form to standard form, we simplify the equation
2(x + 1)2 - 5 = 0 
Here, (x + 1)2 = x2 +2x + 1
So, the equation becomes:
2(x2 + 2x + 1) - 5 = 0
2x2 + 4x + 2 - 5 = 0
2x2 + 4x -3 = 0
Therefore, in intercept form 2x2 - 3x - 2 can be written as 2(x - 2)(x + 1/2) = 0
 

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Problem 4

Convert x2 - 5x + 6 = 0 to intercept form

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 (x - 3)(x -2) = 0
 

Explanation

To convert the quadratic equation from standard form to intercept form, first, we find the roots of the quadratic equation. 
x = (-b ± √b2 - 4ac)/2a
Here, a = 1
b = -5
c = 6
Computing: √b2 - 4ac
√(-5)2 - 4 × 1 × 6 
= √25 - 24 
=√1

Substituting the value of √b2 - 4ac in (-b ± √b2 - 4ac)/2a
x = (-(-5) ± √1)/2
= (5 ± 1)/2
x = (5 + 1)/2 = 6/2 = 3
x = (5 - 1)/2 = 4/2 = 2
So, p = 3 and q = 2
Substituting the value of p and q in the equation:
a(x - p)(x - q) = 0
1(x - 3)(x - 2) = 0
(x - 3)(x -2) = 0
 

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Problem 5

Convert 3(x - 1)(x + 5) = 0 to standard form

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3x2 + 12 x - 15 = 0
 

Explanation

 To convert to standard form, we expand 3(x - 1)(x + 5) = 0
Expanding (x - 1)(x + 5):
(x - 1)(x + 5) = x2 + 5x - x - 5 
= x2 + 4x - 5
Multiplying by 3: 3(x2 + 4x - 5)
3x2 + 12x - 15 = 0
 

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FAQs on Standard Form of Quadratic Equation

1.What is a quadratic equation?

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2.What is the standard form of a quadratic equation?

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3.What are the different forms to represent a quadratic equation?

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4.Can ‘a’ be a zero in the standard form of a quadratic equation?

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5.What are the uses of a quadratic equation?

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6.How does learning Algebra help students in United Kingdom make better decisions in daily life?

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7.How can cultural or local activities in United Kingdom support learning Algebra topics such as Standard Form of Quadratic Equation?

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8.How do technology and digital tools in United Kingdom support learning Algebra and Standard Form of Quadratic Equation?

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9.Does learning Algebra support future career opportunities for students in United Kingdom?

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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