In this article we cover quadratic equations – definitions, formats, solved problems and sample questions for practice.

A quadratic equation is a polynomial whose highest power is the square of a variable (x^{2}, y^{2} etc.)

A

monomialis an algebraic expression with only one term in it.

Example: x^{3}, 2x, y^{2}, 3xyz etc.

A

polynomialis an algebraic expression with more than one term in it.

Alternatively it can be stated as –

A polynomial is formed by adding/subtracting multiple monomials.

Example: x^{3}+2y^{2}+6x+10, 3x^{2}+2x-1, 7y-2 etc.

A polynomial that contains two terms is called a

binomial expression.

A polynomial that contains three terms is called a

trinomial expression.

A standard quadratic equation looks like this:

ax^{2}+bx+c = 0

Where a, b, c are numbers and a≥1.

a, b are called the coefficients of x^{2} and x respectively and c is called the constant.

The following are examples of some quadratic equations:

1) x^{2}+5x+6 = 0 where a=1, b=5 and c=6.

2) x^{2}+2x-3 = 0 where a=1, b=2 and c= -3

3) 3x^{2}+2x = 1

→ 3x^{2}+2x-1 = 0 where a=3, b=2 and c= -1

4) 9x^{2 }= 4

→ 9x^{2}-4 = 0 where a=9, b=0 and c= -4

For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.

For a quadratic equation ax^{2}+bx+c = 0,

the sum of its roots = –b/a and the product of its roots = c/a.

A quadratic equation may be expressed as a product of two binomials.

For example, consider the following equation

x^{2}-(a+b)x+ab = 0

x^{2}-ax-bx+ab = 0

x(x-a)-b(x-a) = 0

(x-a)(x-b) = 0

x-a = 0 or x-b = 0

x = a or x=b

Here, a and b are called the roots of the given quadratic equation.

Now, let’s calculate the roots of an equation x^{2}+5x+6 = 0.

We have to take two numbers adding which we get 5 and multiplying which we get 6. They are 2 and 3.

Let us express the middle term as an addition of 2x and 3x.

→ x^{2}+2x+3x+6 = 0

→ x(x+2)+3(x+2) = 0

→ (x+2)(x+3) = 0

→ x+2 = 0 or x+3 = 0

→ x = -2 or x = -3

This method is called *factoring*.

We saw earlier that the sum of the roots is –b/a and the product of the roots is c/a. Let us verify that.

Sum of the roots for the equation x^{2}+5x+6 = 0 is -5 and the product of the roots is 6.

The roots of this equation -2 and -3 when added give -5 and when multiplied give 6.

Let us solve some more examples using this method.

**Problem 1:** Solve for x: x^{2}-3x-10 = 0

__Solution__:

Let us express -3x as a sum of -5x and +2x.

→ x^{2}-5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0 or x+2 = 0

→ x = 5 or x = -2

**Problem 2:** Solve for x: x^{2}-18x+45 = 0

__Solution__:

The numbers which add up to -18 and give +45 when multiplied are -15 and -3.

Rewriting the equation,

→ x^{2}-15x-3x+45 = 0

→ x(x-15)-3(x-15) = 0

→ (x-15) (x-3) = 0

→ x-15 = 0 or x-3 = 0

→ x = 15 or x = 3

Till now, the coefficient of x^{2} was 1. Let us see how to solve the equations where the coefficient of x^{2} is greater than 1.

**Problem 3:** Solve for x: 3x^{2}+2x =1

__Solution__:

Rewriting our equation, we get 3x^{2}+2x-1= 0

Here, the coefficient of x^{2} is 3. In these cases, we multiply the constant c with the coefficient of x^{2}. Therefore, the product of the numbers we choose should be equal to -3 (-1*3).

Expressing 2x as a sum of +3x and –x

→ 3x^{2}+3x-x-1 = 0

→ 3x(x+1)-1(x+1) = 0

→ (3x-1)(x+1) = 0

→ 3x-1 = 0 or x+1 = 0

→ x = 1/3 or x = -1

**Problem 4:** Solve for x: 11x^{2}+18x+7 = 0

__Solution__:

In this case, the sum of the numbers we choose should equal to 18 and the product of the numbers should equal 11*7 = 77.

This can be done by expressing 18x as the sum of 11x and 7x.

→ 11x^{2}+11x+7x+7 = 0

→ 11x(x+1) +7(x+1) = 0

→ (x+1)(11x+7) = 0

→ x+1 = 0 or 11x+7 = 0

→ x = -1 or x = -7/11.

The factoring method is an easy way of finding the roots. But this method can be applied only to equations that can be factored.

For example, consider the equation x^{2}+2x-6=0.

If we take +3 and -2, multiplying them gives -6 but adding them doesn’t give +2. Hence this quadratic equation cannot be factored.

For this kind of equations, we apply the quadratic formula to find the roots.

The quadratic formula to find the roots,

x = [-b ± √(b^{2}-4ac)] / 2a

Now, let us find the roots of the equation above.

x^{2}+2x-6 = 0

Here, a = 1, b=2 and c= -6.

Substituting these values in the formula,

x = [-2 ± √(4 – (4*1*-6))] / 2*1

→ x = [-2 ± √(4+24)] / 2

→ x = [-2 ± √28] / 2

When we get a non-perfect square in a square root, we usually try to express it as a product of two numbers in which one is a perfect square. This is for simplification purpose. Here 28 can be expressed as a product of 4 and 7.

→ x = [-2 ± √(4*7)] / 2

→ x = [-2 ± 2√7] / 2

→ x = 2[ -1 ± √7] / 2

→ x = -1 ± √7

Hence, √7-1 and -√7-1 are the roots of this equation.

Let us consider another example.

Solve for x: x^{2} = 24 – 10x

__Solution__:

Rewriting the equation into the standard quadratic form,

x^{2 }+10x-24 = 0

What are the two numbers which when added give +10 and when multiplied give -24? 12 and -2.

So this can be solved by the factoring method. But let’s solve it using the new method, applying the quadratic formula.

Here, a = 1, b = 10 and c = -24.

x = [-10 ± √(100 – 4*1*-24)] / 2*1

x = [-10 ± √(100-(-96))] / 2

x = [-10 ± √196] / 2

x = [-10 ± 14] / 2

x = 2 or x= -12 are the roots.

For an equation ax^{2}+bx+c = 0, b^{2}-4ac is called the discriminant and helps in determining the nature of the roots of a quadratic equation.

If b^{2}-4ac > 0, the roots are real and distinct.

If b^{2}-4ac = 0, the roots are real and equal.

If b^{2}-4ac < 0, the roots are not real (they are complex).

Consider the following example:

**Problem:** Find the nature of roots for the equation x^{2}+x+12 = 0.

__Solution__:

b^{2}-4ac = -47 for this equation. So it has complex roots. Let us verify this.

→ [ -1±√(1-48)] / 2(1)

→ [-1±√-47] / 2

√-47 is usually written as i √47 indicating it’s an imaginary number.

Hence verified.

Problem 1

Solve for x: x^{2}-15x+56 = 0

A. x = 14 or x = 4

B. x = 8 or x = 7

C. x = 28 or x = 2

D. All of the above

Answer 1

B.

**Explanation**

Only 8 and 7 satisfy the conditions of adding up to 15 and giving a product of 56.

Problem 2

Find x if 2x^{2}+7x+4 = 0

A. -7 ± √17 / 4

B. -7 ± √7 / 4

C. [-7 ± √17] / 4

D. [-7 ± √17] / 2

Answer 2

C.

**Explanation**:

Applying the quadratic formula and substituting a=2, b=7 and c=4, we get the answer as C.

Problem 3

For what value of k does the equation x^{2}-12x+k = 0 have real and equal roots?

A. 6

B. 35

C. 12

D. 36

Answer 3

D.

**Explanation**

b^{2}-4ac = 0 for the equation to have real and equal roots.

144-4k = 0 → k = 36

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Got queries unrelated to this article? Post them on our General Queries Page.## 8 Comments

36(x)^2-24(2)^1/2 x+8 = 0 solve it

36x^2 – 24(√2)x + 8=0

x =24√2±[√(1152 – 4*36*8)]/2*36

=24√2±[√1152-1152]/72

=24√2±0/72

so, x =24√2/72

you can simplify ahead by dividing further

x=√2 / 3

If p and q are the root of the equation ax2+bx+c=0,than the equation whose roots are p+1/q and q+1 ?

acx^2 + b(a+c)x + a^2 + b^2 +2ac=0

is it p + (1/q) or (p+1)/q?

Find the valu of x and y. Root x +y=7 and x+ root y =11 were x=2

X2-10×+9=0

Solve it x2-20x+a2-b2=0