Solving 2nd Degree Quadratic Equations

Last Updated : 29 Feb, 2024

Second Degree Equations are quadratic equations where the highest power in an equation is 2 and there will be two solutions for the 2^nd Degree Equations. The standard form of a second-degree equation is ax²+bx+c which is a trinomial because the equation consists of three terms. But every second-degree equation need not be a trinomial because it may even consist of two terms where the highest power in that is two. Example:- x²+2x-1, 2x²-4, 3x²+x+3

To solve second-degree equations we can use quadratic formula for the equation ax²+bx+c=0

[Tex]x=\frac{-b±\sqrt{b^2-4ac}}{2a} [/Tex]

Where

b²-4ac is discriminant

if discriminant is positive it indicates there are two real solutions

if zero then only one solution

if negative we get complex solutions

Sample Questions

Question 1: Solve the equation x²+3x-4=0?

Solution:

Given equation:

x²+3x-4=0

Compare the given equation with ax²+bx+c=0 and note a, b, c values

a=1, b=3, c=-4

To solve second degree equation, Quadratic formula is used and before that find the discriminant value to find how many solutions are possible for the equation.

√(b²-4ac)=√(3²-(4×1×(-4)))

=√(9-(-16))

=√(9+16)

=√25

=5>0

So two possible real solutions

[Tex]x=\frac{-b+\sqrt{b^2-4ac}}{2a} [/Tex]

=(-3+5)/(2×1)

=2/2

x=1

[Tex]x=\frac{-b-\sqrt{b^2-4ac}}{2a} [/Tex]

=(-3-5)/(2×1)

=-8/2

x=-4

On solving the equation the possible solutions are x=1,-4

Question 2: Solve the equation x²-3x-10=0?

Solution:

Given equation:

x²-3x-10=0

Compare the given equation with ax²+bx+c=0 and note a, b, c values

a=1, b=-3, c=-10

To solve second degree equation, Quadratic formula is used and before that find the discriminant value to find how many solutions are possible for the equation.

[Tex]\sqrt{(b^2-4ac)}=\sqrt{-3^2-4×1×-10} [/Tex]

=√(9+40)

=√49

=7>0

So two possible real solutions

[Tex]x=\frac{-b+\sqrt{b^2-4ac}}{2a} [/Tex]

=(-(-3)+7)/(2×1)

=10/2

x=5

[Tex]x=\frac{-b-\sqrt{b^2-4ac}}{2a} [/Tex]

=(-(-3)-7)/(2×1)

=(3-7)/2

=-4/2

x=-2

On solving the equation the possible solutions are x=5,-2

Question 3: Solve the second-degree equation 2x²-6=0

Solution:

Given 2x²-6=0

2x²=6

x²=6/2

x²=3

x=±√3

Second-degree equations can also be solved by following the factoring formula of trinomial. As the second-degree equation may have three terms.

Trinomial are of two types. They are

Perfect Square Trinomial
Non perfect Square Trinomial

A trinomial is a Perfect Square Trinomial if it is in the form of a²+2ab+b² or a²-2ab+b²then these can be written into-

a²+2ab+b²=(a+b)²

a²-2ab+b²=(a-b)²

A trinomial is a Non Perfect Square Trinomial if it is not perfect square trinomial and in the form of ax²+bx+c. Below are the steps that need to be followed to find factors.

Steps to Solve

Step 1: Find a, b, c and calculate a × c

Step 2: Find two numbers whose product is ac and the sum is equal to b.

Step 3: Split the middle term as the sum of two numbers that are founded in the above step.

Step 4: Solve the equation.

Sample Questions

Question 1: Solve the equation x²+6x+9=0

Solution:

Given equation

x²+6x+9=0

This can be written into- x²+2(3)(x)+3²=0

Above equation is in the form of a²+2ab+b²

So a=x, b=3

From the formula- a²+2ab+b²=(a+b)²

(x+3)²=0

(x+3)(x+3)=0

So, x=-3,-3

Here we got only one solution.

This can be verified by calculating discriminant which is discussed above.

√(b²-4ac)=√(6²-4(1)(9))

=√(36-36)

=0 indicates there will be only one solution for the equation.

So x=-3 is the solution for the equation x²+6x+9=0

Question 2: Solve the equation x²-10x+21=0?

Solution:

Given x²-10x+21=0

It cannot be written into a²+2ab+b² or a²-2ab+b². So it is non perfect square trinomial.

Compare given equation with ax²+bx+c=0

Where a=1, b=-10, c=21

a × c = 21

Find two number such that product is equal to 21 and sum is equal to -10

Let it be -7,-3

Split the middle term in the given equation into sum of two terms using the above 2 numbers.

x²-7x-3x+21=0

x(x-7)-3(x-7)=0

(x-3)(x-7)=0

x=3,7