Solving 2nd Degree Quadratic Equations
Second Degree Equations are quadratic equations where the highest power in an equation is 2 and there will be two solutions for the 2nd 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
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
=(-3+5)/(2×1)
=2/2
x=1
=(-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.
=√(9+40)
=√49
=7>0
So two possible real solutions
=(-(-3)+7)/(2×1)
=10/2
x=5
=(-(-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+b2 then 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
By these above ways, we can solve any second-degree equation.
Contact Us