How to find the Delta in second degree equations?
A polynomial having degree 2 is considered a second-degree equation and it is also called a quadratic equation. The standard form of the second-degree equation is ax2+bx+c=0. Delta in the second-degree equation is used to find the type of roots that polynomial has. If the Delta value for a polynomial is greater than zero (Delta>0) then the polynomial has two real distinct roots. If the Delta value for a polynomial is equal to zero (Delta=0) then the polynomial has only one root. If the Delta value is less than zero (Delta<0) then the polynomial has two imaginary roots.
If the quadratic equation is of form ax2+bx+c=0 then the formula for finding the delta is given below-
Delta = b2-4ac
Let’s look at the few sample problems on finding the Delta in second-degree equations
Sample Questions
Question 1: Find the delta for the second-degree equation x2 – 10x + 21 = 0
Solution:
Given
x2-10x+21=0
a=1, b=-10, c=21
Delta=b2-4ac
=(-10)2-4(1)(21)
=100-84
=16>0
Given equation has two distinct real roots.
Question 2: Find the delta for the second-degree equation x2 + 5x – 6 = 0
Solution:
Given
x2+5x-6=0
a=1, b=5, c=-6
Delta=b2-4ac
=(5)2-4(1)(-6)
=25+24
=49>0
Given equation has two distinct real roots.
Question 3: Find the delta for the second-degree equation x2 + 4x + 4 = 0
Solution:
Given
x2+4x+4=0
a=1, b=4, c=4
Delta=b2-4ac
=(4)2-4(1)(4)
=16-16
Delta=0
Given equation has only one root.
Question 4: Find the delta for the second-degree equation x2 + 2x + 1 = 0
Solution:
Given
x2+2x+1=0
a=1, b=2, c=1
Delta=b2-4ac
=(2)2-4(1)(1)
=4-4
Delta=0
Given equation has only one root.
Question 5: Find the delta for the second-degree equation x2 + 4x + 5 = 0
Solution:
Given
x2+4x+5=0
a=1, b=4, c=5
Delta=b2-4ac
=(4)2-4(1)(5)
=16-20
=-4<0
Given equation has two imaginary roots.
Question 6: Find the delta for the second-degree equation x2 – 2x + 2 = 0
Solution:
Given
x2-2x+2=0
a=1, b=-2, c=2
Delta=b2-4ac
=(2)2-4(1)(2)
=4-8
=-4<0
Given equation has two imaginary roots.