Question : Prove that x2+6x+15 has no zeroes.
Doubt by Bhavya
Solution :
Method 1
x2+6x+15
on comparing the above equation with standard form of quadratic polynomial i.e. ax²+bx+c
a=1
a=1
b=6
c=15
Discriminant (D) = b²-4ac
If D>0, there are two different zeroes.
If D=0, there are two equal zeroes.
If D=0, there are two equal zeroes.
If D<0, no zeroes exist.
D=b²-4ac
D=(6)²-4(1)(15)
D=36-60
D=36-60
D=-24
D<0
D<0
Hence, no zeroes exist.
But the above method exist in class X only, this is why it is not a recommended method.
Method 2
Let
y=x2+6x+15
Now, lets plot a graph of x2+6x+15
When
x=0 then y=15 i.e. (0,15)
When
x=0 then y=15 i.e. (0,15)
x=-3 then y=6 i.e. (-3,6)
x=-6 then y=15 i.e. (-6,15)
The Graph will look like this
The number of zeroes of the given polynomial in variable x are equal to the number of times it intersect the x-axis.
Clearly this graph is not intersecting the x-axis at all. Hence, it will have no zeroes.
Since, this method is also from class X, this is why it is not a recommended method.
Method 3
x2+6x+15
=[x2+2(x)(3)+(3)²-(3)²]+15
=(x+3)²-(3)²+15 [∵a²+2ab+b²=(a+b)²]
= (x+3)²-9+15
= (x+3)²+6
=[x2+2(x)(3)+(3)²-(3)²]+15
=(x+3)²-(3)²+15 [∵a²+2ab+b²=(a+b)²]
= (x+3)²-9+15
= (x+3)²+6
Clearly, there is no such value of x for which (x+3)² will be negative so (x+3)²+6 can't be equal to zero. Hence, we can conclude that that x2+6x+15 has no possible zeroes.