Question
Question: Find the product of zeroes of \( {u}^{2}+8u \)....
Find the product of zeroes of u2+8u.
Solution
Hint: We will use the concept of the zeroes of the quadratic equation to find the zeros or roots of the equation. Zeros or roots of a Quadratic equation are the values on which the equation becomes zero.
Complete step-by-step solution -
We know that the general equation of a quadratic polynomial is ax2+bx+c.
To find the zeros of a quadratic equation we equate is equal to zero and find the solution of the resultant equation that is ax2+bx+c=0.
We have to find the zeros of u2+8u. So we will equate this to zero and find the solution of the resultant equation.
u2+8u=0 …... (1)
To solve this equation, we will use the method of factorization. In this method, we express the quadratic expression as a multiplication of two factors.
Now, we will take u common in the (1)
u(u+8) =0 …. (2)
Here we have two factors of the quadratic expression as u and u+8.
Now, we know that the product of two factors can be zero only if either of them is zero.
So, we will equate both the factors to zero. This gives:
u=0 or
u+8=0
u=-8
Therefore, we get u=0, -8 as two zeroes of the given quadratic expression.
Therefore, the product of the zeroes is 0*8=0.
Note: Alternative way
We know that a general equation of a quadratic polynomial is ax2+bx+c. The product of zeroes for the general quadratic equation is defined as c/a. Where c is constant term and a is the coefficient of x2
The expression given to us is u2+8u . This is a quadratic expression in variable u. On comparing it with ax2+bx+c
We get,
a=1
b=8
c=0
Now we know that the product of zeroes is c/a.
Since c=0 we get the product of zeroes as 0.
Hence the answer is 0.