Question

Find the zeroes of the quadratic polynomial \[\left( 5{{u}^{2}}+10u \right)\] and verify the relation between the zeroes and the coefficient.

Answer 1

Hint: Compare the given quadratic polynomial to general quadratic expression and equate it to zero. Find the roots of the quadratic polynomial. Then using sum and product of zeroes verify their relationship.

Complete step-by-step answer: -
We have been given the quadratic polynomial, \[\left( 5{{u}^{2}}+10u \right)\]. Now this equation is similar to the quadratic equation, \[a{{x}^{2}}+bx+c=0\], where x is the variable. In \[\left( 5{{u}^{2}}+10u \right)\], u is the variable.
We can write,
\[f\left( u \right)=5{{u}^{2}}+10u=5u\left( u+2 \right)\]
Thus we got, \[f\left( u \right)=5u\left( u+2 \right)\].
Now let us put, \[f\left( u \right)=0\].
\[\therefore \] We can say that, \[5u\left( u+2 \right)=0\].
i.e. \[5u=0\] and \[u+2=0\].
\[u=0\] and \[u=-2\].
Thus we got the zeroes of u as 0 and -2.
Now let us verify the same. For any quadratic equation,
\[{{x}^{2}}\] + (Sum of zeroes) x + (Product of zeroes)
Sum of zeroes = -(Co- efficient of u) / (co –efficient of \[{{u}^{2}}\]) \[=\dfrac{-10}{5}=-2\]
Product of zeroes = constant / coefficient of \[{{u}^{2}}\] \[=\dfrac{0}{5}=0\]
Thus we got the zeroes again as -2 and 0.

Note: Now if we are splitting, \[5{{u}^{2}}+10u=0\].
\[5\left( {{u}^{2}}+2u \right)=0\Rightarrow {{u}^{2}}+2u=0\]
We can split the middle term as (-2 + 0) u.
i.e. \[{{u}^{2}}+2u=u\left( u+2 \right)\]