10th Class Mathematics Polynomials Question Bank Case Based (MCQs) - Polynomials

If \[p(x)=a{{x}^{2}}+bx+c\] and \[a+b+c=0,\] then one zero is:

A) \[\frac{-b}{a}\]

B) \[\frac{c}{a}\]

C) \[\frac{b}{c}\]

D) \[-\frac{c}{a}\]

Correct Answer: B

Solution :

[b] Take \[x=1.\]

\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,p(1)=a{{(1)}^{2}}+b(1)+c=a+b+c=0\]

\[\therefore \] One zero \[(\alpha )=1\]

Now, product of zeroes \[=\frac{c}{a}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\cdot \beta =\frac{c}{a}\Rightarrow \beta =\frac{c}{a}\]

\[\therefore \] Zeroes are 1 and \[\frac{c}{a}\].

So, option [b] is correct .