A) has no linear term and the constant term is negative
B) has no linear term and the constant term is positive
C) can have a linear term but the constant term is negative
D) can have a linear term but the constant term is positive
Correct Answer: A
Solution :
| Let \[p\left( x \right)={{x}^{2}}+ax+b\] |
| Now, product of zeroes\[=\frac{Constant\text{ }term}{Coefficient\,of\,{{x}^{2}}}\] |
| Let \[\alpha \] and \[\beta \] be the zeroes of\[p\left( x \right)\]. |
| \[\therefore \] Product of zeroes \[\left( \alpha \,.\,\beta \right)=\frac{b}{1}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \beta =b\] ...(i) |
| Given that, one of the zeroes of a quadratic polynomial \[p\left( x \right)\]is negative of the other. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \beta <0\] |
| So, \[b<0\] [from Eq. (i)] |
| Hence, b should be negative |
| Put \[a=0\], then, \[p\left( x \right)={{x}^{2}}+b=0\] |
| \[\Rightarrow \,\,\,{{x}^{2}}=-b\] |
| \[\Rightarrow \,\,\,\,\,x=\pm \sqrt{-b}\] \[\left[ \because \,\,\,b<0 \right]\] |
| Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e. a = 0 and the constant term is negative i.e. b < 0. |
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