A) \[\pm 9\]
B) \[\pm 12\]
C) \[\pm 15\]
D) \[\pm 18\]
Correct Answer: D
Solution :
| [d] Given that, \[f(x)={{x}^{2}}+px+45\] |
| Then, \[\alpha +\beta =\frac{-p}{1}=-p\] and \[\alpha \beta =\frac{45}{1}=45\] |
| According to given condition, |
| \[{{(\alpha -\beta )}^{2}}=144\] |
| \[{{(\alpha +\beta )}^{2}}-4\alpha \beta =144\] |
| \[{{(-p)}^{2}}-4(45)=144\] |
| \[{{p}^{2}}=144+180\] |
| \[{{p}^{2}}=324\Rightarrow p=\pm 18\] |
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