A) \[\frac{49}{4}\]
B) 12
C) 3
D) 4
Correct Answer: A
Solution :
Since 4 is one of the roots of equation\[{{x}^{2}}+px+12=0\]. So it must satisfy the equation. \[\therefore \] \[16+4p+12=0\] \[\Rightarrow \] \[4p=-28\] \[\Rightarrow \] \[p=-7\] The other equation is.\[{{x}^{2}}-7x+q=0\]whose roots are equal. Let roots are a and a of above equation. \[\therefore \]Sum of roots\[=\alpha +\alpha =\frac{7}{1}\] \[\Rightarrow \] \[2\alpha =7\] \[\Rightarrow \] \[\alpha =\frac{7}{2}\] and product of roots\[=\alpha .\alpha =p\] \[\Rightarrow \] \[{{\alpha }^{2}}=q\] \[\Rightarrow \] \[{{\left( \frac{7}{2} \right)}^{2}}=q\] \[\Rightarrow \] \[q=\frac{49}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
