MCQ
If one root the equation $2x^2+ kx + 4 = 0$ is $2$, then the other root is :
- A$6$
- B$-6$
- C$-1$
- ✓$1$
✓
Answer
Correct option: D.
$1$
Let $\alpha$ and $\beta$ be the roots of quadratic equation $2x^2+ kx + 4 = 0$ in such a way that $a = 2$
Here $, a = 2, b = k$ and $c = 4$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$2+\beta=\frac{-\text{k}}{2}$
$\beta=\frac{-\text{k}}{2}-2$
$\beta=\frac{-\text{k}-4}{2}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\cdot\beta=\frac{4}{2}$
$\alpha\cdot\beta=2$
Putting the value $\beta=\frac{-\text{k}-4}{2}$ in above
$2\times\frac{(-\text{k}-4)}{2}=2$
$(-\text{k}-4)=2$
$\text{k}=-4-2$
$\text{k}=-6$
Putting the value of $k$ in $\beta=\frac{-\text{k}-4}{2}$
$\beta=\frac{-(-6)-4}{2}$
$\beta=\frac{6-4}{2}$
$\beta=\frac{2}{2}$
$\beta=1$
Therefore, value of other root be $\beta=1$
Thus, the correct answer is $(d)$
Here $, a = 2, b = k$ and $c = 4$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$2+\beta=\frac{-\text{k}}{2}$
$\beta=\frac{-\text{k}}{2}-2$
$\beta=\frac{-\text{k}-4}{2}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\cdot\beta=\frac{4}{2}$
$\alpha\cdot\beta=2$
Putting the value $\beta=\frac{-\text{k}-4}{2}$ in above
$2\times\frac{(-\text{k}-4)}{2}=2$
$(-\text{k}-4)=2$
$\text{k}=-4-2$
$\text{k}=-6$
Putting the value of $k$ in $\beta=\frac{-\text{k}-4}{2}$
$\beta=\frac{-(-6)-4}{2}$
$\beta=\frac{6-4}{2}$
$\beta=\frac{2}{2}$
$\beta=1$
Therefore, value of other root be $\beta=1$
Thus, the correct answer is $(d)$
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