First slide

Theory of equations

Question

$The equations 2 x^{2} + b x + (c + 1) = 0 where b, c \in Q and x^{2} + 3 x - 5 = 0 have a common$ $root, then the value of b - c is$

Moderate

A

9
B

11
C

13
D

17

Solution

$The given equations are 2 x^{2} + b x + (c + 1) = 0 and x^{2} + 3 x - 5 = 0$
$Since, x^{2} + 3 x - 5 = 0 have irrational roots. (\sin ce ∆ is not a perfect square)$
Here one root is common
$\Rightarrow$ the other root also common because coefficients are rational. (The roots of a quadratic equation occurs in conjugate surds if the coefficients are rationals)
$\begin{array}{l} \Rightarrow \frac{2}{1} = \frac{b}{3} = \frac{c + 1}{- 5} \\ \Rightarrow b = 6, c = - 11 \\ \Rightarrow b - c = 6 + 11 = 17 \end{array}$
Therefore, The correct answer is(4).

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