Question

If x²+bx+c=0, x² + cx+b=0(b + c) have a common root, then show that b +c+1=0.( clear explanation )​

Answer 1

For first quadratic equation,

$\sf x^2 + bx + c = 0$

Let $\sf x_1$ and $\sf x_2$ be its roots.

• $\sf x_1 + x_2 = -b\quad\quad\dots(1)$
• $\sf x_1x_2 = c\quad\quad\dots(2)$

For second equation,

$\sf x^2 + cx + b = 0$

Now let $\sf x_1$ be their common root.

So, we can write that,

$\sf (x_1)^2 + bx_1 + c = (x_1)^2 + cx_1 + b = 0$

$\sf \longrightarrow \cancel{(x_1)^2} + bx_1 + c = \cancel{(x_1)^2} + cx_1 + b$

$\sf \longrightarrow bx_1 - cx_1 = b - c$

$\sf \longrightarrow x_1(b - c) = (b - c)$

$\sf \longrightarrow x_1{\cancel{(b - c)}} = \cancel{(b - c)}$

$\sf \longrightarrow x_1 = 1\quad\quad\dots(3)$

From (2) and (3),

$\sf x_1x_2 = c$

$\sf \longrightarrow 1 \times x_2 = c$

$\sf \longrightarrow x_2 = c\quad\quad\dots(4)$

From (1) and (4),

$\sf x_1 + x_2 = -b$

$\sf \longrightarrow 1 + c = -b \quad\quad[From\:(3)\:and\:(4)]$

$\sf \longrightarrow 1 + c + b = 0$

$\longrightarrow \underline{\underline{\sf{\green{b + c + 1 = 0}}}}$

Answer 2

Refer The Attachment ⬆️⬆️

☞ $\fbox\red{1+b+c = 0}$

Hope It Helps You ✌️