First slide

Theory of equations

Question

The equation formed by decreasing each root of $a x^{2} + b x + c = 0$ by $1$ is $2 x^{2} + 8 x + 2 = 0,$ then

Moderate

A

$a = - b$
B

$b = - c$
C

$c = - a$
D

$b = a + c$

Solution

The equation formed by decreasing each root of $a x^{2} + b x + c = 0$ by $1$ , is
$\begin{array}{l} a (x + 1)^{2} + b (x + 1) + c = 0 \\ \Rightarrow a x^{2} + x (2 a + b) + a + b + c = 0 \end{array}$
This is identical to the equation $2 x^{2} + 8 x + 2 = 0 .$
$∴ \frac{a}{2} = \frac{2 a + b}{8} = \frac{a + b + c}{2}$
$\begin{array}{l} \Rightarrow 4 a = 2 a + b, a = a + b + c and 2 a + b = 4 a + 4 b + 4 c \\ \Rightarrow 2 a = b, b + c = 0 and 2 a + 3 b + 4 c = 0 \\ \Rightarrow b = 2 a, b = - c and c = - 2 a \Rightarrow 2 a = b = - c . \end{array}$

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