$E_1: a+b+c=0, \quad$ if 1 is a root of $a x^2+b x+c=0, \quad E_2: b^2-a^2=2 a c$, if $\sin \theta$, $\cos…
$E_1: a+b+c=0, \quad$ if 1 is a root of $a x^2+b x+c=0, \quad E_2: b^2-a^2=2 a c$, if $\sin \theta$, $\cos \theta$ are the roots of $a x^2+b x+c=0$ Which of the following is true?
$E_1$ is true, $E_2$ is true
$E_1$ is true, $E_2$ is false
$E_1$ is false, $E_2$ is true
$E_1$ is false, $E_2$ is false
Solution
Given that, 1 is a root of $a x^2+b x+c=0$
$\begin{aligned}
& \Rightarrow \quad a+b+c=0 \\
& \therefore \quad E_1: a+b+c=0 \text { is true. }
\end{aligned}$
Since $\cos \theta, \sin \theta$ are the roots of
$\begin{aligned}
& a x^2+b x+c=0 \\
\therefore \quad & \sin \theta+\cos \theta=-\frac{b}{a} \\
\text { and } \sin \theta \cos \theta= & \frac{c}{a}
\end{aligned}$
On squaring both sides of equation (i)
$\begin{array}{rlrl}
& (\sin \theta+\cos \theta)^2 & =\frac{b^2}{a^2} \\
\Rightarrow \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta & =\frac{b^2}{a^2} \\
\Rightarrow & 1+2\left(\frac{c}{a}\right) & =\frac{b^2}{a^2} \\
\Rightarrow & 2 \cdot \frac{c}{a} & =\frac{b^2-a^2}{a^2} \\
\Rightarrow & -a^2+b^2 & =2 a c \\
\therefore & E_2: b^2-a^2 & =2 a c \text { is true }
\end{array}$
$\Rightarrow E_1$ and $E_2$ both are true.