Question

$E_1:a+b+c=0$, if $1$ is root of $a{x}^2+bx+c=0$,

$E_2:b^2-a^2=2ac$, if $\sin \left(\theta \right)$ and $\cos \left(\theta \right)$ are the roots of $a{x}^2+bx+c=0$, which of the following is correct ?

• A. $E_1$is correct, $E_2$ is correct.
• B. $E_1$is correct, $E_2$ is incorrect.
• C. $E_1$is incorrect, $E_2$ is correct.
• D. $E_1$is incorrect, $E_2$ is incorrect.

Answer 1

The correct option is A

$E_1$is correct, $E_2$ is correct.

Explanation for the correct option:

Step-1: Statement $E_1$ prove:

$E_1:a+b+c=0$

Given,

$a{x}^2+bx+c=0$

Now, if we put the value $x=1$, then:

$a{\left(1\right)}^2+b\left(1\right)+c=0$

$\Rightarrow$ $a+b+c=0$

Therefore, $E_1$is correct.

Step-2: Statement ${\mathbf{E}}_{\mathbf{2}}\mathbf{}\mathbf{prove}$:

$E_2:b^2-a^2=2ac$

Again, $\sin \left(\theta \right)$ and $\cos \left(\theta \right)$ are the roots of $a{x}^2+bx+c=0$

Therefore,

$$\begin{gathered} \cos \left(\theta \right)+\sin \left(\theta \right)=-\frac{b}{a} \\ \cos \left(\theta \right)\sin \left(\theta \right)=\frac{c}{a} \end{gathered}$$

Now,

$\begin{array}{rcl}{(-\frac{b}{a})}^2& =& {(\cos \left(\theta \right)+\sin \left(\theta \right))}^2\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{b^2}{a^2}& =& {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)+2\cos \left(\theta \right)\sin \left(\theta \right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{b^2}{a^2}& =& 1+2\cos \left(\theta \right)\sin \left(\theta \right)\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{b^2-a^2}{a^2}& =& \frac{2c}{a}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{b}^2-a^2& =& 2ac\end{array}$

$\therefore$$E_2$ is correct.

Hence, Option(A) is the correct answer.