Question

If ${E_1}:a + b + c = 0$, if $1$ is the root of $a{x^2} + bx + c = 0$ , ${E_2}:{b^2} - {a^2} = 2ac$ , if $\sin \theta$ and $\cos \theta$ are the roots of $a{x^2} + bx + c = 0$ .
A. ${E_1}$ is true, ${E_2}$ is true
B. ${E_1}$ is true, ${E_2}$ is false
C. ${E_1}$ is false, ${E_2}$ is true
D. ${E_1}$ is false, ${E_2}$ is false

Answer 1

In the above question we have expressions. We have been given that $1$ is the root of $a{x^2} + bx + c = 0$ .
So we will put, $x = 1$ in the equation and check if this satisfies the given statement or not. Similarly we will use the formula of sum of roots i.e., $\dfrac{{ - b}}{a}$ and we know the formula of product of roots is $\dfrac{c}{a}$ .

Complete step-by-step solution:
Let us take the first statement which is
${E_1}:a + b + c = 0$, We have been given that $1$ is the root of $a{x^2} + bx + c = 0$ .
So let us put $x = 1$ in the equation, so we have:
$a{(1)^2} + b(1) + c = 0$
On simplifying we have
$a + b + c = 0$
So this proves that the given statement is true.
Now let us take the second statement
${E_2}:{b^2} - {a^2} = 2ac$
Here we have two roots i.e. $\sin \theta$ and $\cos \theta$ are the roots of
$a{x^2} + bx + c = 0$ .
Now we know that the sum of the roots is $\dfrac{{ - b}}{a}$ and the product of the roots is $\dfrac{c}{a}$ .
So we can write the sum of the roots as:
$\sin \theta + \cos \theta = \dfrac{{ - b}}{a}$
And the product of the roots can be written as
$\sin \theta \times \cos \theta = \dfrac{c}{a}$
We will square the sum of the roots on both sides of the equation i.e.
${\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\dfrac{{ - b}}{a}} \right)^2}$
On simplifying we have:
${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = \dfrac{{{b^2}}}{{{a^2}}}$
We know the trigonometric identity which is
${\sin ^2}\theta + {\cos ^2}\theta = 1$
So by putting this we can write that
$1 + 2\sin \theta \cos \theta = \dfrac{{{b^2}}}{{{a^2}}}$
Again we have
$\sin \theta \times \cos \theta = \dfrac{c}{a}$
So we can write this as
$1 + 2 \times \dfrac{c}{a} = \dfrac{{{b^2}}}{{{a^2}}}$
By transferring $1$ to the right hand side we have
$\dfrac{{2c}}{a} = \dfrac{{{b^2}}}{{{a^2}}} - 1$
Further we can write this as
$\dfrac{{2c}}{a} = \dfrac{{{b^2} - {a^2}}}{{{a^2}}}$
We can eliminate the same terms from the denominator of the both side of the equation: $2c = \dfrac{{{b^2} - {a^2}}}{a}$
The above expression can also be written as
$2ac = {b^2} - {a^2}$
We can see that this also proves the given statement.
Hence the correct option is (a) ${E_1}$ is true, ${E_2}$ is true .

Note: We should note that in the above question we have expanded the term
${\left( {\sin \theta + \cos \theta } \right)^2}$ with the algebraic formula which is
${(a + b)^2} = {a^2} + {b^2} + 2ab$ .
Here we have
$a = \sin \theta$ and
$b = \cos \theta$ . So we have used this sum of the square formula.