Question

The interval of $a$ for which the expression ${x^2} - ax + 1 - 2{a^2}$ is always positive is
A) $\left( { - 1,1} \right)$
B) $\left( { - 2,2} \right)$
C) $\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$
D) $\left( { - \dfrac{2}{3},\dfrac{2}{3}} \right)$

Answer 1

Here, we will use the formula of discriminant that is calculated ${b^2} - 4ac$ of the standard form of quadratic equation is $a{x^2} + bx + c$. Then we will compare the given expression with the standard form of the quadratic equation to find the value of $a$, $b$ and $c$. Then we will substitute the above value of $a$, $b$ and $c$ in the formula of the discriminant. Since we know that when $y$ is always positive, the discriminant $D$ is less than 0 and then simplifies to find the required interval.

Complete step by step solution:
We are given that $y = {x^2} - ax + 1 - 2{a^2}$ is always positive.
So, we have $y > 0$.
We know that the discriminant is calculated using the formula ${b^2} - 4ac$ of the standard form of quadratic equation is $a{x^2} + bx + c$.
Comparing the given expression with the standard form of a quadratic equation to find the value of $a$, $b$ and $c$, we get
$a = 1$
$b = - a$
$c = 1 - 2{a^2}$
Substituting the above value of $a$, $b$ and $c$ in the formula of discriminant, we get

$$\Rightarrow {\left( { - a} \right)^2} - 4 \cdot 1 \cdot \left( {1 - 2{a^2}} \right) \\\ \Rightarrow {a^2} - 4 + 8{a^2} \\\ \Rightarrow 9{a^2} - 4 \\\$$

Since we know that when $y$ is always positive, the discriminant $D$ is less than 0.

$$\Rightarrow 9{a^2} - 4 < 0 \\\ \Rightarrow {\left( {3a} \right)^2} - {2^2} < 0 \\\$$

Using the rule, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the above equation, we get
$\Rightarrow \left( {3a + 2} \right)\left( {3a - 2} \right) < 0$
Taking $3a + 2 < 0$ and then subtracting the equation by 2 on both sides, we get

$$\Rightarrow 3a + 2 - 2 < 0 - 2 \\\ \Rightarrow 3a < \- 2 \\\$$

Dividing the above equation by 3 on both sides, we get
$\Rightarrow a < \- \dfrac{2}{3}{\text{ ......eq.(1)}}$
Taking $3a - 2 < 0$ and then adding the equation by 2 on both sides, we get

$$\Rightarrow 3a - 2 + 2 < 0 + 2 \\\ \Rightarrow 3a < 2 \\\$$

Dividing the above equation by 3 on both sides, we get
$\Rightarrow a < \dfrac{2}{3}{\text{ ......eq.(2)}}$
Using equation (1) and equation (2) to form a interval, we get
$\left( { - \dfrac{2}{3},\dfrac{2}{3}} \right)$

Hence, option D is correct.

Note:
A quadratic equation represents a parabola graphically. when we talk about roots of a quadratic equation then it’s nothing but the intersection of the parabola with the x-axis. A quadratic equation with a positive coefficient of ${x}^{2}$ will be negative in between in the roots and will be positive outside.