Let the equation x⁴ − ax² + 9 = 0 have four real and distinct roots. Then the least integral value of a is

The correct option is 7.

To find the least integral value of a for which the equation $x^4-a{x}^2+9=0$ has four real and distinct roots, we can use substitution.

Let $y=x^2$.
Since $x$ must have four real and distinct roots, the variable $y$ must have two distinct, positive real roots (since each positive value of $y$ gives two distinct real values for $x$, namely $x=\pm \sqrt{y}$).

Substituting $y$ into the original equation gives the quadratic equation:
$y^2-ay+9=0$

For this quadratic equation to have two distinct positive real roots, the following conditions must be satisfied:

1. The discriminant must be strictly greater than zero (for distinct roots):
$D=b^2-4ac>0$
${(-a)}^2-4\left(1\right)\left(9\right)>0$
$a^2-36>0$
$(a-6)(a+6)>0$
This gives:
$a>6$ or $a<-6$

2. The sum of the roots must be positive:
If $y_1$ and $y_2$ are the roots, then:
$y_1+y_2=a>0$

3. The product of the roots must be positive:
$y_1{y}_2=9>0$ (which is always true)

Combining the conditions $a>0$ and ($a>6$ or $a<-6$), we get:
$a>6$

Therefore, the value of $a$ must be strictly greater than 6. The least integral value of $a$ satisfying $a>6$ is 7.