Let's determine the value of $b$ such that in the equation

$$
x^{2}-b x+12=0
$$

the difference of the squares of the roots is 7.
If the roots are $x_{1}$ and $x_{2}$, then the requirement of the problem can be written as

$$
x_{1}^{2}-x_{2}^{2}=\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)=b\left(x_{1}-x_{2}\right)= \pm 7
$$

Squaring both sides,

$$
b^{2}\left(x_{1}-x_{2}\right)^{2}=b^{2}\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]=b^{2}\left(b^{2}-48\right)=49
$$

[ $_{\text {that is }}$

$$
b^{4}-48 b^{2}-49=0
$$

From this equation, $b_{1}^{2}=49$ and $b_{2}^{2}=1$. The latter does not provide real values for $b$, so it can be ignored. Therefore, only $b^{2}=49$ satisfies, hence $b= \pm 7$.

The equations that meet the requirement are:

$$
x^{2}-7 x+12=0 \quad \text { and } \quad x^{2}+7 x+12=0
$$

The roots of the first equation are: +4 and $+3,\left[4^{2}-3^{2}=16-9=7\right]$.

The roots of the second equation are: -4 and $-3,\left[(-4)^{2}-(-3)^{2}=16-9=7\right]$.[^0]


[^0]:    ${ }^{1} The sign of $x_{1}-x_{2}$ changes if $x_{1}$ and $x_{2}$ are swapped.

    ${ }^{2}$ Since the coefficient of $x^{2}$ is 1, $\left(x_{1}-x_{2}\right)^{2}$ is equal to the discriminant of the equation.