During a lecture on quadratic equations, the instructor told her students that for a particular equation, the difference between the constant term and the coefficient of x is −3. A student randomly wrote the quadratic equation as

x² − (m − 2) x − m − 1 = 0

The minimum value of the sum of the squares of the roots of the given quadratic equation is

Given: x² − (m − 2) x − m − 1 = 0

From the above equation, we get, a = 1, b = −(m − 2), c = −m − 1

Let the roots of the given equation be α and β.

We know that,  $\alpha +\beta =-\frac{b}{a}=-\frac{-(m-2)}{1}=(m-2)$

and  $\alpha \beta =\frac{c}{a}=\frac{-m-1}{1}=-(m+1)$

Now,

α² + β² = (α + β)² − 2αβ

⇒ α² + β² = (m − 2)² + 2 (m + 1)

⇒ α² + β² = m² − 4m + 4 + 2m + 2

⇒ α² + β² = m² − 2m + 6

⇒ α² + β² = (m − 1)² + 5

Now for  (m − 1)² + 5 to be the minimum,  (m − 1) should be equal to 0,

⇒ m =1

If m = 1, then the sum of the squares of the roots = 5

Hence, α² + β² ≥ 5 and hence the minimum value of  α² + β² = 5