Q.Using Sridharacharya's formula, solve $x^2 - 4x + 4 = 0$ and state what this tells us about the number of real roots.

Compute the discriminant first — when it equals zero, the equation has exactly one repeated real root.

Here $a=1, b=-4, c=4$. Compute the discriminant: $D=b^2-4ac=(-4)^2-4(1)(4)=16-16=0$.

Since $D=0$, the equation has exactly one real root (repeated), per the discriminant classification in §6.

Apply Sridharacharya's formula: $x=\dfrac{-b\pm\sqrt D}{2a}=\dfrac{4\pm\sqrt0}{2}=\dfrac{4\pm0}{2}=2$.

Both the $+$ and $-$ branches give the same value $x=2$, confirming a single repeated root. …

Alternative — recognize the expression as a perfect square directly: $x^2-4x+4 = (x-2)^2$, so $(x-2)^2=0 \Rightarrow x=2$ (repeated), confirming the …

Students sometimes report "no solution" when $D=0$, confusing it with the $D<0$ case — $D=0$ means exactly ONE real root (repeated), not zero real roots. Always state the root count pr …