**4. If the roots of the equation x**

^{2}+ px + q = 0 differ by 1, prove that p^{2}= 1 + 4q.**Sol.**

**x**

^{2}+ px + q = 0**Comparing with ax**

^{2}+ bx + c = 0 we have a = 1, b = p, c = q**Let**

**α**

**and**

**β**

**be the roots of given quadratic equation.**

**α – β = 1**

**......(i) [Given]**

**α + β =**

^{-b}/_{a}= -p/1 = -p**Also, α . β = c/a = q/1 = q**

**We know that,**

**(α – β )**

^{2}= (α + β )^{2}– 4αβ**∴ (1)**

^{2}= (– p)^{2}– 4 (q)**∴ 1 = p**

^{2}– 4q**∴ 1 + 4q = p**

^{2}**∴ p**

^{2}= 1 + 4q**Hence proved.**