Question

If the roots of equation x²+px+q=0 differ by 1, prove that p²=1+4q

Answer 1

Answer:

x2 + px + q = 0

Comparing with ax2 + 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 = p2 – 4q

∴ 1 + 4q = p2

∴ p2 = 1 + 4q

Hence proved.

Answer 2

Answer:

p²=1+4q

Step-by-step explanation:

let the zeroes be a, a+1

we know that

$\alpha + \beta = \frac{ - b}{a}$

$\alpha \beta = \frac{c}{a}$

(a)+(a+1)= -p

2a+1= -p

p= -(2a+1) ____________eq 1

a(a+1)=q

a²+a=q

in eq 1 both side squaring

p²=-(2a+1)²

p²= 4a²+4a+1

p²=4(a²+a)+1 (a²+a=q)

p²=4q+1

p²=1+4q