Question

Under what condition the sum of the roots of x^2 - mx + n = 0 is k times their
difference?​

Answer 1

Step-by-step explanation:

Given:-

A quadratic equation is x^2 - mx + n = 0

To find:-

Under what condition the sum of the roots of

x^2 - mx + n = 0 is k times their difference?

Solution:-

Given quadratic equation is x^2 - mx + n = 0

On comparing with the standard quadratic equation ax^2 + bx + c = 0

We have

a = 1

b= -m

c = n

Let α and β be the roots of the given equation

(α > β)

Sum of the roots = -b/a

=>α + β = -(-m)/1

=>α + β = m

Sum of the roots = m-------------(1)

Product of the roots = c/a

=>α × β = n/1

=>αβ = n

Diffrence of the roots = α - β

We know that

(a-b)^2=(a+b)^2-4ab

=>(α - β)^2 = (α + β)^2 - 4αβ

=>(α - β)^2 = m^2-4n

=>α - β = √(m^2 -n)

Difference between the roots = √(m^2 -n)----(2)

Sum of the roots = k× Difference of the roots

From (1)&(2)

=>m =K ×√(m^2 -n) (or)

K = m/[√(m^2 -n)]

Answer:-

The required condition for the given problem is

m =K ×√(m^2 -n) (or) K = m/[√(m^2 -n)]

Used formulae:-

• The standard quadratic equation is

ax^2 + bx + c = 0

• Sum of the roots = -b/a
• Product of the roots = c/a
• (a-b)^2=(a+b)^2-4ab

Answer 2

Answer:

ans is m²(1-k)²=-4k²n

Step-by-step explanation:

solution:-

given equation

x²-mx+n=0

now,

let the zeroes of the given equation be a and b

therefore,

a+b= -(-m)/1 = m

ab = n/1 = n

now, according to the question

(a+b) = k(a-b)

=>(a+b)² = k²(a-b)². {squaring both sides}

=>(m)² = k²(a²+b²-2ab)

=>m² = k² {(a+b)²-2ab-2ab}.

=>m² = k²(m²-4ab)

=>m² = k²(m²-4n)

=>m² = k²m² -4n²

=>m²-k²m² = -4n²

=>m²(1-k²) = -4n²

therefore the required condition is

m²(1-k²)=-4n²