Question

Evaluate "p", if the difference between the roots of the eq. 2x2-px+15=0 is ½.

Answer 1

Given :  difference between the roots of the eq. 2x²-px+15=0 is 1/2

To Find : p

Solution:

Let say roots are  m and m + 1/2   as difference between the roots  is 1/2

2x²-px+15=0

sum of roots = -(-p)/2 = p/2

=> m + m + 1/2 = p/2

=> 4m + 1 = p

Product of roots = 15/2

m( m + 1/2)  = 15/2

=> m(2m + 1) = 15

=> 2m² + m - 15 = 0

=> 2m² + 6m - 5m- 15 = 0

=> 2m(m + 3) - 5(m + 3) = 0

=> (2m - 5) (m + 3) = 0

=> m = 5/2  , m = - 3

(x - 5/2)(x - 3)  =(2x - 5)(x - 3)

2m + 1 = p

m = 5/2  =>p = 4(5/2) +1  = 11

m = - 3 => p = 4(-3) + 1 = - 11

Value of p are  ±11

Another method :

Roots m and n

m - n = 1/2

m + n = p/2

mn= 15/2

(m + n)² = (m - n)² + 4mn

=> p²/4 = (1/2)² + 4(15/2)

=> p² = 121

=> p = ±11

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Answer 2

Given :- Evaluate "p", if the difference between the roots of the eq. 2x² -px + 15 = 0 is (1/2) .

Answer :-

we know that, the relation between the roots of the equation ax² + bx + c = 0 is,

• sum of roots = (-b/a)
• product of roots = c/a .

so, comparing given equation 2x² - px + 15 = 0 with ax² + bx + c = 0 we get,

• a = 2
• b = - p
• c = 15

then,

→ sum of roots = (-b/a) = -(-p/2) = (p/2)

→ product of roots = c/a = (15/2)

and,

→ difference between roots = (1/2) , given .

if we assume roots as d and e,

→ (d - e)² = (d + e)² - 4de

→ (1/2)² = (p/2)² - 4*(15/2)

→ (1/4) + 30 = (p²/4)

→ (121/4) = p²/4

→ p² = 121

→ p = ± 11 (Ans.)

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