Find the value of p, for which one root of the quadratic equation px^2 -14x + 8 = 0 is 6 times the other.

3 answers

The coefficients of this quadratic equation are:

a = p , b = - 14 , c = 8

Note that the second root x₂ is six times larger than the first root x₁

x₂ = 6 x₁

Apply relation between roots and coefficients of a quadratic equation:

Sum of roots:

x₁ + x₂ = - b / a

Produst of roots:

x₁ ∙ x₂ = c / a

Now first condition:

x₁ + x₂ = - b / a

x₁ + 6 x₁ = - ( - 14 ) / p

7 x₁ = 14 / p

Divide both sides by 7

x₁ = 2 / p

Second condition:

x₁ ∙ x₂ = c / a

x₁ ∙ 6 x₁ = 8 / p

6 x₁² = 8 / p

2 ∙ 3 x₁² = 2 ∙ 4 / p

Divide both sides by 2

3 x₁² = 4 / p

Replace x₁ = 2 / p

in this equation

3 ( 2 / p )² = 4 / p

3 ∙ 4 / p² = 4 / p

Divide both sides by 4

3 / p² = 1 / p

Take the reciprocal of both sides

p² / 3 = p

Divide both sides by p

p / 3 = 1

Multiply both sides by 3

p = 3

So your equation is:

3 x² - 14 x + 8

Try to calculate the roots of the equation:

3 x² - 14 x + 8 = 0

The solutions are:

x₁ = 2 / 3

x₂ = 4

Solution check:

x₂ = 6 x₁

4 = 6 ∙ 2 / 3

4 = 12 / 3

4 = 4
or, for
ax^2 + bx + c = 0, the sum of the roots = -b/a, the product of roots = c/a

let the roots be k and 6k
sum: k+6k = 14/p
k = 14/(7p) = 2/p

product: k(6k) = 8/p
k^2 = 4/3p

4/p^2 = 4/3p , since p ≠ 0
1/p = 1/3
p = 3
using the quadratic formula, the roots are
(14±√(196-32p))/(2p)
That means that
(14+√(196-32p))/(2p) = 6(14-√(196-32p))/(2p)
14+√(196-32p) = 84 - 6√(196-32p)
√(196-32p) = 10
196-32p = 100
32p = 96
p = 3