Find the value of p, for which one root of the quadratic equation px^2 -14x + 8 = 0 is 6 times the other.
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
To find the value of p, we need to use the fact that one root of the quadratic equation is 6 times the other.
Let's assume that the roots of the quadratic equation are x and 6x.
We know that the sum of the roots of a quadratic equation can be found by using the formula: sum of roots = -b/a, where a and b are the coefficients of the quadratic equation.
In this case, the sum of roots would be:
x + 6x = 7x
We also know that the sum of the roots is equal to -b/a. So we can equate it with -(-14)/p:
7x = 14/p
Now, let's find the product of the roots of the quadratic equation. The product of the roots can be found by using the formula: product of roots = c/a, where c is the constant term in the quadratic equation.
In this case, the product of roots would be:
x * 6x = 6x^2
We also know that the product of the roots is equal to c/a. So we can equate it with 8/p:
6x^2 = 8/p
Now, we have two equations:
1) 7x = 14/p
2) 6x^2 = 8/p
To solve this system of equations, we can substitute the value of 7x from the first equation into the second equation:
6(14/p)^2 = 8/p
Now, we can simplify and solve for p:
(6 * 14^2) / p^2 = 8/p
Using basic algebraic steps, we can cross-multiply and simplify further:
6 * 14^2 * p = 8 * p^2
Now, divide both sides of the equation by p:
6 * 14^2 = 8 * p
Using the commutative property of multiplication, we can rearrange the equation:
p = (6 * 14^2) / 8
Now, we can evaluate the expression:
p = (6 * 196) / 8
Simplifying further:
p = 1176 / 8
Finally, performing the division:
p = 147
Therefore, the value of p is 147.
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