Quadratic equations are an important topic in mathematics and are used to solve a variety of problems. In particular, finding the roots of a quadratic equation is a common task that arises in many different contexts. The quadratic equation Px^2 – 14x + 8 is a specific example of such an equation. In this article, we will explore how to find the value of P for which one root of this equation is equal to the reciprocal of the other root.
First, let us recall some basic facts about quadratic equations. A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants. The variables x that satisfy the equation are called the roots of the equation. Quadratic equations can have zero, one, or two distinct roots. If the discriminant b^2 – 4ac is negative, then the quadratic equation has no real roots. If the discriminant is zero, then the quadratic equation has one repeated root, and if the discriminant is positive, then the quadratic equation has two distinct roots.
Now, let us consider the quadratic equation Px^2 – 14x + 8. We want to find the value of P for which one root of this equation is equal to the reciprocal of the other root. Let us call the two roots of the equation r1 and r2, where r1 is the reciprocal of r2. That is, r1 = 1/r2.
Using the quadratic formula, we can express the roots of the equation in terms of P as follows:
r1,r2 = (14 ± sqrt(196 – 32P))/2P
Since r1 = 1/r2, we have:
1/r2,r2 = (14 ± sqrt(196 – 32P))/2P
Multiplying both sides by r2, we get:
1, r2^2 = (14r2 ± sqrt(196r2^2 – 32Pr2))/2P
Simplifying the right-hand side, we get:
1, r2^2 = (14r2 ± 4sqrt(49r2^2 – 8Pr2))/2P = 7r2 ± 2sqrt(49r2^2 – 8Pr2)/P
Since r1 = 1/r2, we have r2 = 1/r1. Substituting this into the above equation, we get:
1/r1^2, r1^2 = 7/r1 ± 2sqrt(49/r1^2 – 8P)/P
Multiplying both sides by r1^2, we get:
1,r1^4 = 7r1^3 ± 2r1^2sqrt(49 – 8Pr1^2)/P
Now, we can solve for P by setting the two expressions for r1^4 equal to each other and solving for P. That is,
7r1^3 + 2r1^2sqrt(49 – 8Pr1^2)/P = 1
Multiplying both sides by P and rearranging, we get:
2r1^2sqrt(49 – 8Pr1^2) = P – 7r1^3
Squaring both sides, we get:
4r1^4(49 – 8Pr1^2) = P^2 – 14P r1^3 + 49r1^6
Expanding and simplifying, we get:
4r1^4(49 – 8Pr1^2) = P(P – 14r1^3)(P – r1^3)
This equation is a cubic equation in P, which can be solved using standard techniques. However, to simplify the calculation, we can make use of the fact that r1 = 1/r2. Substituting this into the above equation, we get:
4(1/r2)^4(49 – 8P/r2^2) = P(P – 14/r2^3)(P – 1/r2^3)
Multiplying both sides by r2^6 and simplifying, we get:
4r2^2(49r2^2 – 8Pr2) = P(r2^6 – 14 + P)
This is now a quadratic equation in P, which can be solved using the quadratic formula. Specifically, we have:
P = (4r2^2 ± sqrt((4r2^2)^2 – 4(r2^6 – 14 + P)(-8r2)))/(2(r2^6 – 14 + P))
Simplifying, we get:
P = (2r2^2 ± sqrt(r2^4 + 4r2^2))/
