in the roots of the quadratic equation Ax^2+Bx+C=0 are h and k which of the following are the roots of the equation Cx^2+Bx+A=0

Question

in the roots of the quadratic equation Ax^2+Bx+C=0 are h and k which of the following are the roots of the equation Cx^2+Bx+A=0

Answer 1

Remember that the roots of a quadratic equation can be found using the Quadratic Formula. Here, finding the roots of the first quadratic equation (Ax² + Bx + C) involves plugging in a = A, b = B, and c = C:

 

x = (-B ± √(B² - 4AC)) / 2A

 

We're told we'll get two answers from solving that equation: h and k. We can say that h is the root that comes from using the "plus" in the "plus-or-minus" sign and k is the root that comes from using "minus." In other words:

 

h = (-B + √(B² - 4AC)) / 2A

k = (-B - √(B² - 4AC)) / 2A

 

For each root, multiply both sides of the equation by A (you'll see why later) to get:

 

Ah = (-B + √(B² - 4AC)) / 2

Ak = (-B - √(B² - 4AC)) / 2

 

 

Now, let's look at the second equation (Cx² + Bx + A). Finding its roots will involve plugging a = C, b = B, and c = A into the Quadratic Formula:

 

x = (-B ± √(B² - 4CA)) / 2C

x = (-B ± √(B² - 4AC)) / 2C          (by the Commutative Property of Multiplication, CA = AC)

 

This equation will also have two roots. Let's call them m and n. We can say that m comes from the "plus" and n comes from the "minus." So:

 

m = (-B + √(B² - 4AC)) / 2C

n = (-B - √(B² - 4AC)) / 2C

 

For each root, multiply both sides by C to get:

 

Cm = (-B + √(B² - 4AC)) / 2

Cn = (-B - √(B² - 4AC)) / 2

 

 

Let's review what we know. The roots of Ax² + Bx + C are h and k while the roots of Cx² + Bx + A are m and n. The equations for them are:

 

Ah = (-B + √(B² - 4AC)) / 2
Ak = (-B - √(B² - 4AC)) / 2

 

Cm = (-B + √(B² - 4AC)) / 2
Cn = (-B - √(B² - 4AC)) / 2

 

Notice that Ah and Cm equal the same thing, and so do Ak and Cn! So, we can say:

 

Ah = Cm

Ak = Cn

 

Solving for m and n will give us expressions for the roots of Cx² + Bx + A in terms of h and k:

 

Ah = Cm

(Ah)/C = m

 

Ak = Cn

(Ak)/C = n

 

So, the roots of Cx² + Bx + A are (Ah)/C and (Ak)/C.