if one root of the equation 2x^2 + bx+16=0 is square of other value of b is.

Question

in this question there are two variable but i want to find b.

Answer 1

Let's call one root a, so then the other is a² .

 

So (x - a)(x - a²) = 0   FOIL it:

 

x² - (a² + a)x + a³ = 0  Multiply by 2:

 

2x² - 2(a² + a)x + 2a³ = 0 = 2x² + bx + 16

 

So 2a³ = 16, so a = 2.  Therefore, a² = 4

From 2 lines above, we see b = -2(a² + a) = -2(2² + 2)

 

So b = -12

 

Answer 2

One root is the square of the other, so let's call them r and r².

 

The sum of the roots of Ax² + Bx + C = 0 is -B/A and the product of the roots is C/A.

 

So, for 2x² + bx + 16 = 0, we have:  -b/2 = r + r²

                                                                 16/2 = rr²

 

From the second equation,  r³ = 8.  Therefore  r = 2.

 

Substituting for r in the first equation, we get -b/2 = 2 + 2²

 

                                                                                -b/2 = 6

 

                                                                                     -b = 12

 

                                                                                      b = -12

Answer 3

Let's call the first root a.  So the second root is a².  Let's see if we can come up with a quadratic equation that has these two roots.

 

When you know the roots of a quadratic equation (call them a and b), you know that x-a and x-b are both factors of the quadratic polynomial. So you can say (x - a) (x - b) = 0.

 

But the polynomial can also be multiplied by some constant without changing the roots.

 

In this case, we have roots a and a²

 

That means that the equation can be written as k (x - a) (x - a²) = 0, where k is some constant.

 

That's the same as:

 

k (x² - ax - a²x + a³) = 0

k (x² - (a + a²)x + a³) = 0

kx² - k(a + a²)x + ka³ = 0

 

Let's match this up with the original equation:

 

2x² + bx + 16 = 0

 

Since the coefficient of x² in the original equation is 2, and the coefficient of x² in the equation we derived is k, we can say that it will probably work if we say that k = 2.

 

So now our derived equation is:

 

2x² - 2(a + a²)x + 2a³ = 0

 

Now the constant term in the original equation was 16 and the constant term in our derived equation is 2a³.  So lets set 2a³ equal to 16 and solve for a:

 

 

2a³ =16

a³ = 8

a = 2

 

So now we know that a is 2 and the other root, a², is 4.

 

Finally, in our derived equation, the coefficient of x is -2(a + a²), and the coefficient of x in the original equation was b.  So substitute 2 for a in 2(a + a²), and we will get the value of b:

 

b = -2(a + a²) = -2 (2 + 4) = -12

 

So b is -12, and the original equation was 2x² - 12x + 16 = 0