Question

Given the function f(x)= 14x^2 + 5x - 1 answer the question if you could change the 14 to any other number, but the rest has to stay the same, what value would it have to be so that there is only 1 real root for the function? Show or explain.

Answer 1

The answer is -6.25

There is only 1 real root for the function f(x)= -6.25x^2 + 5x - 1

 

Explanation

The function, f(x)= 14x^2 + 5x - 1 is a quadratic polynomial where,

a = 14

b = 5

c = -1

 

A quadratic polynomial can only have 1 real root if b^2  = 4ac

Therefore, if there will be only 1 real root for the function f(x)= 14x^2 + 5x – 1, and only a can be changed, then,

b^2  = 4ac

5^2  = 4(a)(-1)

25 = -4a

-4a = 25

Divide both sides by -4

-4a/-4 = 25/-4

a = -6.25

 

Therefore, there is only 1 real root for the function f(x)= -6.25x^2 + 5x - 1

So, given f(x)= 14x^2 + 5x – 1; if I could change the 14 to any other number, but the rest has to stay the same, such that there is only 1 real root for the function; the number would be -0.65.

 

SHOWING that f(x)= -6.25x^2 + 5x – 1 has only 1 real root;

f(x)= -6.25x^2 + 5x – 1 = 0

a = -6.25,

b = 5

c = -1

 

Using the quadratic formula, x = [−b ± √(b^2−4ac)]/2a

x = [-5 ± √(5^2−4(-6.25)(-1))]/2(-6.25)

x = [-5 ± √(25−25)]/-12.5

x = [-5 ± √(0)]/-12.5

x = (-5 ± 0)/-12.5

x = -5/-12.5

x = 0.4 (only 1 real root) 

Answer 2

This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 ($14 x^{2}$) by changing the 14 to 0. That way we'll left with $f(x)=5x-1$ which only has one real root:
$0=5x-1$
$5x=1$
$x= \frac{1}{5}$