Let $f(x)=ax^2+bx+a$, where $a$ and $b$ are constants and $a\ne 0$. If one of the roots of the equation $f(x)=0$ is $x=4$, what is the other root? Explain your answer.

michaelcai Dec 20, 2017

#1**+2 **

\(f(x)=ax^2+bx+a\)

If one root is located at x=4, then we can find an equation for this.

\(f(x)=ax^2+bx+a\) | Substitute in \(f(x)=0\) and \(x=4\) and simplify completely. |

\(0=a(4)^2+b*4+a\) | |

\(0=16a+4b+a\) | On the right hand side, the 16a and a are like terms, so they will combine. |

\(0=17a+4b\) | Let's solve for b now by subtracting 17a from both sides. |

\(-17a=4b\) | Divide by 4 to completely isolate b. |

\(-\frac{17a}{4}=b\) | |

Since we now know what *b *equals in terms of a, we can simplify the original function to \(f(x)=ax^2-\frac{17a}{4}x+a\). We can now begin to solve this and find the other root.

\(f(x)=ax^2-\frac{17a}{4}x+a\) | First off, factor out the GCF from the right hand side. In this case, the GCF will be a. Let's actually divide by a and get rid of it. The problem states that a is both a nonzero constant, so this is a valid operation. Of course, we only care about the zeros of this particular function, so let's set f(x) equal to 0. | ||

\(0=x^2-\frac{17}{4}x+1\) | It may be to difficult to work with fractions, so it will be beneficial to rid the problem of any and all fractions. Eliminating the fraction can be accomplished by mutliplying both sides by 4. | ||

\(0=4x^2-17x+4\) | Break up the b-term, -17, into parts that can be factored. | ||

\(0=(4x^2-16x)+(-x+4)\) | Complete factoring by grouping. This requires finding the GCF of both groups (indicated by the parentheses). | ||

\(0=4x(x-4)-1(x-4)\) | 4x and -1 are being multiplied by the same quantity, so we can combine this into a factored form. | ||

\(0=(4x-1)(x-4)\) | Set each factor equal to zero and solve. | ||

| |||

| |||

| It should be reassuring that we found that x=4 is a zero of this polynomial function because that information was given in the problem. | ||

Therefore, x=1/4 is the other zero of this polynomial function.

TheXSquaredFactor Dec 20, 2017

#2**+3 **

ax^2 + bx + a

The product of the roots = a/a = 1

So....let r be the other root

So

4r = 1 ⇒ r = 1/4

CPhill Dec 20, 2017

#3**+2 **

This method is significantly more elegant than my long-winded answer.

TheXSquaredFactor
Dec 20, 2017