+0  
 
-1
70
1
avatar+193 

The quadratic $x^2+\frac{3}{2}x-1$ has the following unexpected property: the roots, which are $\frac{1}{2}$ and $-2$, are one less than the final two coefficients. Now find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one more than these coefficients.

 Jan 30, 2021
 #1
avatar
0

Let's see if we can find this elusive quadratic with the strange property that the roots are one more than the coefficient of the linear term and one more than the constant term.

 

A general quadratic of this type must be of the equation \(f(x)=x^2+bx+c\), where b is the coefficient of the linear term and c is the constant term. Let's also take note that the problem states that \(b\neq0\text{ and } c\neq0\).

 

A quadratic of this type would also be factorable as follows: \(f(x)=(x-z_1)(x-z_2)\) where z_1 is the first zero and z_2 is the second zero of the quadratic. We also know a special property of these zeroes. That special property, as stated in the problem, is that \(z_1=b+1\text{ and }z_2=c+1\). Putting all this information in an algebraic format is the key to solving this question.

 

\(f(x)=(x-z_1)(x-z_2)\\ f(x)=(x-(b+1))(x-(c+1))\\ f(x)=(x-b-1)(x-c-1)\\ f(x)=x^2-cx-x-bx+bc+b-x+c+1\\ f(x)=x^2+{\color{red}(-b-c-2)}x+{\color{blue}(bc+b+c+1)}\\ f(x)=x^2+{\color{red}b}x+{\color{blue}c}\)

 

Admittedly, this form looks quite ugly, but this will allow us to create a system of equations and solve for b and c.

 

\(-b-c-2=b\\ \fbox{1}\;-2b-c=2\\ bc+b+c+1=c\\ \fbox{2}\;bc+b=-1 \)

 

Now, let's solve this system of equations! I will start by solving for c in equation 1.

 

\(\fbox{1}\;-2b-c=2\\ -c=2b+2\\ c=-2b-2 \)

 

After solving for c, I will substitute this expression for c in the equation 2.

 

\(\fbox{2}\;bc+b=-1\\ b(-2b-2)+b=-1\\ -2b^2-2b+b+1=0\)

 

While it is certainly possible to combine like terms here, I noticed that I could factor by grouping right now.

 

\(-2b^2-2b+b+1=0\\ -2b(b+1)+1(b+1)=0\\ (-2b+1)(b+1)=0\\ -2b+1=0\text{ or }b+1=0\\ b_1=\frac{1}{2}\text{ or }b_2=-1\)

 

Wow! There are two possible answers for the coefficient of the linear term. Let's find the corresponding constant terms. I will substitute into equation 1.

 

\( \fbox{1}\;c=-2b-2\\ c_1=-2*\frac{1}{2}-2\text{ or }c_2=-2*-1-2\\ c_1=-3\text{ or } {c_2=0}\)

 

Recall that we restricted b and c so that they were nonzero, so we should eliminate the solution b_2 and c_2 because c_2 is 0.

 

Therefore, this is the quadratic that satisfies the inital conditions: \(f(x)=x^2+\frac{1}{2}x-3\)

 Jan 30, 2021

57 Online Users

avatar
avatar
avatar
avatar
avatar