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The quadratic polynomial p(x) leaves a remainder of 1 on division by x - 1 or x - 2.  The product of the roots of p(x) is 1.  What is the sum of the roots of p(x)?

 May 15, 2020
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The quadratic polynomial p(x) leaves a remainder of 1 on division by x - 1 or x - 2.  
The product of the roots of p(x) is 1.  
What is the sum of the roots of p(x)?

 

The quadratic polynomial is \( p(x)=ax^2+bx+c\)

\(\begin{array}{|lrcll|} \hline & p(x) &=& q_1(x)(x-1)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=1: & ax^2+bx+c &=& q_1(x)(x-1)+1 \\ & a+b+c &=& q_1(x)\cdot 0 +1 \\ &\mathbf{ a+b+c} &=& \mathbf{1} \qquad (1) \\ \hline & p(x) &=& q_2(x)(x-2)+1 \quad | \quad p(x)=ax^2+bx+c \\ x=2: & p(x)=ax^2+bx+c &=& q_2(x)(x-2)+1 \\ & 4a+2b+c &=& q_2(x)\cdot 0+1 \\ & \mathbf{4a+2b+c} &=& 1 \qquad (2) \\ \hline \end{array}\)

 

We have roots \(r_1\) and \(r_2\)

\(\begin{array}{|rcll|} \hline p(x)=ax^2+bx+c &=& 0 \\ ax^2+bx+c &=& 0 \quad | \quad : a \\ x^2+\underbrace{\frac{b}{a}}_{=-(r_1+r_2)}x+\underbrace{\frac{c}{a}}_{=r_1r_2} &=& 0 \\ \hline r_1r_2= 1 &=& \frac{c}{a} \\ 1 &=& \frac{c}{a} \\ \mathbf{c} &=& \mathbf{a} \\ \hline -(r_1+r_2) &=& \frac{b}{a} \\ \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}} \\ \hline \end{array}\)

 

 

\(\begin{array}{|lrcll|} \hline (1): &\mathbf{ a+b+c} &=& \mathbf{1} \quad | \quad c= a \\ & a+b+a &=& 1 \\ & 2a+b &=& 1 \\ & \mathbf{b} &=& \mathbf{1-2a} \qquad (3) \\ \\ (2): & \mathbf{4a+2b+c} &=& \mathbf{1} \quad | \quad c= a \\ & 4a+2b+a &=& 1 \\ & 5a+2b &=& 1 \quad | \quad \mathbf{b=1-2a} \\ & 5a+2(1-2a) &=& 1 \\ & 5a+2-4a &=& 1 \\ & a+2 &=& 1 \\ & a &=& 1-2 \\ & \mathbf{ a } &=& \mathbf{-1} \\\\ (3): & \mathbf{b} &=& \mathbf{1-2a} \quad | \quad \mathbf{ a =-1} \\ & b &=& 1-2(-1) \\ & b &=& 1+2 \\ & \mathbf{ b } &=& \mathbf{3} \\\\ & \mathbf{c} &=& \mathbf{a} \quad | \quad \mathbf{ a =-1} \\ & \mathbf{c} &=& \mathbf{-1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{r_1+r_2} &=& \mathbf{-\frac{b}{a}}\quad | \quad b=3,\ a=-1 \\ r_1+r_2 &=& -\frac{3}{-1} \\ \mathbf{r_1+r_2} &=& \mathbf{3} \\ \hline \end{array}\)

 

The sum of the roots of \(p(x)= -x^2+3x-1\) is 3

 

laugh

 May 15, 2020
edited by heureka  May 15, 2020

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