Chapter 4, Problem 27CT

Solution

(a.)

The number of possible solutions for a system of two quadratic equations.

It has been determined that a system of two quadratic equations can have at most two solutions.

Given:

Two quadratic equations; $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2=0$ .

Concept used:

The solution(s) for a system of equations are obtained by solving them simultaneously.

Calculation:

The given quadratic equations are $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2=0$ .

Subtracting the second equation from the first equation,

  $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)=0$

Now, the solutions for this equation are the solutions for the system.

Note that the above equation is again a quadratic equation, which can have at most two solutions.

This implies that a system of two quadratic equations can have at most two solutions.

Conclusion:

It has been determined that a system of two quadratic equations can have at most two solutions.

(b.)

The number of possible solutions for a system of two quadratic inequalities.

It has been determined that a system of two quadratic inequalities can have infinite solutions.

Given:

Two quadratic inequalities; $a_1{x}^2+b_{1}x+c_1<0$ and $a_2{x}^2+b_{2}x+c_2<0$ , where $a_i,b_i,c_i\in ℝ$ .

Concept used:

The solution(s) for a system of inequalities are obtained by solving them simultaneously.

Calculation:

The given quadratic inequalities are $a_1{x}^2+b_{1}x+c_1<0$ and $a_2{x}^2+b_{2}x+c_2<0$ , where $a_i,b_i,c_i\in ℝ$ .

Note that any $>$ inequality can also be transformed into a given form of inequality by changing the signs of the coefficients.

Subtracting the second inequality from the first inequality,

  $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)<0$

Now, the solutions for this inequality are the solutions for the system.

Note that the above equation is again a quadratic inequality.

Let the roots of the equation, $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)=0$ be $p$ and $q$ .

Then, $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)=\left(x-p\right)\left(x-q\right)$ .

Put $\left(a_1-a_2\right)x^2+\left(b_1-b_2\right)x+\left(c_1-c_2\right)=\left(x-p\right)\left(x-q\right)$ in the obtained inequality to get,

  $\left(x-p\right)\left(x-q\right)<0$

Assuming that $p<q$ , the solution for the above inequality is the interval $\left(p,q\right)$ , which gives infinite solutions.

This implies that a system of two quadratic inequalities can have infinite solutions.

Conclusion:

It has been determined that a system of two quadratic inequalities can have infinite solutions.