Chapter 1.8, Problem 58E

Solution

$\left(a\right)$

Evaluate $\left(c\right)$ with the help of discriminant for which the equation $-x^2+16x+c=0$ has two real solutions.

The value of $c>-64$ for which the equation has two real and distinct solutions.

Given:

The quadratic equation is $-x^2+16x+c=0$ .

Concept used:

If the quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has discriminant $\left(D\right)=b^2-4ac$ greater than zero, then it has two different real solutions.

Calculation:

The quadratic equation has two real and distinct solutions if discriminant $D>0$ of the quadratic equation $-x^2+16x+c=0$ .

  $\begin{array}{c}D>0\text{ gives,}\\ {\left(16\right)}^2-4\left(-1\right)\left(c\right)>0\mathrm{......}\left(-x^2+16x+c=0\right)\\ 256+4c>0\\ \text{or, }4c>-256\\ c>-64\end{array}$

Hence, the value of $c>-64$ for which the equation has two real and distinct solutions.

$\left(b\right)$

Evaluate $\left(c\right)$ with the help of discriminant for which the equation $-x^2+16x+c=0$ has one real solution.

The value of $c=-64$ for which the equation has one real solution.

Given:

The quadratic equation is $-x^2+16x+c=0$ .

Concept used:

If the quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has discriminant $\left(D\right)=b^2-4ac$ equal to zero, then the quadratic equation has one real solution.

Calculation:

The quadratic equation has one real solution if discriminant $D=0$ of the quadratic equation $-x^2+16x+c=0$ .

  $\begin{array}{c}D=0\text{ gives,}\\ {\left(16\right)}^2-4\left(-1\right)\left(c\right)=0\mathrm{......}\left(-x^2+16x+c=0\right)\\ 256+4c=0\\ 4c=-256\\ c=-64\end{array}$

Hence, the value of $c=-64$ for which the equation has one real solution.

$\left(c\right)$

Evaluate $\left(c\right)$ with the help of discriminant for which the equation $-x^2+16x+c=0$ has two imaginary solutions.

The value of $c<-64$ for which the equation has two imaginary solutions.

Given:

The quadratic equation is $-x^2+16x+c=0$ .

Concept used:

If the quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has discriminant $\left(D\right)=b^2-4ac$ less than zero, then it has two different imaginary solutions.

Calculation:

The quadratic equation has two imaginary solutions if discriminant $D<0$ of the quadratic equation $-x^2+16x+c=0$ .

  $\begin{array}{c}D<0\text{ gives,}\\ {\left(16\right)}^2-4\left(-1\right)\left(c\right)<0\mathrm{......}\left(-x^2+16x+c=0\right)\\ 256+4c<0\\ \text{or, }4c<-256\\ c<-64\end{array}$

Hence, the value of $c<-64$ for which the equation has two imaginary solutions.