Chapter 1.8, Problem 61E

Solution

$\left(a\right)$

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

The value of $c<\frac{1}{4}$ for which the equation has two real and distinct solutions.

Given:

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

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 $x^2-x+c=0$ .

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

Hence, the value of $c<\frac{1}{4}$ 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 $x^2-x+c=0$ has one real solution.

The value of $c=\frac{1}{4}$ for which the equation has one real solution.

Given:

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

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 $x^2-x+c=0$ .

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

Hence, the value of $c=\frac{1}{4}$ for which the equation has one real solution.

$\left(c\right)$

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

The value of $c>\frac{1}{4}$ for which the equation has two imaginary solutions.

Given:

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

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 $x^2-x+c=0$ .

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

Hence, the value of $c>\frac{1}{4}$ for which the equation has two imaginary solutions.