Chapter 7.3, Problem 36WE

To determine

The value of $\left(k\right)$ for which the equation $3x^2-6kx+12=0$ has a real double root.

Answer to Problem 36WE

For $k=\pm 2$ there exists a real double root condition.

Explanation of Solution

Given information: The given quadratic equation is,

  $3x^2-6kx+12=\mathrm{0....}\left(\text{1}\right)$

Formula used: The discriminant $\left(D=b^2-4ac\right)$ of a quadratic equation $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ should be equal to zero for the real double root condition.

Mathematically,

  $b^2-4ac=\mathrm{0....}\left(2\right)$

Calculation: From (1), evaluate discriminant $\left(D\right)$ and equate that to zero according to formula (2).

  $\begin{array}{c}D={\left(-6k\right)}^2-4\left(3\right)\left(12\right)\mathrm{....}\left\{\begin{array}{l}a=3,b=-6k\\ \text{and }c=12\end{array}\right\}\\ 0=36k^2-\mathrm{144....}\left(\begin{array}{l}\text{From }\left(2\right)\\ D=0\end{array}\right)\\ \text{or, }{k}^2=4\\ k=\pm \sqrt{4}\\ k=\pm 2\end{array}$

To determine

The value of $\left(k\right)$ for which the equation $3x^2-6kx+12=0$ has two different real roots.

Answer to Problem 36WE

For $k<-2\text{ or }k>2$ the equation has two different real roots

Explanation of Solution

Given information: The given quadratic equation is $3x^2-6kx+12=0$ .

Formula used: The discriminant

  $\left(D=b^2-4ac\right)$ of a quadratic equation $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ should be greater than zero for two different real root condition.

Mathematically,

  $b^2-4ac>\mathrm{0....}\left(3\right)$

Calculation: From (1), evaluate discriminant $\left(D\right)$ and apply the formula (3).

  $\begin{array}{c}{\left(-6k\right)}^2-4\left(3\right)\left(12\right)>\mathrm{0....}\left\{\begin{array}{l}a=3,b=-6k\\ \text{and }c=12\end{array}\right\}\\ 36k^2-144>0\\ \text{or, }{k}^2-4>0\\ \left(k-2\right)\left(k+2\right)>0\\ k<-2\text{ or }k>2\end{array}$

To determine

The value of $\left(k\right)$ for which the equation $3x^2-6kx+12=0$ has the imaginary roots.

Answer to Problem 36WE

For $-2<k<2$ the equation has imaginary roots.

Explanation of Solution

Given information: The given quadratic equation is $3x^2-6kx+12=0$ .

Formula used: The discriminant $\left(D=b^2-4ac\right)$ of a quadratic equation $a{x}^2+bx+c=0$ where $\left(a\ne 0\right)$ should be less than zero for imaginary roots.

Mathematically,

  $b^2-4ac<\mathrm{0....}\left(4\right)$

Calculation: From (1), evaluate discriminant $\left(D\right)$ and apply formula (4).

  $\begin{array}{c}{\left(-6k\right)}^2-4\left(3\right)\left(12\right)<\mathrm{0....}\left\{\begin{array}{l}a=3,b=-6k\\ \text{and }c=12\end{array}\right\}\\ 36\left(k^2-4\right)<\mathrm{0....}\left(\text{Simplified}\right)\\ \text{or, }\left(k-2\right)\left(k+2\right)<0\\ -2<k<2\end{array}$