Question

Find the value or sets of values of k so that the equation $ {{x}^{2}}+kx-3k=0 $ has no real roots.

Answer 1

Hint: We have been given a quadratic equation of $ x $ as $ {{x}^{2}}+kx-3k=0 $ . We use the quadratic formula to solve the value of the $ x $ . we have the solution in the form of $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ for general equation of $ a{{x}^{2}}+bx+c=0 $ . We put the values and find the solution for which the discriminant is less than 0.

Complete step by step solution:
We know for a general equation of quadratic $ a{{x}^{2}}+bx+c=0 $ , the value of the roots of $ x $ will be $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . This is the quadratic equation solving method. The root part $ {{b}^{2}}-4ac $ of $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ is called the discriminant of the equation. If the discriminant is less than 0 then the roots are imaginary which means $ {{b}^{2}}-4ac<0 $ .
In the given equation we have $ {{x}^{2}}+kx-3k=0 $ . The values of a, b, c is $ 1,k,-3k $ respectively.
The roots are imaginary which means $ {{b}^{2}}-4ac<0 $ .
We put the values and get $ x $ as
 $ \begin{align}
  & {{k}^{2}}-4\times 1\times \left( -3k \right)<0 \\
 & \Rightarrow {{k}^{2}}+12k<0 \\
 & \Rightarrow k\left( k+12 \right)<0 \\
\end{align} $
The multiplication of two numbers is negative.
This means their signs are opposite to each other.
This gives -12 < k < 0.
The domain is $ k\in \left( -12,0 \right) $ .

Note: We can take the equation as $ {{x}^{2}}-x+3=0 $ where we have $ k=-1 $ .
In the given equation we have $ {{x}^{2}}-x+3=0 $ . The values of a, b, c is $ 1,-1,3 $ respectively.
We put the values and get $ x $ as \[x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 3\times 1}}{2\times 1}=\dfrac{1\pm \sqrt{-11}}{2}=\dfrac{1\pm i\sqrt{11}}{2}\]
The roots of the equation are imaginary numbers.