Question

Find the value of $ k $ for which $ x = 3 $ is a solution of the quadratic equation, $ \left( {k + 2} \right){x^2} - kx + 6 = 0 $ . Thus, find the other root of the equation.

Answer 1

Hint: If $ a{x^2} + bx + c = 0 $ is a quadratic equation and $ x = a,b $ are the roots of the equation, then $ x = a $ and $ x = b $ satisfies the quadratic equation. If there is any unknown term in the quadratic equation, we can find the unknown term by substituting $ x = a $ and $ x = b $ in the quadratic equation.

Complete step-by-step answer:
The given quadratic equation is $ \left( {k + 2} \right){x^2} - kx + 6 = 0 $ .
  $ x = 3 $ is a solution of the given quadratic equation.
Now, we can substitute the value of $ x = 3 $ in the equation $ \left( {k + 2} \right){x^2} - kx + 6 = 0 $ to find the value of $ k $ .
  $
\Rightarrow \left( {k + 2} \right){\left( 3 \right)^2} - kx + 6 = 0\\
\Rightarrow 9\left( {k + 2} \right) - k\left( 3 \right) + 6 = 0\\
\Rightarrow 9k + 18 - 3k + 6 = 0
  $
On simplifying this equation further, we get,
  $
\Rightarrow 6k + 24 = 0\\
\Rightarrow6k = - 24
  $
Divide the entire equation by 6.
  $
\Rightarrow k = - \dfrac{{24}}{6}\\
 = - 4
  $
The quadratic equation is $ \left( {k + 2} \right){x^2} - kx + 6 = 0 $ , then substitute $ k = - 4 $ .
  $
\Rightarrow \left( { - 4 + 2} \right){x^2} - \left( { - 4} \right)x + 6 = 0\\
 - 2{x^2} + 4x + 6 = 0
  $
After multiplying the entire equation by -1, we get,
\[\Rightarrow 2{x^2} - 4x - 6 = 0\]
Now, we can divide the entire equation by 2.
\[\Rightarrow {x^2} - 2x - 3 = 0\]
The quadratic equation can be factorised in the form of
  $ \left( {x - a} \right)\left( {x - b} \right) $ when $ {x^2} - \left( {a + b} \right)x + ab = 0 $ is the quadratic equation.
Now, we can compare \[{x^2} - 2x - 3 = 0\] with $ {x^2} - \left( {a + b} \right)x + ab = 0 $ .
We get $ a + b = 2 $ and $ ab = - 3 $ .
  $ a + b = 2 $ , $ a \cdot b = - 3 $ happens only when $ a = 3,{\rm{ }}b = - 1 $ .

Therefore, the other root of the equation \[{x^2} - 2x - 3 = 0\] is $ x = - 1 $ and the value of $ k = - 4 $.

Additional Information
Factorisation of polynomials is done by separating it in the product of factors which cannot be reduced further. In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials. It is one of the ways to find the solution of a quadratic equation.

Note: To find the roots of a quadratic equation, there is an alternative method. If there is a quadratic equation of the form $ a{x^2} + bx + c = 0 $ , then the roots of the equation can be calculated by $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .