Question

Find the value of k for which the quadratic equations \[k{x^2} - 14x + 8 = 0\]

Answer 1

Hint: To solve the given quadratic equation, first we should know the factorisation,
Sum of roots and product of roots, square roots of number, perfect squares.
 We should also be aware of cross checking the final answer by substituting to the given question.

Complete step-by-step answer:
Let us consider the given question \[k{x^2} - 14x + 8 = 0\]
 To find the value of k
Consider, \[k{x^2} - 14x + 8 = 0\] \[ \to \](1)
Let \[\alpha \]and \[\beta \]are the roots of the equation
Let \[\beta \]=6\[\alpha \]
We know that, sum of the roots=\[\dfrac{{ - b}}{a}\] and product of the roots =\[\dfrac{c}{a}\]
Here a=k b=-14 and c=8
\[ \Rightarrow \]\[\alpha \]+\[\beta \]=\[\dfrac{{14}}{k}\] \[ \to \](2)
 \[\alpha \].\[\beta \]=\[\dfrac{8}{k}\] \[ \to \](3)
Now consider (2)
\[ \Rightarrow \]\[\alpha \]+6\[\alpha \]=\[\dfrac{{14}}{k}\] [Here substituting value of\[\beta \]]
\[ \Rightarrow \]7\[\alpha \]=\[\dfrac{{14}}{k}\] [adding LHS]
\[ \Rightarrow \]\[\alpha \]=\[\dfrac{{14}}{{7k}}\]=\[\dfrac{2}{k}\]
\[ \Rightarrow \]\[\alpha \]=\[\dfrac{2}{k}\] \[ \to \](4)
Now consider product of the root equation (3) \[\alpha \].\[\beta \]=\[\dfrac{8}{k}\]
\[ \Rightarrow \]\[\alpha \](6\[\alpha \])=\[\dfrac{8}{k}\] [here substituting value of \[\beta \]]
\[ \Rightarrow \] \[6{\alpha ^2}\]=\[\dfrac{8}{k}\]
\[ \Rightarrow \]\[3{\alpha ^2}\]=\[\dfrac{4}{k}\]
\[ \Rightarrow \]3\[{\left( {\dfrac{2}{k}} \right)^2}\]=\[\dfrac{4}{k}\] [here substituting value of \[\alpha \]from equation (4)]
\[ \Rightarrow \]3\[\left( {\dfrac{4}{{{k^2}}}} \right)\]=\[\dfrac{4}{k}\]
\[ \Rightarrow \]k=3 \[ \to \](5)
Now consider equation (1) \[k{x^2} - 14x + 8 = 0\]
Now substitute value of k=3 from equation (5)
We get, \[3{x^2} - 14x + 8 = 0\]
\[ \Rightarrow \]\[3{x^2} - 12x - 2x + 8\] [ By factorisation]
\[ \Rightarrow \]\[(3{x^2} - 12x) - (2x + 8)\] [grouping]
\[ \Rightarrow \]\[3x(x - 4) - 2(x - 4)\] [taking common factors]
\[ \Rightarrow \]\[(3x - 2)(x - 4)\]=0
\[ \Rightarrow \]x-4=0 [here we can also consider (3x-2) to find the value of x]
\[ \Rightarrow \]x=4
Next substitute value of x=4 in \[3{x^2} - 14x + 8 = 0\]
We get LHS=RHS
\[\therefore \]the obtained value k=3 is correct

Note: Definition of factorisation is given as writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind
However, this factorisation is unique to find order of factors.
A perfect square is a number from a given number system that can be expressed as the square of a number from the same number system.