Question

$ y{x^2} - 12x + k $ is a perfect square find the numerical nature of $ k $

Answer 1

Hint: In this question we use algebraic identity to find the nature of $ k $ ,selecting the value of y is very important so that the equations becomes a perfect square and accordingly k can be obtained.

Complete step-by-step answer:
Considering equation to be: $ y{x^2} - 12x + k, $
We know, the algebraic identities
 $ {(a + b)^2} = {a^2} - 2ab + {b^2} $
Put, the value of $ y $ in a such a way the given quadratic equation is a perfect square,
Put $ y = 4 $
We get,
 $ 4{x^2} - 12x + k $
Compare the given quadratic equation with algebraic identities.
 $ 4{x^2} - 12x + k = {a^2} - 2ab + {b^2} $
 $ \therefore 4{x^2} = {a^2} $
 $ a = 2x $
 $ {b^2} = k $
 $ - 2ab = - 12x $
 $ \Rightarrow ab = 6x $
Put the value of $ a = 2x $
 $ 2x\times b=6x$
$b=3$
We know that,
 $ {b^2} = k $
 $ k = {3^2} $
 $ \therefore k = 9 $
The numerical value of $ k $ is $ 9 $

Note: In this solution we have used the comparison of two equations to find the nature of $ k $ . We have also used algebraic identities, $ {(a + b)^2} = {a^2} - 2ab + {b^2} $ in the above solution.