Question

How do you find the values of $k$ that will make $9{x^2} + 6x + k$ a perfect square?

Answer 1

Hint: In this question, we have to find the value of $k$ that will make a perfect square.
The given equation is of the form $a{x^2} + bx + c = 0$ .
In case the given equation is of quadratic form, then we can use the formula of discriminant $D = {b^2} - 4ac$ and find the values of $k$ that will make the equation a perfect square. In the case of perfect squares, the value of discriminant should be $0$ .
Finally, we will get the required answer.

Formula used: $D = {b^2} - 4ac$
If $D > 0,$ then there will be two real and distinct roots.
If $D = 0,$ then there will be two real but same roots.
If $D < 0,$ then there will be two complex roots.

Complete step-by-step solution:
A perfect square trinomial will have a discriminant ${b^2} - 4ac = 0$
Let us write the given equation $9{x^2} + 6x + k$ …. $\left( 1 \right)$
The given equation is of the form $a{x^2} + bx + c = 0$
Here, we get $a = 9\,,\,b = 6\,,\,c = k$ from the given equation.
${b^2} - 4ac = 0$
Substitute the value of $a = 9\,,\,b = 6\,,\,c = k$ in the above equation
 We can get the required solution.
${b^2} - 4ac = 0$
$\Rightarrow$${\left( 6 \right)^2} - 4\left( 9 \right)\left( k \right) = 0$
$\Rightarrow$$36 - 36k = 0$
$\Rightarrow$$36 = 36k$
$\Rightarrow$$k = \dfrac{{36}}{{36}}$
$\Rightarrow$$k = 1$
Hence $k = 1$ .

Hence the value of k is equal to 1.

Note: Here, we have to find the same question by using another method as follows.
Let us consider a generic case, say:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ …. $\left( 2 \right)$
Now let us make a direct comparison between the two equation $\left( 1 \right)$ and $\left( 2 \right)$
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ $ = 9{x^2} + 6x + k$
So, ${a^2} = 9{x^2}$
$ \Rightarrow a = \sqrt {9{x^2}} \,$
$ = \pm 3x$
$a = + 3x, - 3x$
Also, $2ab = 6x$
$ab = \dfrac{{6x}}{2}$
So, $ab = 3x$
If we substitute $a = \pm 3x$ in the above equation, we can get the value of $b$
$ \pm 3xb = 3x$
$b = \dfrac{{3x}}{{3x}}$
$b = 1$
As $a = \pm 3x$ , then $b = 1$
Let’s try this out with $a = + 3x$ and $b = 1$ in equation $\left( 2 \right)$
${\left( {3x + 1} \right)^2} = $${\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 1 \right) + {\left( 1 \right)^2}$
$\left( {3x + 1} \right)\left( {3x + 1} \right) = 9{x^2} + 6x + 1$
Thus $k = 1$.
We can also use the formula to find the third term of the given equation.
Let $k \in R$ be the third term to complete the square.
That is $9{x^2} + 6x + k$
In the LHS we have,
First term = $9{x^2}$
Second term =$6x$
Third term = $k$
We have formula for third term
Third term $ = \dfrac{{{{\left( {\sec ond\,term} \right)}^2}}}{{4 \times first\,term}}$
$ \Rightarrow k = \dfrac{{{{\left( {6x} \right)}^2}}}{{4 \times 9{x^2}}}$$ = \dfrac{{36{x^2}}}{{36{x^2}}}$
$ \Rightarrow k = 1$
From $\left( 1 \right)$ we get,
$9{x^2} + 6x + 1 = {\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 1 \right) + {\left( 1 \right)^2}$$ = {\left( {3x + 1} \right)^2}$
We can also use the above third term formula to find the third term for any equation without any doubt.