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

Jul 3, 2018

The value of $k = 1$

#### Explanation:

The perfect square is

$9 {x}^{2} + 6 x + k = {\left(3 x + \sqrt{k}\right)}^{2}$

$= 9 {x}^{2} + 6 x \sqrt{k} + {k}^{2}$

Comparing the $2$ sides

$6 \sqrt{k} = 6$

$\sqrt{k} = 1$

$k = 1$

Jul 3, 2018

${\left(3 x + a\right)}^{2}$ will give a perfect square

${\left(3 x + a\right)}^{2} = \left(3 x + a\right) \left(3 x + a\right)$

$= 9 {x}^{2} + 3 a x + 3 a x + {a}^{2}$

$= 9 {x}^{2} + 6 a x + {a}^{2}$

So $9 {x}^{2} + 6 a x + {a}^{2} = 9 {x}^{2} + 6 x + k$

$6 a x = 6 x \implies a = 1$

$k = {a}^{2} , {1}^{2} = 1$

$k = 1$

Jul 3, 2018

$k = 1$

#### Explanation:

Lets consider a generic case. Say: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Now lets make a direct comparison between the two:

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
$\textcolor{w h i t e}{\text{dddddddd}} 9 {x}^{2} + 6 x + k$

So ${a}^{2} = 9 {x}^{2} \implies a = \sqrt{9 {x}^{2}} = \pm 3 x$

Also $2 a b = 6 x \text{ so } a b = 3 x$

As $a = \pm 3 x$ then $b = 1$

Lets try this out with:$a = + 3 x$

(3x+1)(3x+1) = 9x^2+6x+1 color(red)(larr" As required"

Thus $k = 1$

Jul 3, 2018

$k = 1$

#### Explanation:

Let , color(violet)(kinRR  be the ${3}^{r d} t e r m$ to complet square.

$i . e . 9 {x}^{2} + 6 x + \textcolor{v i o \le t}{k} \ldots \to \left(1\right)$

In the $L H S$ we have ,

color(blue)(diamond 1^(st)term=9x^2

color(blue)(diamond2^(nd)term=6x

color(blue)(diamond3^(rd)term)=color(violet)(k

$\text{We have "color(orange)"formula}$ for ${3}^{r d} t e r m :$

$\textcolor{red}{{3}^{r d} t e r m = {\left({2}^{n d} t e r m\right)}^{2} / \left(4 \times {1}^{s t} t e r m\right)} \ldots \to \left(A\right)$

$\implies \textcolor{v i o \le t}{k} = {\left(6 x\right)}^{2} / \left(4 \times 9 {x}^{2}\right) = \frac{36 {x}^{2}}{36 {x}^{2}}$

=>color(violet)(k=1

From $\left(1\right)$,we get

$9 {x}^{2} + 6 x + \textcolor{v i o \le t}{1} = {\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(1\right) + {\left(1\right)}^{2} = {\left(3 x + 1\right)}^{2}$

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Note :

Formula $\left(A\right) : \textcolor{red}{{3}^{r d} t e r m = {\left({2}^{n d} t e r m\right)}^{2} / \left(4 \times {1}^{s t} t e r m\right)}$ can be use

to find THIRD TERM for any eqn. without any doubt .

WHY ??? $\to$Please see below.

$\diamond \mathmr{if} , {a}^{2} + 2 a b + k = 0$ ,then [use $\left(A\right)$]

$k = {\left(2 a b\right)}^{2} / \left(4 \times {a}^{2}\right) = \frac{4 {a}^{2} {b}^{2}}{4 {a}^{2}} = {b}^{2}$

$\implies {a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$