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Hint:- Try to make complete square from given equation by using the transformation of the quadratic equation from \[{x^2} + {\text{ }}2ax{\text{ }} + {\text{ }}{a^2} = {\text{ }}0\;\] to \[{\left( {x{\text{ }} + {\text{ }}a} \right)^2}\].

As we know that any quadratic equation written in the form,

\[{x^2} + {\text{ }}2ax{\text{ }} + {\text{ }}{a^2} = {\text{ }}0\;\] can also be written as \[{\left( {x{\text{ }} + {\text{ }}a} \right)^2}\]. (statement 1)

\[{x^2} + {\text{ }}2ax{\text{ }} + {\text{ }}{a^2} = {\text{ }}0\;\] (Eq 1)

So, we will write the given equation in the form of equation 1.

So, comparing equation 1 and the given equation. We get,

\[2a{\text{ = 6}}\]

So, \[a{\text{ = 3}}\].

As we can see that there is no term of \[{a^2}\] (i.e. \[{3^2}\]) in the given equation.

So, adding \[{3^2}\] to both sides of the given equation. We will get,

\[\;{x^{2\;}} + {\text{ }}6x{\text{ }} + {\text{ }}{3^{2\;}}-{\text{ }}7{\text{ }} = {\text{ }}{3^2}\]

Now, using statement 1 above equation can also be written as,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^{2\;}}-{\text{ }}7{\text{ }} = {\text{ }}9\]

Solving above equation. We get,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^2} = {\text{ }}16{\text{ }} = {\text{ }}{4^2}\]

Now, subtracting \[{4^2}\] both sides of the above equation. We will get,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^2}-{\text{ }}{4^2} = {\text{ 0}}\] (Eq 2)

As we know that according that,

\[{a^2}-{\text{ }}{b^2} = {\text{ }}\left( {a{\text{ }} + {\text{ }}b} \right)\left( {a{\text{ }} - {\text{ }}b} \right)\] (Eq 3)

So, manipulating equation 2 using formula at equation 3. We will get,

\[

\left( {x{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}4} \right)\left( {x{\text{ }} + {\text{ }}3{\text{ }}-{\text{ }}4} \right){\text{ }} = {\text{ }}0 \\

\left( {x{\text{ }} + {\text{ }}7} \right)\left( {x{\text{ }}-{\text{ }}1} \right){\text{ }} = {\text{ }}0 \\

\]

Using the above equation. We will get,

\[x{\text{ }} + {\text{ }}7{\text{ }} = {\text{ }}0\;\] or \[x{\text{ }}-{\text{ }}1{\text{ }} = {\text{ }}0\]

\[x{\text{ }} = {\text{ }}1,{\text{ }} - 7\]

Hence, the solution of quadratic equation \[{x^{2\;}} + {\text{ }}6x{\text{ }}-{\text{ }}7{\text{ }} = {\text{ }}0\;\] will be \[x{\text{ }} = {\text{ }}1,{\text{ }} - 7\].

Note:- Whenever we came up with this type of problem then to solve the given quadratic equation efficiently. First, we will manipulate the given equation such that it becomes a perfect square. And then we can get the value of x easily from the perfect square. But if the method to solve the equation is not specified in question then we can also use formula of roots of quadratic equation that is, if \[a{x^2} + {\text{ }}bx{\text{ }} + {\text{ }}c\] will be any quadratic equation then its roots will be \[x{\text{ }} = {\text{ }}\dfrac{{ - b{\text{ }} \pm {\text{ }}\sqrt {{b^2}{\text{ }} - {\text{ }}4ac} }}{{2a}}\].

As we know that any quadratic equation written in the form,

\[{x^2} + {\text{ }}2ax{\text{ }} + {\text{ }}{a^2} = {\text{ }}0\;\] can also be written as \[{\left( {x{\text{ }} + {\text{ }}a} \right)^2}\]. (statement 1)

\[{x^2} + {\text{ }}2ax{\text{ }} + {\text{ }}{a^2} = {\text{ }}0\;\] (Eq 1)

So, we will write the given equation in the form of equation 1.

So, comparing equation 1 and the given equation. We get,

\[2a{\text{ = 6}}\]

So, \[a{\text{ = 3}}\].

As we can see that there is no term of \[{a^2}\] (i.e. \[{3^2}\]) in the given equation.

So, adding \[{3^2}\] to both sides of the given equation. We will get,

\[\;{x^{2\;}} + {\text{ }}6x{\text{ }} + {\text{ }}{3^{2\;}}-{\text{ }}7{\text{ }} = {\text{ }}{3^2}\]

Now, using statement 1 above equation can also be written as,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^{2\;}}-{\text{ }}7{\text{ }} = {\text{ }}9\]

Solving above equation. We get,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^2} = {\text{ }}16{\text{ }} = {\text{ }}{4^2}\]

Now, subtracting \[{4^2}\] both sides of the above equation. We will get,

\[{\left( {x{\text{ }} + {\text{ }}3} \right)^2}-{\text{ }}{4^2} = {\text{ 0}}\] (Eq 2)

As we know that according that,

\[{a^2}-{\text{ }}{b^2} = {\text{ }}\left( {a{\text{ }} + {\text{ }}b} \right)\left( {a{\text{ }} - {\text{ }}b} \right)\] (Eq 3)

So, manipulating equation 2 using formula at equation 3. We will get,

\[

\left( {x{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}4} \right)\left( {x{\text{ }} + {\text{ }}3{\text{ }}-{\text{ }}4} \right){\text{ }} = {\text{ }}0 \\

\left( {x{\text{ }} + {\text{ }}7} \right)\left( {x{\text{ }}-{\text{ }}1} \right){\text{ }} = {\text{ }}0 \\

\]

Using the above equation. We will get,

\[x{\text{ }} + {\text{ }}7{\text{ }} = {\text{ }}0\;\] or \[x{\text{ }}-{\text{ }}1{\text{ }} = {\text{ }}0\]

\[x{\text{ }} = {\text{ }}1,{\text{ }} - 7\]

Hence, the solution of quadratic equation \[{x^{2\;}} + {\text{ }}6x{\text{ }}-{\text{ }}7{\text{ }} = {\text{ }}0\;\] will be \[x{\text{ }} = {\text{ }}1,{\text{ }} - 7\].

Note:- Whenever we came up with this type of problem then to solve the given quadratic equation efficiently. First, we will manipulate the given equation such that it becomes a perfect square. And then we can get the value of x easily from the perfect square. But if the method to solve the equation is not specified in question then we can also use formula of roots of quadratic equation that is, if \[a{x^2} + {\text{ }}bx{\text{ }} + {\text{ }}c\] will be any quadratic equation then its roots will be \[x{\text{ }} = {\text{ }}\dfrac{{ - b{\text{ }} \pm {\text{ }}\sqrt {{b^2}{\text{ }} - {\text{ }}4ac} }}{{2a}}\].

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