Question

How do you solve by completing the square $2{x^2} + 7x - 1 = 0$?

Answer 1

Hint: We have to find the solution of a given quadratic equation by creating a trinomial square on the left side of the equation and using algebraic identity. First, add $1$ to both sides of the equation. Then, divide each term in $2{x^2} + 7x = 1$ by $2$. Next, create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $b$. Next, add the term to each side of the given equation. Next, simplify the right-hand side of the equation by simplifying each term. Next, combine the numerator over the common denominator and simplify the numerator. Next, factor the perfect trinomial using (i). Next, take the square root of each side of the equation to set up the solution for $x$. Then remove the perfect root factor $x + \dfrac{7}{4}$ under the radical to solve for $x$. Next, simplify the right side of the equation by pulling terms out from under the radical, assuming positive real numbers. Next, subtract $\dfrac{7}{4}$ from both sides of the equation and get the desired result.

Complete step by step solution:
First of all, we have to add $1$ to both sides of the equation, we get
$2{x^2} + 7x = 1$
Divide each term in $2{x^2} + 7x = 1$ by $2$.
${x^2} + \dfrac{7}{2}x = \dfrac{1}{2}$
Now, we have to create a trinomial square on the left side of the equation. For this we have to find a value that is equal to the square of half of $b$.
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{7}{4}} \right)^2}$
Now, we have to add the term to each side of the given equation.
${x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{1}{2} + {\left( {\dfrac{7}{4}} \right)^2}$
Now, we have to simplify the right-hand side of the equation by simplifying each term. For this first apply the product rule to $\dfrac{7}{4}$. Then raise $7$ to the power of $2$. Then raise $4$ to the power of $2$.
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{1}{2} + \dfrac{{{7^2}}}{{{4^2}}}\]
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{1}{2} + \dfrac{{49}}{{{4^2}}}\]
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{1}{2} + \dfrac{{49}}{{16}}\]
Now, we have to write $\dfrac{1}{2}$ as a fraction with a common denominator, i.e., multiply by $\dfrac{8}{8}$. Then combine 7 and $\dfrac{4}{4}$. Then combine the numerator over the common denominator and simplify the numerator.
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{1}{2} \cdot \dfrac{8}{8} + \dfrac{{49}}{{16}}\]
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{{1 \cdot 8}}{{16}} + \dfrac{{49}}{{16}}\]
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{{8 + 49}}{{16}}\]
\[ \Rightarrow {x^2} + \dfrac{7}{2}x + {\left( {\dfrac{7}{4}} \right)^2} = \dfrac{{57}}{{16}}\]
Now, we have to factor the perfect trinomial into ${\left( {x + \dfrac{7}{4}} \right)^2}$ using (i).
${\left( {x + \dfrac{7}{4}} \right)^2} = \dfrac{{57}}{{16}}$
Now, we have to take the square root of each side of the equation to set up the solution for $x$. Then remove the perfect root factor $x + \dfrac{7}{4}$ under the radical to solve for $x$.
$ \Rightarrow {\left( {x + \dfrac{7}{4}} \right)^{2 \cdot \cdot \dfrac{1}{2}}} = \pm \sqrt {\dfrac{{57}}{{16}}} $
$ \Rightarrow x + \dfrac{7}{4} = \pm \sqrt {\dfrac{{57}}{{16}}} $
Now, we have to simplify the right side of the equation by pulling terms out from under the radical, assuming positive real numbers.
$ \Rightarrow x + \dfrac{7}{4} = \pm \dfrac{{\sqrt {57} }}{{\sqrt {16} }}$
$ \Rightarrow x + \dfrac{7}{4} = \pm \dfrac{{\sqrt {57} }}{{\sqrt {{4^2}} }}$
$ \Rightarrow x + \dfrac{7}{4} = \pm \dfrac{{\sqrt {57} }}{4}$
Now, we have to subtract $\dfrac{7}{4}$ from both sides of the equation.
$x = - \dfrac{7}{4} \pm \dfrac{{\sqrt {57} }}{4}$

Hence, we obtain the solution to the quadratic equation $2{x^2} + 7x - 1 = 0$, which are $x = - \dfrac{7}{4} \pm \dfrac{{\sqrt {57} }}{4}$.

Note: We can also find the solution of given quadratic equation by quadratic formula:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(ii)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
Step by step solution:
First, we have to compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $2{x^2} + 7x - 1 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 2$, $b = 7$ and $c = - 1$
Now, we have to substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 7 \right)^2} - 4\left( 2 \right)\left( { - 1} \right)$
After simplifying the result, we get
$ \Rightarrow D = 49 + 8$
$ \Rightarrow D = 57$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - 7 \pm \sqrt {57} }}{{2 \times 2}}$
$ \Rightarrow x = - \dfrac{7}{4} \pm \dfrac{{\sqrt {57} }}{4}$