Question

How do you solve \[{x^2} + 5x - 7 = 0\]?

Answer 1

Hint: First compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$……(i)
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.

Complete 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 \[{x^2} + 5x - 7 = 0\] with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = 5$ and $c = - 7$
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.
$D = {\left( 5 \right)^2} - 4\left( 1 \right)\left( { - 7} \right)$
After simplifying the result, we get
$ \Rightarrow D = 25 + 28$
$ \Rightarrow D = 53$
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{{ - 5 \pm \sqrt {53} }}{{2 \times 1}}$
$ \Rightarrow x = - \dfrac{5}{2} \pm \dfrac{{\sqrt {53} }}{2}$

Hence, with the help of formula (i) we obtain the solution to the quadratic equation \[{x^2} + 5x - 7 = 0\], which are $x = - \dfrac{5}{2} \pm \dfrac{{\sqrt {53} }}{2}$.

Note:
We can also find the solution of given quadratic equation by creating a trinomial square on the left side of the equation and using below algebraic identity:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$……(ii)
Where, $a$and $b$are any two numbers.
Step by step solution:
First of all, we have to add $7$to both sides of the equation, we get
${x^2} + 5x = 7$
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 $b$.
${\left( {\dfrac{b}{2}} \right)^2} = {\left( {\dfrac{5}{2}} \right)^2}$
Now, we have to add the term to each side of the given equation.
${x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = 7 + {\left( {\dfrac{5}{2}} \right)^2}$
Now, we have to simplify the right-hand side of the equation by simplifying each term. For this first apply the product rule to $\dfrac{5}{2}$. Then raise $5$ to the power of $2$. Then raise $2$ to the power of $2$.
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = 7 + \dfrac{{{5^2}}}{{{2^2}}}\]
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = 7 + \dfrac{{25}}{{{2^2}}}\]
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = 7 + \dfrac{{25}}{4}\]
Now, we have to write 7 as a fraction with a common denominator, i.e., multiply by $\dfrac{4}{4}$. Then combine 7 and $\dfrac{4}{4}$. Then combine the numerator over the common denominator and simplify the numerator.
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = 7 \cdot \dfrac{4}{4} + \dfrac{{25}}{4}\]
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{7 \cdot 4}}{4} + \dfrac{{25}}{4}\]
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{28 + 25}}{4}\]
\[ \Rightarrow {x^2} + 5x + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{53}}{4}\]
Now, we have to factor the perfect trinomial into ${\left( {x + \dfrac{5}{2}} \right)^2}$ using (i).
${\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{53}}{4}$
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 $x + \dfrac{5}{2}$ under the radical to solve for $x$.
$ \Rightarrow {\left( {x + \dfrac{5}{2}} \right)^{2 \cdot \cdot \dfrac{1}{2}}} = \pm \sqrt {\dfrac{{53}}{4}} $
$ \Rightarrow x + \dfrac{5}{2} = \pm \sqrt {\dfrac{{53}}{4}} $
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{5}{2} = \pm \dfrac{{\sqrt {53} }}{{\sqrt 4 }}$
$ \Rightarrow x + \dfrac{5}{2} = \pm \dfrac{{\sqrt {53} }}{{\sqrt {{2^2}} }}$
$ \Rightarrow x + \dfrac{5}{2} = \pm \dfrac{{\sqrt {53} }}{2}$
Now, we have to subtract $\dfrac{5}{2}$ from both sides of the equation.
$x = - \dfrac{5}{2} \pm \dfrac{{\sqrt {53} }}{2}$
Final solution: Hence, with the help of formula (ii) we obtain the solution to the quadratic equation \[{x^2} + 5x - 7 = 0\], which are $x = - \dfrac{5}{2} \pm \dfrac{{\sqrt {53} }}{2}$.