Question

How do you solve using the completing the square method $2{x^2} + 8x - 10 = 0$ ?

Answer 1

In this problem, we have to find the value of $x$ by solving the given quadratic equation in completing the square method. For this, we can first convert the given quadratic equation into a complete square equation by algebraic whole square formula. We can add the missing term on both the sides to get a complete square equation on the left-hand side, then we will get a whole square form, for which we can take square root on both the sides to get the value of $x$.

Complete step by step solution:
(i)
We know that the given quadratic equation to be solved is,
$2{x^2} + 8x - 10 = 0$
Now, we can add $10$ on both sides in the above equation and we will get:
$2{x^2} + 8x - 10 + 10 = 10 \\\ 2{x^2} + 8x = 10 \\\$
Now we can divide both the sides by $2$ to get a perfect square equation:
$\dfrac{{2{x^2}}}{2} + \dfrac{{8x}}{2} = \dfrac{{10}}{2} \\\ {x^2} + 4x = 5 \\\$
(ii)
We can now take the LHS ${x^2} + 4x$
We know that,
${\left( {x + a} \right)^2} = {x^2} + 2ax + {a^2}$
We can see that,
$2a = 4$
i.e.,
$a = \dfrac{4}{2} \\\ a = 2 \\\$
And, therefore
${a^2} = {2^2} = 4$
(iii)
We can add the above value of ${a^2}$ in both the LHS and RHS of the equation we obtained in the first step to get a perfect square equation. Therefore, we will get:
${x^2} + 4x + 4 = 5 + 4 \\\ {x^2} + 4x + 4 = 9 \\\$
Since, we know that LHS is the expansion of ${\left( {x + 2} \right)^2}$, we can write our equation as:
${\left( {x + 2} \right)^2} = 9$
(iv)
Now we can take square root on both sides, we get
$\sqrt {{{\left( {x + 2} \right)}^2}} = \sqrt 9$
On simplifying, it becomes
$x + 2 = \pm 3$
Now, to get two separate roots of the given quadratic equation, we will segregate the above equation
$x + 2 = 3$ and $x + 2 = - 3$
Solving both of them, we will get:
$x = 3 - 2 \\\ x = 1 \\\$
And,
$x = - 3 - 2 \\\ x = - 5 \\\$
Hence, for $2{x^2} + 8x - 10 = 0$ the value of $x$ is $1$ and $- 5$.

Note: We can now verify to check whether the values we got are correct or not. We can substitute the value of $x$ in the equation to check.
We can now take the given quadratic equation and substitute the value of $x$ as $1$ and $- 5$, we get:
When $x = 1$,
$2{\left( 1 \right)^2} + 8\left( 1 \right) - 10 = 0 \\\ 2 + 8 - 10 = 0 \\\ 10 - 10 = 0 \\\$
When $x = - 5$,
$2{\left( { - 5} \right)^2} + 8\left( { - 5} \right) - 10 = 0 \\\ 50 - 40 - 10 = 0 \\\ 50 - 50 = 0 \\\$
Therefore, we successfully verified our answer.