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# How do you solve it? $2{x^2} - 5 = 0$

Last updated date: 25th Jul 2024
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Hint: The general form of a quadratic equation is $a{x^2} + bx + c = 0$, where $a \ne 0$ . The values of $x$ which satisfy the equation are called the roots or the solutions. These roots can be real or imaginary. There are two ways to find solutions/roots of $x$ – comparing method and discriminant method.

Under comparing methods, we keep terms both on the left-hand and right-hand side and keep on comparing them until and unless the left-hand side is equal to the right-hand side. Let us try to solve the given question by comparing methods.
First, we have to add $5$ to both the sides of the equation.
$\Rightarrow 2{x^2} - 5 + 5 = 0 + 5 \\ \Rightarrow 2{x^2} = 5 \;$
Next, we are supposed to divide both the sides of the equation by $2$ .
$\Rightarrow \dfrac{{2{x^2}}}{2} = \dfrac{5}{2} \\ \Rightarrow {x^2} = \dfrac{5}{2} \;$
Now, taking the square root of both sides of the equation in order to solve for $x$ .
$\Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{5}{2}} \\ \Rightarrow x = \pm \dfrac{{\sqrt 5 }}{{\sqrt 2 }} \;$
We can further simplify this solution by rationalizing the denominator. In order to do that, we have to multiply and divide the solution with $\sqrt 2$ .
$\Rightarrow x = \pm \dfrac{{\sqrt 5 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \\ \Rightarrow x = \pm \dfrac{{\sqrt {10} }}{2} \;$
This gives us the two required solutions and that are $x = \dfrac{{\sqrt {10} }}{2}$ and $x = - \dfrac{{\sqrt {10} }}{2}$ .
So, the correct answer is “ $x = \dfrac{{\sqrt {10} }}{2}$ and $x = - \dfrac{{\sqrt {10} }}{2}$ ”.

Note: Let us try to solve this question by discriminant method. As we know that the quadratic formula is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Here, $a = 2,b = 0,c = - 5$ .
Now by putting values of $a,b$ and $c$ in the quadratic formula we get,
$\Rightarrow x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \times 2 \times \left( { - 5} \right)} }}{{2 \times 2}}$
After simplifying, we get,
$\Rightarrow x = \dfrac{{0 \pm \sqrt {0 + 40} }}{4} \\ \Rightarrow x = \dfrac{{ \pm \sqrt {40} }}{4} \\ \Rightarrow x = \dfrac{{ \pm 2\sqrt {10} }}{4} \\ \Rightarrow x = \dfrac{{ \pm \sqrt {10} }}{2} \;$
Hence, the two values of $x$ are $x = \dfrac{{\sqrt {10} }}{2}$ and $x = - \dfrac{{\sqrt {10} }}{2}$ .
There are other ways to solve quadratic equations such as factorization or completing the square. These methods/ways couldn’t be used in the given question because the value of $x$ was missing. In order to solve questions by these methods, the question should be in standard form i.e., $a{x^2} + bx + c = 0$ where $a \ne 0$ and also $b \ne 0$ . Every quadratic equation has two solutions.