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

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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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.

Complete step-by-step answer:
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.
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