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Solve $2{x^2} + x - 6$ by completing the square method.

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: In the above question you have to solve the equation by completing the square method. In a square method, you have to make a perfect square by modifying the equation. You have to modify the equation in such a manner that its value remains the make and it can become a square. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given question we have solved $2{x^2} + x - 6$ by completing the square method.
 $=2{x^2} + x - 6$
Applying the completing square method,
On dividing the equation with 6 we get,
 $\Rightarrow {x^2} + \dfrac{x}{2} - 3 = 0$
On adding and subtracting $\dfrac{1}{{16}}$ in the equation we get,
 $\Rightarrow {x^2} + \dfrac{x}{2} + \dfrac{1}{{16}} - \dfrac{1}{{16}} - 3 = 0$
If we look closely then we have ${x^2} + 2.x.\dfrac{1}{4} + \dfrac{1}{{16}} = 0$ which is equal to ${(x + \dfrac{1}{4})^2}$ , so
 $\Rightarrow {(x + \dfrac{1}{4})^2} - \dfrac{{49}}{{16}} = 0$
49 is square of 7 and 16 is square of 4, so we have
 $\Rightarrow {(x + \dfrac{1}{4})^2} - {(\dfrac{7}{4})^2} = 0$
On applying the formula of ${a^2} - {b^2}$ on the above equation we get,
 $\Rightarrow (x + \dfrac{1}{4} - \dfrac{7}{4})(x + \dfrac{1}{4} + \dfrac{7}{4}) = 0$
On solving the above expression we get,
 $\Rightarrow (x - \dfrac{3}{2})(x + 2) = 0$
On equating $x - \dfrac{3}{2}$ and $x + 2$ with 0 we get,
 $\Rightarrow x = \dfrac{3}{2}, - 2$

On solving $2{x^2} + x - 6$ with completing the square method we get $x = \dfrac{3}{2}$ and $x = - 2$.

Note:
In the above solution we tried to separate ${x^2}$ from its coefficient, afterwards we made it a complete square. And finally by applying the formula of ${a^2} - {b^2}$ that is $(a + b)(a - b)$ we get two values of x. Also, we used the square formula of ${(a + b)^2}$ that is ${a^2} + 2ab + {b^2}$ to complete the square formula.
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