Answer
Verified
409.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE