Question

Solve the quadratic equation \[2{x^2} - x - 6 = 0\]

Answer 1

Hint:
According to the question, use splitting the middle term method. In this method you can factorise the given equation using multiplication of coefficients and taking out the common factors and hence find the values of x.

Complete step by step solution:
 Factorising \[2{x^2} - x - 6 = 0\]

As first term is, \[2{x^2}\] its coefficient is 2.
Then the middle term is, \[ - x\] its coefficient is \[ - 1\] .
And the last term is the constant, that is \[ - 6\]
Step1: Multiply the coefficient of the first term by the constant that is last term.
So, we get
  \[2 \times - 6 = - 12\]
Step2: Find the two factors of \[ - 12\] whose sum is equal to the coefficient of the middle term, which is \[ - 1\] .
Step3: Rewrite the polynomial splitting the middle term using the two factors that are observed in step2 above, which are \[ - 4\] and 3
Here, we get \[2{x^2} - 4x + 3x - 6 = 0\]
Step4: Add up the first 2 terms, then take out like factors: \[\;2x\; \times \;\left( {x - 2} \right)\]
 Then add up the last 2 terms, taking out common factors: \[3\; \times \left( {x - 2} \right)\]
Step5: Add up the four terms that are in step 4: \[\left( {2x + 3} \right)\; \times \;\left( {x - 2} \right)\] Which is the resultant factorization.
So, the equation come out to be \[\left( {2x + 3} \right)\; \times \;\left( {x - 2} \right) = 0\]
A product of several terms is equal to zero. When a product of two or more terms are equal to zero, then at least one of the terms must be equal to zero. Here, we will solve each term equals to zero separately.
 Here, we will Solve: \[x - 2 = 0\]
Taking \[ - 2\] on the right hand side we get; \[x = 2\]:
 Now, we will Solve: \[2x + 3 = 0\]
By, taking 3 on the right hand side of the equation:
\[2x = - 3\]
Dividing both sides of the equation by 2:
So,
\[x = \dfrac{{ - 3}}{2}\] or \[x = - 1.5\]

Hence, factors come out to be \[x = 2, - 1.5\]

Note:
To solve these types of questions, you can also use the formula to solve quadratic equation \[a{x^2} + bx + c = 0\] that is \[d = {b^2} - 4ac,x = \dfrac{{ - b \pm \sqrt d }}{{2a}}\] . Here, you can also use the completing the square method also to calculate for x.