Question

Which of the following are roots of the quadratic equation $6x^2-x-2=0$?

• A. $\frac{1}{3}$ and $-\frac{2}{3}$
• B. $4$ and $-3$
• C. $\frac{2}{3}$ and $-\frac{1}{2}$
• D. $\frac{1}{2}$ and $-\frac{2}{3}$

Hint:

For quadratic equations of the form: \[ ax^2+bx+c=0 \] while factorization:
• Multiply \(a\) and \(c\)
• Find two numbers whose product is \(ac\)
• Their sum should equal \(b\) This makes middle-term splitting very easy.

Answer 1

The Correct Option is C

Concept: A quadratic equation is an equation of degree 2 and has the general form: \[ ax^2+bx+c=0 \] The values of \(x\) satisfying the equation are called its roots. We can solve quadratic equations by:
• Factorization
• Completing square
• Quadratic formula Here we use the factorization method.

Step 1: Write the given equation.
\[ 6x^2-x-2=0 \] We need to factorize this expression.

Step 2: Multiply the coefficient of \(x^2\) and the constant term.
Coefficient of \(x^2 = 6\) Constant term \(= -2\) Their product: \[ 6 \times (-2)=-12 \] Now we need two numbers whose:
• Product is \(-12\)
• Sum is \(-1\) The required numbers are: \[ -4 \quad \text{and} \quad 3 \] because \[ (-4)(3)=-12 \] and \[ -4+3=-1 \]

Step 3: Split the middle term.
Replace \(-x\) by \(-4x+3x\): \[ 6x^2-4x+3x-2=0 \]

Step 4: Group the terms.
Group terms pairwise: \[ (6x^2-4x)+(3x-2)=0 \] Take common factors from each group. From the first group: \[ 2x(3x-2) \] From the second group: \[ 1(3x-2) \] Thus, \[ 2x(3x-2)+1(3x-2)=0 \]

Step 5: Factor out the common binomial.
\[ (3x-2)(2x+1)=0 \]

Step 6: Find the roots.
For a product to be zero, at least one factor must be zero. So, \[ 3x-2=0 \] or \[ 2x+1=0 \] Solving the first equation: \[ 3x=2 \] \[ x=\frac{2}{3} \] Solving the second equation: \[ 2x=-1 \] \[ x=-\frac{1}{2} \] Therefore, the roots are: \[ \boxed{\frac{2}{3} \text{ and } -\frac{1}{2}} \]