Question

Find the possible value or values of n in the quadratic equation 2n2 – 7n + 6 = 0

Answer 1

Answer:

$n=\frac{3}{2},\,\,n=2$.

Step-by-step explanation:

The equation you have is a quadratic equation because the polynomial $2n^{2}-7n+6$ has degree 2. One of the methods available to solve kind of equations is  to factorize the polynomial  on the left hand side. To factorize you can do the following:

(1) $2n^2-7n+6$. The given polinomial

(2) $\frac{2\times(2n^2-7n+6)}{2}=\frac{(2n)^{2}-7(2n)+12}{2}$.  Multiply and divide by 2, because it is the coeficient of $n^{2}$

(3)  $\frac{(2n)^{2}-7(2n)+12}{2}=\frac{(2n-\_\_)(2n-\_\_)}{2}$. Separate the polynomial in two factors, each one with $2n$ as a first term. The sign in the first factor is equal to the sign in the second term of the polynomial, that is to say, $-7n$. The sign in the second factor is the sign of the second term multiplied by the sign of the third term, that is to say $(-)\times(+)=(-)$ . In the blanks you should select two numbers whose sum is 7 and whose product is 12. Those numbers must be 3 and 4.

(4)The polynomial factorized is $\frac{(2n-4)(2n-3)}{2}$

(5)Use the common factor in the numerator to cancel the number 2 in the denominator to obtain $(n-2)(2n-3)$

Then the given equation can be written as:

${(2n-3)(n-2)=0$

The product of two expression equals zero if and only if one of the expression is zero. From here we have that

$2n-3=0$ or $n-2=0$

From the first equality we obtain that $n=\frac{3}{2}$. From the second equality we obtain that $n=2$.

Answer 2

2n^2-7n+6=0

2n^2-4n-3n+6=0

2n(n-2)-3(n-2)=0

(2n-3)(n-2)=0

n=3/2, n=2