Chapter ISG, Problem 8.5.8P

To determine

Solve the equation

Answer to Problem 8.5.8P

Value of x is $x=\frac{-7}{2}$ and $3$

Explanation of Solution

Given information:

  $2x^2+qx+7x=21$

Calculation:

The given expression is $2x^2+qx+7x=21$

Rearranging,

  $2x^2+(q+7)x-21=0$

For grouping we know that $21\times 2$ , $7x+qx$ should be the result to take out common from the given equation,

Taking out LCM,

42 can be written as,

  $\begin{array}{l}42\times 1\\ 21\times 2\\ 14\times 3\\ 12\times 4\\ 7\times 6\end{array}$

7 is the first number as coefficient of x, value of q is 6. The sign is of minus as the sign of c is also minus. So, q=-6

Putting in equation,

  $2x^2-6x+7x-21=0$

Solving,

  $2x(x-3)+7(x-3)=0$

Taking common,

  $(2x+7)(x-3)=0$

For zeroes,

  $(2x+7)=0,(x-3)=0$

Solving,

  $2x=-7$ and $x=3$

Hence, value of x is $x=\frac{-7}{2}$ and $3$