Question

The product of a number and the one obtained by decreasing it by 7 is -12. Then, the possible values for the number are .

• A. -3 and -4
• B. 4 and 3
• C. -3 and 2
• D. 2 and -4

Answer 1

The correct option is B 4 and 3

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Let the required number be $x.$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Then, the number obtained by decreasing it by 7 is given by$x-7.$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Given, the product of these two numbers is -12.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow x(x-7)=-12$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow x^2-7x=-12$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\Rightarrow x^2-7x+12=0$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> The roots of a quadratic equation $a{x}^2+bx+c=0,$ where $a,b$ and $c$ are constants ($a\ne 0$) are given by
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Since $a=1,b=-7$ and $c=12,$ we have
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $x=\frac{7\pm \sqrt{(-7{)}^2-4\times 1\times 12}}{2\times 1}.$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $⟹x=4,3$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Thus, 3 and 4 are the roots of the given quadratic equation.
Therefore, the possible values of the number are 3 and 4.