Question

In solving a problem, one student makes a mistake in the coefficient of the first degree term and obtains $-9$ and $-1$ for the roots. Another student makes a mistake in the constant term of the equation and obtains $8$ and $2$ for the roots. The correct equation was?

• A. $x^2+10x+9=0$
• B. $x^2-10x+16=0$
• C. $x^2-10x+9=0$
• D. None of the above

Answer 1

The correct option is C $x^2-10x+9=0$

$x^{2}10x+9=0$

$x^2+9x+x+9=0$

x(x+9)+(x+9)=0

(x+1)(x+9)=0

x= - 1,- 9 $\to$ obtained by $1^{st}$ student Hence this cannot be the equation

$x^2-10x+16=0$

$x^2-8x-2x+16=0$

x(x-8)-2(x-8)=0

x= 2,8 $\to$obtained by $I{I}^{nd}$ student, Hence cannot be the equation

$x^2-10x+9=0$

$x^2-9x-x+9=0$

x(x-9)-(x-9)=0

(x-1)(x-1)=0

x=1,9 can be a Answer

option C is correct.