Question

For what integer value/s of s, the quadratic equation $(s+2)x^2+1=(s+3)x$ has co-incident roots.

• A. $-1,-2$
• B. $-1,$
• C. $-1,-2$
• D. $2,-3$

Answer 1

The correct option is B $-1,$

$(s+2)x^2+1=(s+3)x$

$\text{Converting into standard form}$

$(s+2)x^2-(s+3)x+1=0$

$a=s+2,b=-(s+3),c=1$

Since the quadratic equation has real and equal roots,

$D=b^2-4ac=0$

$(-(s+3){)}^2-4\times (s+2)\times 1=0$

$s^2+6s+9-4s-8=0$

$s^2+2s+1=0$

$(s+1)(s+1)=0$

$s=-1,-1$

$\text{(b) is correct}$