Question

show that the quadratic equation 9x^+6x+1=0 has equal roots​

Answer 1

Step-by-step explanation:

$Rewrite 9x^{2}+6x+1 \\ \left(9x^{2}+3x\right)+\left(3x+1\right).$

$3x\left(3x+1\right)+3x+1 \\ \\\left(3x+1\right)\left(3x+1\right) \\ \\ \left(3x+1\right)^{2} \\ To find equation solution, solve 3x+1=0$

$x=-\frac{1}{3}$

Answer 2

Answer:

9x^2+6x+1=0

is a quadratic equation

then,

x=( -b plus or minus root of (b^2-4ac))/2a

Here, a=9

b=6

c=1

x=( -6 +/-√(6^2-4*9*1))/2*9

=( -6+/-√(36-36))/18

=( -6+/-0)/18

= -6/18

= -1/3