Determine the number of solutions for each quadratic equation.

(a) $9x^2-6x+1=0$

(b) $3y^2-8y+1=0$

(c) $7m^2+12m+4=0$

(d)$5n^2-n+1=0$

The required equation is$9x^2-6x+1=0$.

Comparing the given equation with the standard form we have $a=9,b=-6,c=1$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-6\right)}^2-4\times 9\times 1\\ & =& 36-36\\ & =& 0\end{array}$

The discriminant is zero.

So, there is one real solution of the equation.

The given equation is$3y^2-8y+1=0$.

Comparing the given equation with the standard form we have $a=3,b=-8,c=1$.

role="math" localid="1645687180314" $\begin{array}{rcl}{b}^2-4ac& =& {\left(-8\right)}^2-4\times 3\times 1\\ & =& 64-12\\ & =& 52>0\end{array}$

The discriminant is positive.

So, there are two real solutions of the equation.

The given equation is$7m^2+12m+4=0$.

Comparing the given equation with the standard form we have $a=7,b=12,c=4$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(12\right)}^2-4\times 7\times 4\\ & =& 144-112\\ & =& 32>0\end{array}$

The discriminant is positive.

So, there are two real solutions.

The given equation is $5n^2-n+1=0$.

Comparing the given equation with the standard form we have $a=5,b=-1,c=1$.

$\begin{array}{rcl}{b}^2-4ac& =& {\left(-1\right)}^2-4\times 5\times 1\\ & =& 1-20\\ & =& -19<0\end{array}$

The discriminant is negative.

So, there is no real solution.

Hence, two complex solutions exists.