91Ó°ÊÓ

Consider the given information,

$$\begin{gathered} a)r^2+12r+36=0 \\ b)8t^2-11t+5=0 \\ c)4u^2-12u+9=0 \\ d)3v^2-5v-1=0 \end{gathered}$$

The given equation is $a)r^2+12r+36=0$. Using the quadratic, which is, $x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$.

Where a is the coefficient of x², b is the coefficient of x and c is the constant value.

The roots are real and equal. Therefore, there is only one root.

Given equation, $b)8t^2-11t+5=0$.

Using the quadratic formula,

role="math" localid="1654513332007" $x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a},\mathrm{where}\mathrm{a}=\mathrm{co}-\mathrm{efficient}\mathrm{of}{\mathrm{x}}^2,\mathrm{b}=\mathrm{co}-\mathrm{efficient}\mathrm{of}\mathrm{x},\mathrm{c}=\mathrm{consatnt}\mathrm{value}.$

Now discriminant,

Therefore, the number of solutions is zero.

Given the equation is,$c)4u^2-12u+9=0$

Using the quadratic formula,

role="math" localid="1654513496261" $x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a},\mathrm{where}\mathrm{a}=\mathrm{co}-\mathrm{efficient}\mathrm{of}{\mathrm{x}}^2,\mathrm{b}=\mathrm{co}-\mathrm{efficient}\mathrm{of}\mathrm{x},\mathrm{c}=\mathrm{consatnt}\mathrm{value}.$

Now discriminant,

$\begin{array}{rcl}D& =& \sqrt{b^2-4ac}\\ & =& \sqrt{{(-12)}^2-4\left(4\right)\left(9\right)}\\ & =& \sqrt{0}\\ & =& 0\end{array}$

Roots are real and equal.

Therefore, the number of solutions is one.

Given the equation is, $d)3v^2-5v-1=0$

Using the quadratic formula,

$x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a},\mathrm{where}\mathrm{a}=\mathrm{co}-\mathrm{efficient}\mathrm{of}{\mathrm{x}}^2,\mathrm{b}=\mathrm{co}-\mathrm{efficient}\mathrm{of}\mathrm{x},\mathrm{c}=\mathrm{consatnt}\mathrm{value}.$

Now discriminant,

$$\begin{gathered} D=\sqrt{b^2-4ac} \\ =\sqrt{{(-5)}^2-4(-1)\left(3\right)}>0 \end{gathered}$$

The roots are real and different.

Therefore, there is two solutions.