Use the Discriminant to Predict the Number of Solutions of a Quadratic

$$\begin{gathered} a)9v^2-15v+25=0 \\ b\hspace{0.17em})100w^2+60w+9=0 \\ c)5c^2+7c-10=0 \\ d)15d^2-4d+8=0 \end{gathered}$$

$$\begin{gathered} a)9v^2-15v+25=0 \\ b\hspace{0.17em})100w^2+60w+9=0 \\ c)5c^2+7c-10=0 \\ d)15d^2-4d+8=0 \end{gathered}$$

Given equation is. $a)9v^2-15v+25=0$

using quadratic formula, which is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, discriminant

$$\begin{gathered} D=\sqrt{b^2-4ac} \\ =\sqrt{{(-15)}^2-4\left(9\right)\left(25\right)}= \end{gathered}$$

Therefore, the number of solution is Zero.

Given equation is. $100w^2+60w+9=0$

using quadratic formula, which is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, discriminant

$$\begin{gathered} D=\sqrt{b^2-4ac} \\ =\sqrt{{\left(60\right)}^2-4\left(9\right)\left(100\right)}=\sqrt{0}=0 \end{gathered}$$

roots are real and equal .

Therefore, the number of solution is One .

Given equation is. $5c^2+7c-10=0$

using quadratic formula, which is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, discriminant

$$\begin{gathered} D=\sqrt{b^2-4ac} \\ =\sqrt{{\left(7\right)}^2-4\left(5\right)(-10)}=\sqrt{249}>0 \end{gathered}$$

roots are real and different.

Therefore, the number of solution is Two.

Given equation is. $15d^2-4d+8=0$

using quadratic formula, which is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, discriminant

$$\begin{gathered} D=\sqrt{b^2-4ac} \\ =\sqrt{{(-4)}^2-4\left(15\right)\left(8\right)} \end{gathered}$$

= not a real number .

Therefore, the number of solution is Zero.