Question

Determine the number of solutions to each quadratic equation:

(a) $b^2+7b-13=0$

(b) $5a^2-6a+10=0$

(c) $4r^2-20r+25=0$

(d)$7t^2-11t+3=0$

Short answer

(a) The equation $b^2+7b-13=0$has two solutions.

(a) The equation$5a^2-6a+10=0$has no solution.

(c) The equation $4r^2-20r+25=0$has one solution.

(d) The equation $7t^2-11t+3=0$has two solutions.

Step-by-step solution

Part (a) Step 1: Given Information 

The given equation is $b^2+7b-13=0$.

Part (a) Step 2: Determine the discriminant.

$b^2+7b-13=0$

Here, $a=1,b=7,c=-13$

So, Discriminant is

$$\begin{gathered} D=b^2-4ac \\ D=7^2-4\left(1\right)(-13) \\ D=49+52 \\ D=101 \end{gathered}$$

Discriminant is greater than zero, the equation has two solutions.

Part (b) Step 1: Given Information  

The given equation is$5a^2-6a+10=0$.

Part (b) Step 2: Determine the discriminant.  

$5a^2-6a+10=0$

Here, $a=5,b=-6,c=10$

So, the discriminant is

$D=b^2-4ac$

$$\begin{gathered} D={(-6)}^2-4\left(5\right)\left(10\right) \\ D=36-200 \\ D=-164 \end{gathered}$$

Since discriminant is negative, the equation has no solution.

Part (c) Step 1: Given Information 

The given equation is$4r^2-20r+25=0$.

Part (c) Step 2: Find the discriminant. 

$4r^2-20r+25=0$

Here, $a=4,b=-20,c=25$

So, the discriminant is

$$\begin{gathered} D=b^2-4ac \\ D={(-20)}^2-4\left(4\right)\left(25\right) \\ D=400-400 \\ D=0 \end{gathered}$$

Since the discriminant is zero, the equation has one solution.

Part (d) Step 1: Given Information. 

The given equation is$7t^2-11t+3=0$.

Part (d) Step 2: Find the discriminant.

$7t^2-11t+3=0$

Here, $a=7,b=-11,c=3$

The discriminant is

$$\begin{gathered} D=b^2-4ac \\ D={(-11)}^2-4\left(7\right)\left(3\right) \\ D=121-84 \\ D=37 \end{gathered}$$

Since the discriminant is more than zero, the equation has two solutions.