Question

How to determine the number and nature of the solutions of each?

Answer 1

we know that

Use the value of the discriminant to determine the nature of the solutions to the quadratic equation.

so

The discriminant is equal to

`D=b^2-4ac`

Part 23

we have

`\begin{gathered} 49c^2+4=-28c \\ \text{equate to zero the quadratic equation} \\ 49c^2+28c+4=0 \\ a=49 \\ b=28 \\ c=4 \\ \text{substitute in the equation of discriminant} \\ D=(28^2)-4(49)(4) \\ D=784-784 \\ D=0 \end{gathered}`

that means -----> the roots are equal and real.

part 24

we have

`\begin{gathered} (3x+1)^2=5x-1 \\ 9x^2+6x+1=5x-1 \\ 9x^2+6x-5x+1+1=0 \\ 9x^2+x+2=0 \\ a=9 \\ b=1 \\ c=2 \\ \text{substitute} \\ D=(1^2)-4(9)(2) \\ D=1-72 \\ D=-71 \\ \end{gathered}`

that means -----> The discriminant is negative, so the equation has two non-real solutions.

Part 26

we have

`\begin{gathered} 5z^2+2z-4=0 \\ a=5 \\ b=2 \\ c=-4 \\ \text{substitute} \\ D=(2^2)-4(5)(-4) \\ D=4+80 \\ D=84 \end{gathered}`

that means -----> The discriminant is positive, so the equation has two distinct real solutions.