Question

How many and of what type are the solutions of a quadratic equation when the value of the radicant is 25?

a. No real solutions
b. Two identical rational solutions
c. Two different rational solutions
d. Two irrational solutions

Answer 1

A Quadratic Equation\[ax^2+bx+c=0\]
The Quadratic Formula\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

The radicand of the quadratic formula is called
the Discriminant \(\Delta =b^2-4ac\)

if \(\Delta=0\) there is only one solution to the quadratic equation
if \(\Delta>0\) there are two solutions
if \(\Delta <0\) the solutions are complex

Answer 2

another way of putting this

if \(Δ=0\) there is two equal solutions
if \(Δ>0\) there are two different solutions
if \(Δ<0\) there are no real solutions

Answer 3

if the discriminant is a square number
say\[\Delta = d^2\]
then taking the square root will not leave any irrational component
\[x_{1,2}=\frac{-b\pm \sqrt\Delta}{2a}=\frac{-b\pm \sqrt d^2}{2a}=x_{1,2}=\frac{-b\pm d}{2a}\]

Answer 4

so I still don't get the answer im confused

Answer 5

ok well you have \[\Delta =25\] right?

Answer 6

no \[\sqrt{25}\]

Answer 7

(the terminology is a little confusing)

but the radicand is the bit Under the square root

Answer 8

ok so whats the answer

Answer 9

first check is the radicand equal to, greater than, or less that, zero ?

Answer 10

greater?

Answer 11

good, so the must be two different solutions,

Answer 12

ok thx

Answer 13

but are they rational or irrational?

is 25 a square number?

Answer 14

do you remember what a square number is ?

Answer 15

yes

Answer 16

5

Answer 17

right. so the quadratic formula would have simplified ,
the solutions are
\[x_{1,2}=\frac{-b\pm \sqrt{25}}{2a}=\frac{-b\pm \sqrt 5^2}{2a}=\frac{-b\pm 5}{2a}\]

which means \(x=\frac{-b+5}{2a}\) or \(x=\frac{-b-5}{2a}\)

which are both rational because the radical symbol has gone

Answer 18

ok

Answer 19

does it make any sense?

Answer 20

yes thanks!

Answer 21

Say if a quadratic equation was

\[x^2+5x=0\]

\(a\) would be 1
\(b\) would be 5
and \(c\) would be 0

so the discriminant would equal 25 as we have just had

we found that there would be two different rational solutions

this is consistent with if we had factored

\[x^2+5x=0\\
x(x+5)=0\]

where we find the roots as \(x=0\) or \(x=-5\)