Question

Help with quadratics? Picture attached. WILL MEDAL AND FAN!

Answer 1

\[\huge\rm Ax^2+Bx+C=0\]standard form of quadratic equation

Answer 2

\[(-b+\sqrt{b^2-4ac})/2a\] or \[(-b-\sqrt{b^2-4ac})/2a\]

Answer 3

where, \(A\)= 1
\(B\)= 5
\(C\)= 3

Answer 4

equation is already in standard form
so move on to 2nd step
list a b c values
a=leading coefficient
b= coefficient of x term
c=constant term

Answer 5

As per your equation, use \(-b....\) formula to find \(D\)

Answer 6

What?

Answer 7

alright he already listed a , b ,c values so just plug them into the formula

Answer 8

Right... so \[(-5+\sqrt{5^2-4*1*3})/2\]

or \[(-5-\sqrt{5^2-4*3*1})/2\]

Answer 9

good ! :=))

Answer 10

When I plug it into my calculator, though, it gets goofy. :(

Answer 11

well you need to put parentheses

Answer 12

\[\huge\rm \frac{ - 5 \pm \sqrt{ \color{ReD}{(5)^2 -4(1)(3)}} }{ 2 }\]
first solve the discriminant

Answer 13

25-12=13

Answer 14

right

Answer 15

\[\frac{ -5\pm \sqrt{13} }{ 2 }\]

Answer 16

looks good

Answer 17

step4) they just want to separate that into 2 fractions

Answer 18

like \[\rm a \pm b \rightarrow a -b ~~~and~~~a+b\]

Answer 19

Okay... I get it. Thank you so much!

Answer 20

The online thing says the answer is right! :)

Answer 21

good with part 5 ?

Answer 22

Yup. Thanks again!

Answer 23

alright np :=))