Question

5x^2+x-2+0 quadratic formula

Answer 1

You mean
\( 5x^2 + x - 2 = 0\)?
If so, use a = 5, b = 1, and c = -2 in the quadratic formula.

Answer 2

no i meant 3x^2=2x+7

Answer 3

\[ax^2+bx+c=0 \] is the quadratic formula.

Answer 4

first i put in other form of....
4x^2-2x-7=0

Answer 5

\[3x^2=2x+7\] So subtract 2x and 7... giving you \[3x^2-2x-7=0\]

Answer 6

hey its 4x^2..sorry
4x^2-2x-7=0

Answer 7

That's fine. Her is a website that will make all of this much easier for you now and in the future. http://www.mathportal.org/calculators/solving-equations/quadratic-equation-solver.php

Answer 8

*Here* My bad, typo.

Answer 9

Okay, sorry. It gives you an explanation. Just though it might help.

Answer 10

i need someone to walk through with me to see where i go wrong

Answer 11

@mathstudent55 0

Answer 12

you have -(-2) but next line as 4.....how?

Answer 13

-(-2) = -1(-2)
The product of 2 negative numbers is positive.

Answer 14

yes but how did it double to 4?

Answer 15

Sorry. My mistake. It's 2. You're right.

Answer 16

also

Answer 17

how did 116 go to 29?

Answer 18

I was thinking of the (-2)^2 inside the radical which does turn to 4.

Answer 19

you can factour the three remaining numbers if necessary and leave in division form

Answer 20

Let me do it agian to correct that mistake and to explain better.

Answer 21

4x^2=2x+7

Answer 22

\(4x^2 - 2x - 7 = 0\)

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}} {2a}\)

\(x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-7)}} {2(4)} \)

Ok?

Answer 23

got that as well

Answer 24

2(+,-) sq rt 4-4(-28) divided by 8

Answer 25

\(x = \dfrac{2 \pm \sqrt{4 - (-112)}} {8}\)

\(x = \dfrac{2 \pm \sqrt{4 + 112)}} {8}\)

\(x = \dfrac{2 \pm \sqrt{116)}} {8}\)

Still good?

Answer 26

i see, i have the same

Answer 27

Now we need to take care of the 116 indside the radical.

\( x = \dfrac{2 \pm \sqrt{116}} {8} \)

I have the same.

Answer 28

\( x = \dfrac{2 \pm \sqrt{4 \times 29}} {8} \)

\( x = \dfrac{2 \pm \sqrt{4}\sqrt{29}} {8} \)

\( x = \dfrac{2 \pm 2\sqrt{29}} {8} \)

Answer 29

yup

Answer 30

i dont get that

Answer 31

how 116 splits up by four and why

Answer 32

116 = 2 * 2 * 29 = 4 * 29
2 is prime and 29 is prime

Answer 33

ok

Answer 34

The way to simplify a radical is to factor out the largest perfect square you can.
The perfect squares that help us here are the squares of the natural numbers:
1, 4, 9, 16, 25, 36, etc.

Answer 35

give me one to try to split

Answer 36

Since 116 = 4 * 29, 4 is the largest perfect square that can be factored out.
The you separate the root into :
\(\sqrt{116} = \sqrt{4 \times 29} = \sqrt{4}\sqrt{29} = 2\sqrt{29} \)

Answer 37

I will soon. Let's just finsh this problem now. We're almost done.

Answer 38

k

Answer 39

We are here:

\(x = \dfrac{2 \pm 2\sqrt{29}} {8} \)

Now we factor out a 2 from the numerator and a 2 from the denominator.

Answer 40

1(+,-)sqrt 29 divided by 4

Answer 41

\( x = \dfrac{2(1 \pm \sqrt{29})} {2(4)} \)

\( x = \dfrac{2}{2} \times \dfrac{1 \pm \sqrt{29}} {4} \)

\( x = \dfrac{1 \pm \sqrt{29}} {4} \)

\( x = \dfrac{1 + \sqrt{29}} {4} \) or \( x = \dfrac{1 - \sqrt{29}} {4} \)

Answer 42

Ok?

Answer 43

yes im stuck on one its6 (+,-)sq rt 24 divided by 6

Answer 44

You mean just to simplify it?

Answer 45

This?

\( \dfrac{6 \pm \sqrt{24}}{4} \)

Answer 46

no all divided by 6

Answer 47

\(\dfrac{6 \pm \sqrt{24}}{6} \)

\(=\dfrac{6 \pm \sqrt{4 \times 6}}{6}\)

Answer 48

is it 3(+.-)sq rt 6 divided by 3?

Answer 49

\(=\dfrac{6 \pm \sqrt{4} \sqrt{6}}{6} \)

\(=\dfrac{6 \pm 2 \sqrt{6}}{6} \)

\(=\dfrac{3 \pm \sqrt{6}}{3} \)

You are correct.

Answer 50

ok the original problem is 3x^2=6x-1 will you check, sorry

Answer 51

\(3x^2=6x-1\)

\(3x^2 - 6x + 1 = 0\)

Answer 52

\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}} {2a}\)

\(x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(1)}} {2(3)}\)

Ok?

Answer 53

\(x = \dfrac{6 \pm \sqrt{36 - 12}} {6}\)

\(x = \dfrac{6 \pm \sqrt{24}} {6}\)

This is what you had, so you are correct.