quadratic formula solution

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Find approximate solutions for equation:

2x^2-5x-4=0

My book says the answers are -0.64, 3.14.

I don't understand how they got this.
 
I would suspect they plugged a = 2, b = -5, and c = -4 into the Quadratic Formula, and plugged the results into a calculator.

Are you supposed to be using some other method to find solutions?

Eliz.
 
Yes, they plugged those numbers in, but I got a different answer then them.
 
The book is correct. Are you familiar with the quadratic formula? If you show what you did, we can pinpoint your error.
 
angelasloan7038 said:
Find approximate solutions for equation:

2x^2-5x-4=0

My book says the answers are -0.64, 3.14.

I don't understand how they got this.
"approximate" solutions...

If f(x) = 2x<sup>2</sup> - 5x - 4

A linear approximatiom

f(-1) = 3
f(0) = -4

x, such that f(x) = 0, is approximately -1+(3/7) = -0.571

f(3) = -1
f(4) = 8

x, such that f(x) = 0, is approximately 3+(1/9) = 3.111
 
Hello, angela!

Could you be using your calculator incorrectly?

Yes, they plugged those numbers in, but I got a different answer than them.
We have: .\(\displaystyle 2x^2\,-\,5x\,-\,4\:=\:0\qquad\Rightarrow\qquad a=2,\;b=-5,\;c=-4\)

Quadratic Formula: . \(\displaystyle x\;=\;\frac{-(-5)\,\pm\,\sqrt{(-5)^2\,-\,4(2)(-4)}}{2(2)}\)

If you try to do all that on your calculator, you're bound to make errors.

I suggest you do a little simplifying first: . \(\displaystyle x\;=\;\frac{5\,\pm\,\sqrt{57}}{4}\)

Even then, you can enter the problem incorrectly.


Let's calculate: \(\displaystyle \frac{5\, +\, \sqrt{57}}{4}\)

It looks like: \(\displaystyle [5]\;[+]\;[\sqrt{x}]\;[5][7]\;[\div]\;[4]\;[=]\) . . . but this is wrong!

What we did was: .\(\displaystyle 5\,+\,\frac{\sqrt{57}}{4}\)
. . (Order of operations: Division comes before Addition.)


We must divide the <u>entire</u> \(\displaystyle (5 + \sqrt{57})\) by \(\displaystyle 4.\)

We can add parentheses: . \(\displaystyle [(]\:[5]\,[+]\,[\sqrt{x}]\,[5][7]\:[)]\,[\div]\,[4]\;[=]\)
. . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\) . . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\)

Or we can press \(\displaystyle [=]\) after the addition:

. . . \(\displaystyle [5]\,[+]\,[\sqrt{x}]\,[5][7]\,[=]\,[\div]\,[4]\;[=]\)
. . . . . . . . . . . . . . . . . . . . .\(\displaystyle \Uparrow\)
 
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