Quadratic equations ( solving for zeros/roots) trickey

In summary, the conversation is about solving the equation -2x^2+3x+20 and finding the zeros. The equation is first rewritten in standard form and then factored down to -1(2x+5)(x-4), which gives the correct answers of x=4 and x=-5/2. The conversation also discusses the process of factoring by grouping and recognizing when to use it.
  • #1
miller1991
2
0
−2x^2+3x+20why is this equation so special it is in standard form, and when i solve for zeros
i factor down to

-1( 2x^2-3x-20)

factor
( 2x^2-3x-20)

factor by grouping 2(-20)=-40 what numbers multiple to equal -40 and add to equal -3

-8(5) = -40
-8+5= -3

and then just to find zeros the next step would be
(x-8) (x+5)
x=8 x=-5 BUT
thats not right apparently the answer is

x=4 x = -5/2

the answer i got here

(x-8) (x+5)
x=8 x=-5

is meant to be -1(2x+5)(x-4)

which gives
x=4 x = -5/2 but like 5+-4 =1 don't even equal -3

apparently if i use the huge equation to solve for zeros that works but
why did (2x+5)(x-4) using this work, and how to i recognize when to use it like that
and i don't even know where they pulled the 5 and -4 fromdoes anyone know why when solving for zeros
y=-2x^2 +3x+20
is a tricky question
p.s
originally the question was solve zeros for
y=3x+20 -2x^2
but putting it in standard form was easy and i don't think that messes with the final answer
thanks !
 
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  • #2
miller1991 said:
factor by grouping 2(-20)=-40 what numbers multiple to equal -40 and add to equal -3

-8(5) = -40
-8+5= -3

and then just to find zeros the next step would be
(x-8) (x+5)
x=8 x=-5

You have the right idea!
You are trying to factor
$-2x^2+3x+20$
Factor out the negative (like you did).
$-(2x^2-3x-20)$
Then you determined the values you are working with are 8 and -5. That tells you how to "re-write" your equation so that is set up in a way that you can factor by grouping.
$2x^2-3x-20 = 2x^2+5x-8x-20$
Now factor by grouping.
$2x^2+5x-8x-20 = x(2x+5)-4(2x+5)=(x-4)(2x+5)$
Giving,
$x=4, x=-\frac{5}{2}$
 

FAQ: Quadratic equations ( solving for zeros/roots) trickey

1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains a variable raised to the power of two. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.

2. How do you solve a quadratic equation?

To solve a quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. We can also use factoring or completing the square methods.

3. What are the zeros/roots of a quadratic equation?

The zeros or roots of a quadratic equation are the values of x that make the equation equal to 0. These can be found by solving the equation using the methods mentioned above.

4. What is the discriminant of a quadratic equation?

The discriminant of a quadratic equation is the value b^2 - 4ac, which is used to determine the nature of the solutions. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. And if it is negative, there are no real solutions.

5. Can a quadratic equation have imaginary solutions?

Yes, a quadratic equation can have imaginary solutions if the discriminant is negative. In this case, the solutions will involve the imaginary number i, which is equal to √(-1).

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