10x^2 + 21x - 10=0 steps to how to solve it?

Question

10x^2 + 21x - 10=0 I need to know ow to solve it?

Answer 1

Hi Heather;

10x² + 21x - 10=0

For the FOIL...

FIRST must be (10x)(x) or (5x)(2x) to produce 10x².

OUTER and INNER must add-up to 21x.

LAST must be (10)(1) or (1)(10) or (5)(2) or (2)(5) and one number must be negative to produce the negative result of 10.

(5x-2)(2x+5)=0

Let's FOIL...

FIRST...(5x)(2x)=10x²

OUTER...(5x)(5)=25x

INNER...(-2)(2x)=-4x

LAST..(-2)(5)=-10

10x²+25x-4x-10=0

10x²+21x-10=0

 

(5x-2)(2x+5)=0

Either or both parenthetical equations must equal zero.

5x-2=0

5x=2

x=2/5

 

2x+5=0

2x=-5

x=-5/2

Answer 2

There are some tricks to finding solutions to homework problems.  Look at the a=10, b=21, c=-10 coefficients and try and figure out the zeros of the function.  This takes great practice and is worth perfecting.  If/when you fail....you are left with the standard quadratic formula X=[-b ± sqrt(b^2-4ac) ]/2a

 

It's easier to pattern recognize the factors if they exist.  All experienced math people try and find "nice factors" by eye.  I just spent 5 seconds and found:

 

(5X - 2)(2X + 5)

 

The zeros of the eqtn are: X=2/5 or X= -5/2

 

It is difficult to put into words how we do this.  The best I can offer you is the following for

 

aX^2+bX+c:

 

1)Find the factors of a

2)Fnd the factors of c

3)Juggle these factors to achieve b

 

Think that over and over and over until you get it.