Can I get some help with quadratic functions and real solutions

[letoatreides3508]

I'm having a really hard time. I'm being asked to find the real solution of

10x^2-11x-6 = 0

It then reads

To find the real solutions of the equation, first factor the left side of the equation

To factor the expression, find two numbers whose product is the same as the product of the co-efficient of x^2 and the constant term
and whose sum is the same as the co-efficient of x.

The factor of the co-efficient of x^2 and the constant term is -60, but the answer reads (-15, 4) How is -15 the product of 10*-6? Where is it getting these numbers? And what
am I being asked to do?!? Please help me.
,

 

[stapel]

...find the real solution of 10x^2-11x-6 = 0

It then reads: To find the real solutions of the equation, first factor the left side of the equation

To factor the expression, find two numbers whose product is the same as the product of the co-efficient of x^2 and the constant term and whose sum is the same as the co-efficient of x.

The factor of the co-efficient of x^2 and the constant term is -60, but the answer reads (-15, 4) How is -15 the product of 10*-6? Where is it getting these numbers?

Try following the step-by-step instructions. If you do so, you'll see where the -15 come into play. (They've skipped showing the results of some of the steps they've given you to do.)

And what am I being asked to do?

You are being asked to factor the quadratic. To learn the method they're referencing, try here. You exercise is an example of what that lesson calls "the hard case".

 

[Deleted member 4993]

I'm having a really hard time. I'm being asked to find the real solution of

10x^2-11x-6 = 0

It then reads

To find the real solutions of the equation, first factor the left side of the equation

To factor the expression, find two numbers whose product is the same as the product of the co-efficient of x^2 and the constant term
and whose sum is the same as the co-efficient of x.

The factor of the co-efficient of x^2 and the constant term is -60, but the answer reads (-15, 4) How is -15 the product of 10*-6? Where is it getting these numbers? And what
am I being asked to do?!? Please help me.
,

You are asked to find factors of (10 * -6 =) -60 such that the sum of two factors is -11.

So factors of (-60)

..1.........-1.........2.........-2.........3.........-3.........4.........-4.........5.........-5.........6.........-6...........

-60.........60......-30........30.......-20........20.......-15........15.......-12........12........10........10...........

-15 & 4

when multiplied with each other results in -60 and

when added to each other results in -11

 

[srmichael]

I'm having a really hard time. I'm being asked to find the real solution of

10x^2-11x-6 = 0

It then reads

To find the real solutions of the equation, first factor the left side of the equation

To factor the expression, find two numbers whose product is the same as the product of the co-efficient of x^2 and the constant term
and whose sum is the same as the co-efficient of x.

The factor of the co-efficient of x^2 and the constant term is -60, but the answer reads (-15, 4) How is -15 the product of 10*-6? Where is it getting these numbers? And what
am I being asked to do?!? Please help me.
,

Hi Leto.

Here is another way that one can factor a quadratic (provided it's factorable).

Starting with \(\displaystyle 10x^2-11x-6=0\), multiply the leading coefficient times the constant and rewrite the quadratic with a 1 as the leading coefficient and the constant as the product of the original leading coefficient and the original constant.

\(\displaystyle (10)(-6)=-60\)

\(\displaystyle x^2-11x-60=0\)

Then, find two numbers whose product is -60 and whose sum is -11. You may need to hunt a little bit when there are many factors but for our problem the two numbers are -15 and +4 since (-15)(4) = -60 and -15 + 4 = -11. Once you have found these two numbers write out the quadratic in factored form using these two numbers like so:

\(\displaystyle (x - 15)(x + 4) = 0\)

Then, divide each number by the original leading coefficient to bring us back to the original problem at hand that we are trying to solve.

\(\displaystyle \left (x - \dfrac{15}{10} \right) \left (x + \dfrac{4}{10} \right)=0\)

Simplfy the fractions if necessary:

\(\displaystyle \left (x - \dfrac{3}{2} \right) \left (x + \dfrac{2}{5} \right)=0\)

At this point, since we are solving for x, we can simply set each factor equal to 0 and get \(\displaystyle x=\dfrac{3}{2}, x = -\dfrac{2}{5}\)

If you wish to just finish the factoring of the quadratic, you can then multiply the x in each factor by the denominator of each fraction within the separate factors like so:

\(\displaystyle (2x - 3)(5x + 2)\)

 

[Deleted member 4993]

From where d'heck doya'll get 60?
That equation fators thusly: (2x - 3)(5x + 2) ..... Correct - but to get there through the method described, you need to find "heck"

Perhaps equation should be one of:
(x + 15)(x - 4) = x^2 + 11x - 60
(x - 15)(x + 4) = x^2 - 11x - 60

10x^2 -11x - 6 = 0

New address of heck is (10) * (-6)

10x^2-11x-6 = 0

10x^2 - 15x + 4x - 6 = 0

5x(2x - 3) + 2(2x - 3) = 0

(2x - 3)(5x + 2) = 0

 

[lookagain]

10x^2 - 11x - 6 = 0

10x^2 - 15x + 4x - 6 = 0

5x(2x - 3) + 2(2x - 3) = 0

(2x - 3)(5x + 2) = 0

Also, interchanging the two middle terms works equally as well:


10x^2 + 4x - 15x - 6 = 0

2x(5x + 2) - 3(5x + 2) = 0

(5x + 2)(2x - 3) = 0