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# Help Me With Solving This Equation Please.

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I have to solve this equation either by factorising or using the quadratic equation formula.

6x^2-5x-4=0

math algebra
Guest Aug 17, 2014

#4
+93311
+10

6x2 - 5x - 4 factors nicely into (2x + 1)(3x - 4)

$$6x^2-5x-4$$

To factor this you have to look for 2 numbers.

They have to multiply to give 6*-4=-24

Since this is negative, one of the numbers will be negative and the other one is positive.

They have to add to -5  this means the larger digit will be the negative one.

6*4=24 but there is no way these will add or subtract to give -5

2*12=24 but thes can't make -5 either.

8*3=24 and 8-3 is 5 they might be helpful.  How about 3-8=-5 excellent and 3*-8=-24 So the numbers are 3 and -8

We are going to split -5 up into -8x+3x  like this

$$\\6x^2-5x-4=0\\ 6x^2-8x+3x-4=0\\$$

NOW we will factorise in pairs.  Like this:

$$\\2x(3x-4)+1(3x-4)=0\qquad so\\ (2x+1)(3x-4)=0\\$$

That is how I would do it.

HOWEVER

If the factors cannot be found easily it is easier to use the quadratic formula.

$$\\6x^2-5x-4=0\\ a=\;6\quad b=\;-5\quad and \quad c=\;-4 \\$$

$$\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\ x=\frac{5\pm \sqrt{(-5)^2-4*6*-4}}{2*6}\\\\ x=\frac{5\pm \sqrt{25+96}}{12}\\\\ x=\frac{5\pm \sqrt{121}}{12}\\\\ x=\frac{5\pm 11}{12}\\\\ x=\frac{5+ 11}{12}\qquad or \qquad x=\frac{5- 11}{12}\\\\ x=\frac{16}{12}\qquad or \qquad x=\frac{-6}{12}\\\\ x=1\frac{1}{3}\qquad or \qquad x=\frac{-1}{2}\\\\$$

Melody  Aug 18, 2014
#1
+26971
+5

6x2 - 5x - 4 factors nicely into (2x + 1)(3x - 4)

For this to be zero we must have x = -1/2  or  x = 4/3

Alan  Aug 17, 2014
#2
0

I'm sorry to ask, but the part I'm most confused about is how does it actually factorise into  (2x + 1)(3x - 4)?

Guest Aug 17, 2014
#3
+26971
+5

You can start by assuming that, if it factors nicely, it will be in the form

(ax + b)(cx + d) where a, b, c and d are nice whole numbers (positive or negative).

Expanding these brackets we get:

(ax + b)(cx + d)  =  a*c*x2 + a*d*x + b*c*x + b*d = a*c*x2 + (a*d + b*c)*x + b*d

Now compare this with the left-hand side of your equation 6x2 - 5x - 4

A. From the constant term we must have b*d = -4, so the magnitudes of b and d must be 1 and 4 or 2 and 2.

B. From the x2  term a*c = 6 so the magnitudes of a and c must be 1 and 6 or 2 and 3

C. From the x term we must have a*d + b*c = -5

Now you have to try various combinations of a, b, c and d from A and B until you get C correctly (you'll have to include the possiblity that the signs are positive or negative as well).  This sounds like a lot of combinations, but actually there aren't that many to try and you'll find the right values fairly quickly.

If no combinations fit it means the equation doesn't factorize nicely and you would have to use the quadratic formula.

Alternatively, you could use the quadratic formula right away!

Alan  Aug 17, 2014
#4
+93311
+10

6x2 - 5x - 4 factors nicely into (2x + 1)(3x - 4)

$$6x^2-5x-4$$

To factor this you have to look for 2 numbers.

They have to multiply to give 6*-4=-24

Since this is negative, one of the numbers will be negative and the other one is positive.

They have to add to -5  this means the larger digit will be the negative one.

6*4=24 but there is no way these will add or subtract to give -5

2*12=24 but thes can't make -5 either.

8*3=24 and 8-3 is 5 they might be helpful.  How about 3-8=-5 excellent and 3*-8=-24 So the numbers are 3 and -8

We are going to split -5 up into -8x+3x  like this

$$\\6x^2-5x-4=0\\ 6x^2-8x+3x-4=0\\$$

NOW we will factorise in pairs.  Like this:

$$\\2x(3x-4)+1(3x-4)=0\qquad so\\ (2x+1)(3x-4)=0\\$$

That is how I would do it.

HOWEVER

If the factors cannot be found easily it is easier to use the quadratic formula.

$$\\6x^2-5x-4=0\\ a=\;6\quad b=\;-5\quad and \quad c=\;-4 \\$$

$$\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\ x=\frac{5\pm \sqrt{(-5)^2-4*6*-4}}{2*6}\\\\ x=\frac{5\pm \sqrt{25+96}}{12}\\\\ x=\frac{5\pm \sqrt{121}}{12}\\\\ x=\frac{5\pm 11}{12}\\\\ x=\frac{5+ 11}{12}\qquad or \qquad x=\frac{5- 11}{12}\\\\ x=\frac{16}{12}\qquad or \qquad x=\frac{-6}{12}\\\\ x=1\frac{1}{3}\qquad or \qquad x=\frac{-1}{2}\\\\$$