Homeschool. Algebra Trouble?

Both of these are problems that can be solved with the quadratic formula, as long as you get them in the form:

Ax² + Bx + C = 0

QUESTION 1:

Add y² to both sides:

y² + 5y - 14 = 0

You could easily solve this with factoring:

(y - 2)(y + 7) = 0

y = 2

or

y = -7

The smaller solution is -7. However, the problem specifically asks you to use the quadratic formula.

a = 1

b = 5

c = -14

...... -5 ± √(5² - 4(1)(-14))

y = -----------------------------

............... 2(1)

...... -5 ± √(25 +56)

y = ----------------------

............... 2

...... -5 ± √81

y = ---------------

............... 2

...... -5 ± 9

y = ----------

.......... 2

y = (-5 - 9)/2 = -14/2 = -7

or

y = (-5 + 9)/2 = 4/2 = 2

The smaller answer is:

y = -7

P.S. This matches with the answer found using factoring.

QUESTION 2:

2x² + x - 6 = 0

a = 2

b = 1

c = -6

Now plug that into the quadratic formula:

...... -b ± √(b² - 4ac)

x = ----------------------

............... 2a

...... -1 ± √(1² - 4(2)(-6))

x = ----------------------

............... 2(2)

...... -1 ± √(1 + 48)

x = ----------------------

............... 4

...... -1 ± √49

x = --------------

............. 4

...... -1 ± 7

x = ----------

.......... 4

x = -1/4 ± 7/4

x = -8/4 or 6/4

Answer:

x = -2

or

x = 1½

Q2. The quadratic formula states two solutions:

They are (-b+ sqrt(b^2-4ac))/2a and

(-b- sqrt(b^2-4ac))/2a.

a stands for the coefficient of the squared term,

b stands for the coefficinet of the first degree term,

and c stands for the constant term.

Thus in this case a=2. b=1, and c=-6

Thus the solutions are ( -1+sqrt(1+24))/4 = ( -1 +5)/4 = 1

and (-1 -sqrt(1+24))/4 = ( -1 -5)/4 = -3/2

Q1. Here use the same formula (after you write the equation in the correct form, that is y^2+5y-14=0).

Hi,

1) Add y^2 to both sides:

y^2 + 5y - 14 = 0

Now you have a quadratic which you can factorise:

(y + 7)(y - 2) = 0

So, either y + 7 = 0, therefore y = -7

OR

y - 2 = 0, therefore y = 2

These are your solutions :)

2) Factorise again, so:

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

So, either 2x - 3 =0, therefore x = 1.5

OR x + 2 = 0, therefore x = - 2

These are your solutions

Hope this helps :)

Add y^2 to both sides to get

y^2 + 5y - 14 = 0

This quadratic factors as

(y + 7)(y - 2) = 0

y + 7 = 0 -> y = -7

y - 2 = 0 -> y = 2

The smallest solution is therefore -7

I do not understand the second question

Well, for the first question let's reorder the expresion

y^2+5y-14=0 Now using factoring

(y+ 7)(y-2)=0 so y=-7 and y=2 the smallest will be y=-7

For the second one

Cuadratic formula

x=(-b+-sqrt (b^2-4ac))/2a

a=2

b=1

c=-6

so,

x=(-1+- sqrt (1^2-4(2)(-6))/2(2)

x=(-1+- sqrt (49))/4

x=(-1+-7)/4

x=3/2 and x= -2

I hope this can be useful

I haven't really done Quadrtics before but I'll give it shot

1) Move the y^2 to the left (don't forget to change its sign) to get

y^2 + 5y - 14 = 0

Factorise

(y + 7)(y - 2) = 0

Therefore y is either equl to -7 or +2 (1 of the 2 parnthesis have to equl 0 for their product to be 0)

Since they asked for the smaller answer, the answer would be -7.

or, using the quadratic formula,

ay^2 + by + c

where a = 1, b = 5 and c = -14

y = -b +- sqrt(b ^ 2 - 4ac)

-----------------------------

2a

= -5 +- sqrt((-5)^2 - 4.1.-14)

-------------------------------------

2.1

= -5 +- sqrt(25 - (-56))

----------------------------

2

= -5 +- (sqrt(81))

---------------------

2

= -5 +- 9

----------

2

is either = to -14/2 or +4/2

which is = to -7 and +2 respectively

the smaller one of the two is -7

I'll post the answer to the second question when i can

u kno the the quadratic equation is

x or y = (-b +/- sqrt (b^2 -4ac)) /2a

so use it to solve ur equations

1. bring the y^2 to the other side

y^2 + 5y - 14 = 0

y = (-5 +/- sqrt( 25 -4*1*-14))/2

y= (-5 + 9)/2 or y= (-5 -9)/2

y= 2 , y= -7

2. 2x^2+x-6=0

x = (-1 +/- sqrt( 1 -4*2*-6))/4

x= (-1 + 7)/4 or x= (-1 -7)/4

x= 1.5 , x= -2

5y-14+y^2 = y^2+5y-14 =

y^2+2y -7y-14 =0

(y-7)(y+2)

the solution using the quadratic formula.

2x^2+x-6=0

a=2

b=1

c=-6

-b+(sqrt(b^2 - 4ac)) /2a or -b-(sqrt(b^2 - 4ac)) /2a

-1+(sqrt(1 +48)) /4 or -1-(sqrt(1 +48)) / 4

(-1+7)/4 or (-1-7)/4

1.5 or -2

5y-14= -y^2

y^2+5y-14= 0

( y+7)( y-2)= 0

y+7= 0

y= -7

y-2= 0

y= 2

2.

2x^2+x-6= 0

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

2x-3= 0

x= 3/2

x+2= 0

x= -2