Homeschool. Algebra Trouble?

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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...Show more

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)....Show more

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 :)...Show more

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...Show more

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...Show more