Please help?

Question

If someone can explain how to do this I can do the rest of my problems. Thanks
solve by using the quadratic formula
4x^2+7x+5=0

and also x^2=7x-9
I got this far on the second one
x^2-7x+9=0 (but do not know what to do after)

Answer 1

The quadratic formula is x = -b +/- [square root(4a - bc) /2a]
I believe that is it...
so in your equation
1. 4x^2+7x+5=0
a = 4
b = 7
c = 5
Just plug those into the formula and you will get the answer.

2. Yes you are right so far. So you have
x^2-7x+9=0
a = 1
b = -7
c = 9
Once again just plug those into the formula and you will get the answer. When I say plug those into the formula what I mean is where if you see an a b or c plug in the value for a b and c. That is where ever you an a plug in 1 and where you see b plug in -7 and where ever you see c plug in a 9

if you need further help or clarification email me or im me. Good Luck.

Answer 2

the quadratic formula is:

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

a, b, c are the coefficient in the quadratic equations you have written above.

In the first problem, a=4, b=7 and c=5. These values should be plugged into the quadratic formula. First with -b + "the rest" in the numerator, then with -b - "the rest" in the numerator.

for the first problem, you end up with imaginary roots. You should get:

(-1/4) + i*(sqrt (31)/16 and (-1/4) - i*(sqrt (31)/16)

In the second problem your first step is correct, a = 1, b = -7 and c = 9.

You should get two real roots:

(7 + sqrt (13))/2 and (7 - sqrt (13))/2

Good Luck.

Answer 3

My teacher told me a song to sing for the quadratic formula. It goes to the tune of Pop goes the weasel.

X equals the opposite of B plus or minus the square root, of Bsquared minus 4 A C. All over 2A

Your problem is 4x^2 + 7x +5 =0
Ax^2 + Bx +C =0

Substitute the 4 in for the A's and the 7 in for the B's and 5 in for the C's.

Then just solve the problem in the quadratic equation. My teacher was a fun teacher and she was always showing us these little ways to help us, so i hope it helps you.

Answer 4

1. 4x^2+7x+5 = 0

Quadtratic Formula: x = [-b +/- V`(b^2 - 4ac)] / 2a

First: you have three terms for three variables which, are:
a = 4; b = 7; & c = 5

Sec: replace/substitute the terms with the corresponding variables.

x = [-7 +/- V`(7^2 - 4(4)(5))] / 2(4)

x = [-7 +/- V`(49 - 4(20))] / 8

x = [-7 +/- V`(49 - 80)] / 8

x = [-7 +/- V`(-31)] / 8

Third: you have 2 solutions, one has addition - the other has subtraction.

a. x = [-7 + V`(-31)] / 8
b. x = [-7 - V`(-31)] / 8

Try the 2nd problem with the same format as problem 1 :-)

Answer 5

remember the square equation:
Ax^2 + Bx = C
so plug this in,
A is equal to 4
B is equal to 7
and
C is equal to 5

We start by simplifying this, change "-5" to the other side...
4x^2+7x=0+5
4x^2+7x=5

divide both sides of the equation by 4...
(4x^2+7x)/4=5/4

simplify:
x^2+7/4x=5/4

divide 7/4x by 2 to get 7/8x and square that,
add to the other side to make this equation balanced..

x^2+7/4x+(7/8)^2=5/4+(7/8)^2

simplify

x^2+7/4x+49/64=(5/4+49/64)

simplify again by finding the a common denominator:

x^2+7/4x+49/64=129/64

ok so substitute the first x value by factorization

(x+7/8)^2=129/64

take the square root of both sides (to even it out)

sqrt((x+7/8)^2)=sqrt(129/64)

simplify

x+7/8=sqrt(129)/8
x+7/8=-sqrt(129)/8

change the 7/8's side:

x=-7/8+sqrt(129)/8
x=-7/8-sqrt(129)/8

simplify

x=(7+sqrt(129))/8
x=(7-sqrt(129))/8

so x = around 0.544727 and -2.294727

voila!

Answer 6

4x² +7x+5=0

Quadratic equation
ax² + bx + c = 0

Solve for x
x = (-b +- √(b² - 4ac))/2a

Your first problem
x = (-7 +- √(49 - 80))/2

x = (-7 +- √(-31))/2

x = -7/2 +(½)i√(31) and x = -7/2 - (½)i√(31)

Your second problem
Solve for x
x = (7 +- √(49 - 36))/2
x = (7 +- √(13))/2
x = 7/2 + ½√(13), x = 7/2 - ½√(13),

Answer 7

You have formulas in the form of ax² + bx + c. In your two cases, b = 7, and b = -7. c = 5, and c = 9.

Find the quadratic formula in your chapter, and make the substitutions. Plug in. "Plug and chug."

Answer 8

x=-b+sqrt(b^2-4ac)/2a
or
x=-b-sqrt(b^2-4ac)/2a
in this equation a*x^2+b*x+c=0
just substitute