i am having trouble solving these quadratic equations and need help?

(x+5) (x-2)
and
the second one is impossible i think

x^2 + 3x - 10 = 0
(x + 5)(x - 2) = 0
x = -5 OR x = 2

x^2 + 3x - 5 = 0
(x + 3/2)^2 + 3/2 - 5 = 0
(x + 3/2)^2 = 7/2
x + 3/2 = sqrt 7/2
x = 1.87 - 1.5
x = 0.37
...OR...
x = -1.87 - 1.5
x = -3.37

Question 1
(x + 5)(x - 2) = 0
x = - 5 , x = 2

Question 2
x = [ - b ± √ (b ² - 4 a c ) ] / 2 a
x = [ - 3 ± √ (9 + 20) ] / 2
x = [ - 3 ± √ (29) ] / 2
x = 1.19 , x = - 4.19

1.x squared +3x – 10 = 0
x squared +5x – 2x – 10 = 0
x(x+5) – 2(x+5) = 0
(x-2)(x+5)=0
x = 2 or -5
2.x squared +3x – 5 = 0
comparing this with the equation
ax squared + bx + c=0
a=1, b=3, c=-5.
so, b square - 4ac = 9 -4*1*(-5) = 29
so x= (-3- √29)/2 or (-3+ √29)/2

1)
x^2 + 3x - 10 = 0 (solve using factoring)
x^2 - 2x + 5x - 10 = 0
x(x - 2) + 5(x - 2) = 0
(x - 2)(x + 5) = 0

x - 2 = 0
x = 2

x + 5 = 0
x = -5

∴ x = 2 , -5

= = = = = = = =

2)
x^2 + 3x - 5 = 0 (solve using quadratic discriminant)
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 3
c = -5

x = [-3 ±√(9 + 20)]/2
x = [-3 ±√29]/2
x = [-3 ±5.38]/2 (approx.)

x = [-3 + 5.38]/2
x = 2.38/2
x = 1.19

x = [-3 - 5.38]/2
x = -8.38/2
x = -4.19

∴ x = 1.19 , -4.19

Complete the square:

x² + 3x - 10 = 0

x² + 3x = 10

Take half of three, square it and add it to both sides.

x² + 3x + 2.25 = 10 + 2.25

x² + 3x + 2.25 = 12.25

(x + 1.5)² = 12.25

x + 1.5 = ±√(12.25)

x = ±√(12.25) - 1.5

x = 2, x = -5



x² + 3x - 5 = 0

x² + 3x = 5

x² + 3x + 2.25 = 5 + 2.25

x² + 3x + 2.25 = 7.25

(x + 1.5)² = 7.25

x + 1.5 = ±√(7.25)

x = ±√(7.25) - 1.5

use the quadratic formula

http://www.intmath.com/Quadratic-equations/Image2063.gif

the top most equation is the formula. ignore the rest.

where the quadratic eq'n your solving is ax^2 + bx - c

ie sub in the values of a, b and c. This is the fool proof way to solve the quadratic equations.

You can factorise the eq's into bracket form eg. (x+a)(x-b) where a and b can be any number, and then solve = 0, but not all eq's can be factorised, so the quad formula is better :D


over all the answer for the 1st one is x=2,-5

2nd eq'n is x= (-3 + or - square root of 29) / 2

sorry for 2nd answer, cant write math on a pc, and there is no whole answer for the root of 29, so it has to be a surd :P it may not even be correct, but im pre sure. cbf checking

1) x^2 + 3*x - 10 = 0

The first step is to factorize. Just think: what two numbers can I have such that if I add them, I get 3 and multiply them i get -10?

How about 5 and -2?

So now the equation becomes

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

Now you can see the values of x which will result in the product being zero is -5 and 2.

2) x^2 + 3*x -5 = 0

There exist no two whole numbers that add to 3 and multiply to -5. Now you have to use the quadratic formula

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

the values of a, b and c respectively are 1, 3 and -5. Just sub it in.

x^2 + 3x - 10 = 0 can be rewritten as
(x + 5)(x - 2) = 0
Solving, we get x = -5 or x = 2

For x^2 + 3x - 5 = 0, we use the quadratic formula, x = [-b +/- sqrt(b^2 - 4ac)] / 2a.
The equation is of the form ax^2 + bx + c = 0, so a = 1, b = 3, and c = -5.
Plugging into the equation, we get x = [-3 +/- sqrt(29)] / 2

u know that u are going to have (x + something)(x + something) = 0

you need to look at the -10 and think 'what two numbers multiply to give me -10 and a difference of 3? +5 and -2

So it becomes (x + 5)(x - 2)

With next one you have to use quadratic formula. I'm guessing u dont konw what that is

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

http://www.teacherschoice.com.au/Maths_Library/Algebra/Alg_6.htm
for more info!

ok for the first one,
x^2 + 3x - 10 = 0
you need to work out what times to (-10) and adds to 3
so (-2) + (5)
therefor x^2 - 2x + 5x - 10 = 0
x(x-2) + 5(x-2) = 0
therefor (x + 5)(x - 2)
x = -5, + 2