Solve equations using any solution?

Question

1) x(sq)+x-42=0

2) x(sq)+8x-48=0

3) x(sq)+12x+11=0

4) 10x(sq)+3x=1

5) 3x(sq)+7x-31=0

If you can answer any that would be greatly appreciated!

Answer 1

1= (x+7)(x+6) so x= -7, -6

2=(x+12)(x-4) so x= -12, 4

3= (x+11)(x+1) so x= -11, -1

4= (2x+1)(5x-1) so x= -1/2, 1/5

5= use quadratic formula

Answer 2

The easiest way to solve a quadratic equation is to factor it out, which takes some time, but is very worthwile. When that doesn't work, use the quadratic formula:
(-b±√(b²-4ac))/2a

1:
x²+x-42=0
(x+7)(x-6)=0
X can equal -7 or +6 to set the equation equal to 0.

2:
x²+8²-48=0
(x+12)(x-4)=0
x equals -12 or +4

3:
x²+12x+11=0
(x+11)(x+1)=0
X equals -11 or -1

4:
10x²+3x=1 (1 must be subtracted)
10x²+3x-1=0
(2x+1)(5x-1)=0
X=-½ or 1/5

5:
3x²+7x-31=0 - this cannot be factored
a=3
b=7
c=-31
Plug into the quadratic formula (-b±√(b²-4ac))/2a:
(-7±√(7²-(4(3)(-31))))/2(3) Simplify next
(-7±√(49-­­(-372)))/6 Simplify more...remember that subtracting a negative number is really adding a positive number.
(-7±√(421))/6
Your roots/solutions are:
(-7+√421)/6
(-7-√421)/6
√421 is not reducible so leave it as I left it above. Hope this helps and that you can follow the math. Let me know if you need any more help. Warning: Be VERY careful with those negative signs. Other answerers are even messing up with them. The 4ac part of the quadratic equation should be negative, and it will add up to 421.

Answer 3

1. x^2 + x - 42 = 0
(x + 7)(x - 6) = 0
x = -7 OR x = 6

2. x^2 + 8x - 48 = 0
(x + 12)(x - 4) = 0
x = -12 OR x = 4

3. x^2 + 12x + 11 = 0
(x + 11)(x + 1) = 0
x = -11 OR x = -1

4. 10x^2 + 3x - 1 = 0
(5x - 1)(2x +1)
x = 1/5 OR x = -1/2

5. 3x^2 + 7x - 31 = 0
x = [-7 +/- sq. root(49 + 372)]/6
x = (-7 +/- sq. root(421))/6

Answer 4

1. x^2 + x - 42 = 0
(x + 7)(x - 6) = 0
x = -7 or x = 6

2. x^2 + 8x - 48 = 0
(x + 12)(x - 4) = 0
x = -12 or x = 4

3. x^2 + 12x + 11 = 0
(x + 11)(x + 1) = 0
x = -1 or x = -11

4. 10x^2 + 3x = 1
10x^2 + 3x - 1 = 0
(5x - 1)(2x + 1) = 0
x = 1/5 or x = -1/2

5. 3x^2 + 7x + 31 = 0
Using quadratic equation
x = (-7 +/- sqrt(49 - 4(3)(31)) / 2(3)
x = (-7 +/- sqrt(49 - 12(31)) / 6
No solutions for x.

Answer 5

geee a lot of question, you should give me best answer or i will not help you again.... lol :)

x^2 + x - 42 = 0
(x + 7) (x - 6) = 0
x = 6 and x=-7

x^2 + 8x - 48 = 0
(x + 12) (x - 4) = 0
x = 4 and x=-12

x^2 + 12x + 11 = 0
(x + 11) (x + 1) = 0
x=-11 and x =-1

10x^2 + 3x = 1
10x^2 + 3x - 1 = 0
(5x - 1)(2x + 1) = 0
x = 1/5 or x = -1/2

3x^2 + 7x - 31 = 0
x^2 + (7/3)x - (31/3) + (7/6)^2 - (7/6)^2 = 0
(x - 7/6)^2 - (31/3) - (49/36) = 0
(x - 7/6)^2 = 323/36
x = 4.1 or x = -1.8

Answer 6

confident, they're from Sridharacharya of the eleventh Century. The declare that algebra, trigonometry and calculus have been given right here from India is nice. "SREEDHARACHARYA'S RULE" (a.ok.a. "Sridharacharya's formula") is shopper-friendly indoors the west as "the quadratic formula". i'm constructive the quadratic formula has many different names in many different places too... There are 3 hardship-unfastened techniques to locate roots of quadratic equations a million. Factorization 2. formula 3. ending up sq. on account that the a quadratic equation is a 2nd degree polynomial equation, then the customary theorem of algebra states that 2 complicated roots exist, counting multiplicity. Ax^2 + Bx + C = o Ax^2 + Bx = -C (4A)(Ax^2 + Bx) = -4AC 4A^2x^2 + 4ABx + B^2 = -4AC + B^2 (2Ax + B)^2 = -4AC + B^2 2Ax + B = Sqrt( B^2 - 4AC) x = -B Sqrt(B^2 - 4AC) / 2A there are particularly some analytical techniques used for finding the roots of quadratic equations, between the substantial shopper-friendly techniques is the so-spoke of as quadratic formula and is derived by ending up the sq. on the marvelous expression shown above. The quadratic formula could suitable be written subsequently, The term below the sq. root is shopper-friendly because of the fact of fact the discriminant and could suitable be used to make certain the kind of the roots of the quadratic equation. If then there are 2 different actual roots. as nicely if the discriminant is a proper sq., then the two roots are additionally rational. If then there is one repeated actual root. If then there are 2 different non-actual roots. those 2 roots are the complicated conjugate of one yet yet another. SREEDHARACHARYA'S RULE is greater perfect shopper-friendly as "ending up the sq.". according to threat it would be instructive to incorporate a derivation of the quadratic formula using ending up the sq..

Answer 7

1) x^2+x-42=0
factor it first...
(x+7)(x-6)... there will be 2 values of x.... x=-7 nd x=6
2)x^2+8x-48=0... factor..
(x+12)(x-4)... x=-12 and x=4
3)x^2+12x+11=0
(x+11)(x+1)..... x=-11 and x=-1
4)10x^2+3x=1... transpose 1 first...
10x^2+3x-1=... factor again..
(5x-1)(2x+1)...
5x-1=0... 5x=1.. divide both sides by 5... x=1/5
2x+1=0... 2x=-1... divide both sides by 2.. x=-1/2
x=1/5 and x=-1/2...
5) I thnk #5 is not factorable...

I hope this helps u...

Answer 8

all of your problems can be solved by the quadratic equation

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

problem 4 needs to be in the form:
ax^2 + bx + c = 0

just plug in the variables and solve for x twice because of the + or -

Answer 9

#1---------------------------------------------------

GIVEN #1:
x(sq)+x-42=0

Multiply the 1st and 3rd terms together to get the product of: 42x(sq)
Next, factor into two factors whose sum = middle term of 1x

1x times 42x
2x times 21x
3x times 14x
6x times 7x .........BINGO! -6x + 7x = 1x

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

SOLUTION #1:
First, bring down the 1st term: x(sq)
Next, write down the smaller of the two factors: -6x
Followed, by the LARGER of the two factors: 7x
Finally, bring down the 3rd term: -42

x(sq)-6x+7x-42=0

Now let's Factor by Grouping!
[x(sq)-6x] + [7x-42] = 0
GCF#1=x GCF#2=7

x(x-6) + 7(x-6) = 0
GCF#3=(x-6)
(x-6)(x+7)=0

Set each factor = 0 and solve for "x".
x-6=0
x-6+6=0+6
x+0=6
x = 6

AND

x+7=0
x+7-7=0-7
x + 0 = -7
x = -7
-------------------------------------

ANSWER #1:
x = -7 and 6

#2---------------------------------------

GIVEN#2:
x(sq)+8x-48=0

Multiply the 1st and 3rd terms together to get the product of: -48x(sq)
Next, factor into two factors whose sum = middle term of 8x

1x times 48x
2x times 24x
3x times 16x
4x times 12x .........BINGO! -4x + 12x = 8x

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

SOLUTION #2:
First, bring down the 1st term: x(sq)
Next, write down the smaller of the two factors: -4x
Followed, by the LARGER of the two factors: 12x
Finally, bring down the 3rd term: -48

x(sq)-4x+12x-48=0

Now let's Factor by Grouping!
[x(sq)-4x] + [12x-48] = 0
GCF#1=x GCF#2=12

x(x-4) + 12(x-4) = 0
GCF#3=(x-4)
(x-2)(x+12)=0

Set each factor = 0 and solve for "x".
x-2=0
x-2+2=0+2
x+0=2
x = 2

AND

x+12=0
x+12-12=0-12
x + 0 = -12
x = -12

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

ANSWER#2:
x = -12 and 2

#3---------------------------------------------------

GIVEN#3:
x(sq)+12x+11=0

Multiply the 1st and 3rd terms together to get the product of: 11x(sq)
Next, factor into two factors whose sum = middle term of 12x

1x times 11x .........BINGO! 1x + 11x = 12x

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

SOLUTION#3:
First, bring down the 1st term: x(sq)
Next, write down the smaller of the two factors: 1x
Followed, by the LARGER of the two factors: 11x
Finally, bring down the 3rd term: +11

x(sq)+1x+11x+11=0

Now let's Factor by Grouping!
[x(sq)+1x] + [11x+11] = 0
GCF#1=x GCF#2=11

x(x+1) + 11(x+1) = 0
GCF#3=(x+1)
(x+1)(x+11)=0

Set each factor = 0 and solve for "x".
x+1=0
x+1-1=0-1
x+0 = -1
x = -1

AND

x+11=0
x+11-11=0-11
x + 0 = -11
x = -11
-------------------------------------

ANSWER #3:
x = -11 and -1

#4--------------------------------------------------

GIVEN#4:
10x(sq)+3x=1

First make sure the equation is set = 0

10x(sq)+3x-1=1-1
10x(sq)+3x-1=0

Multiply the 1st and 3rd terms together to get the product of: -10x(sq)
Next, factor into two factors whose sum = middle term of +3x

1x times 10x
2x times 5x .........BINGO! -2x +5x = +3x

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

SOLUTION#4:
First, bring down the 1st term:10x(sq)
Next, write down the smaller of the two factors: -2x
Followed, by the LARGER of the two factors: +5x
Finally, bring down the 3rd term of -1

10x(sq)-2x+5x-1=0

Now let's Factor by Grouping, again!
[10x(sq)-2x] + [5x-1] = 0
GCF#1=2x GCF#2=1

2x(5x-1) + 1(5x-1) = 0
GCF#3=(5x-1)
(5x-1)(2x+1) = 0

Set each factor = 0 and solve for "x".
5x-1=0
5x-1+1=0+1
5x+0=1
5x = 1
5x/5 =1/5
x=1/5

AND

2x+1=0
2x+1-1=0-1
2x+0 = -1
2x = -1
2x/2 = -1/2
x = -1/2

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

ANSWER #4:
x = -1/2 and 1/5

#5---------------------------------------------------

GIVEN#5:
3x(sq)+7x-31=0

Multiply the 1st and 3rd terms together to get the product of: -93x(sq)
Next, factor into two factors whose sum = middle term of 7x

1x times 93x
3x times 31x
There are no more factors for 93.

Therefore, the EQN is prime and we must use the quadratic formula to solve for "x".

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

STANDARD FORM OF A QUADRATIC EQUATION:
ax(sq) + bx + c = 0

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

THE QUADRATIC FORMULA:
x = { -b +/-(sqrt)[b(sq)-4ac] } / 2a

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

SOLUTION#5:
ax(sq) + bx + c = 0
3x(sq) + 7x - 31 = 0

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

LET:
a = 3
b = 7
c = -31

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

SUBSTITUTE for "a", "b", and "c" to solve for "x":

x = { -b (+/-)(sqrt)[b(sq)-4ac] } / 2a
x = { -(7) (+/-)(sqrt)[(7)(sq)-4(3)(-31)] } / [2(3)]
x = { -7 (+/-)(sqrt)[49-12(-31)] } / 6
x = { -7 (+/-)(sqrt)[49+372] } / 6

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

ANSWER#5:
x = { -7 (+/-)(sqrt)[421] } / 6

OR

"x" is approximately -4.586 and 2.253