04.06.2020

Factor quadratic equation. 6x squared +10-56

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24.06.2023, solved by verified expert
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Factor quadratic equation. 6x squared +10-56, №17886464, 04.06.2020 00:12

Step-by-step explanation:

Factor using the Quadratic formula

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Mathematics
Step-by-step answer
P Answered by PhD

x=\frac{±\sqrt{69} }{3}

Step-by-step explanation:

Factor using the Quadratic formula

Mathematics
Step-by-step answer
P Answered by Master

Part 1) x=3

Part 2) x = −1.11 and x = 1.11

Part 3) 105

Part 4) a = −6, b = 9, c = −7

Part 5) x equals 5 plus or minus the square root of 33, all over 2

Part 6) In the procedure

Part 7) -0.55

Part 8) The denominator is 2

Part 9) a = −6, b = −8, c = 12

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

Part 1)

in this problem we have

x^{2} -6x+9=0  

so

a=1\\b=-6\\c=9

substitute in the formula

x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(9)}} {2(1)}

x=\frac{6(+/-)\sqrt{0}} {2}

x=\frac{6} {2}=3

Part 2) in this problem we have

49x^{2} -60=0  

so

a=49\\b=0\\c=-60

substitute in the formula

x=\frac{0(+/-)\sqrt{0^{2}-4(49)(-60)}} {2(49)}

x=\frac{0(+/-)\sqrt{11,760}} {98}

x=(+/-)1.11

Part 3) When the solution of x2 − 9x − 6 is expressed as 9 plus or minus the square root of r, all over 2, what is the value of r?

in this problem we have

x^{2} -9x-6=0  

so

a=1\\b=-9\\c=-6

substitute in the formula

x=\frac{-(-9)(+/-)\sqrt{-9^{2}-4(1)(-6)}} {2(1)}

x=\frac{9(+/-)\sqrt{105}} {2}

therefore

r=105

Part 4) What are the values a, b, and c in the following quadratic equation?

−6x2 = −9x + 7

in this problem we have

-6x^{2}=-9x+7  

-6x^{2}+9x-7=0  

so

a=-6\\b=9\\c=-7

Part 5) Use the quadratic formula to find the exact solutions of x2 − 5x − 2 = 0.

In this problem we have  

x^{2} -5x-2=0  

so

a=1\\b=-5\\c=-2

substitute in the formula

x=\frac{-(-5)(+/-)\sqrt{-5^{2}-4(1)(-2)}} {2(1)}

x=\frac{5(+/-)\sqrt{33}} {2}

therefore  

x equals 5 plus or minus the square root of 33, all over 2

Part 6) Quadratic Formula proof

we have

ax^{2} +bx+c=0  

Divide both sides by a

x^{2} +(b/a)x+(c/a)=0  

Complete the square

x^{2} +(b/a)x=-(c/a)  

x^{2} +\frac{b}{a}x+\frac{b^{2}}{4a^{2}} =-\frac{c}{a}+\frac{b^{2}}{4a^{2}}

Rewrite the perfect square trinomial on the left side of the equation as a binomial squared

(x+\frac{b}{2a})^{2}=-\frac{4ac}{a^{2}}+\frac{b^{2}}{4a^{2}}

Find a common denominator on the right side of the equation

(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}

Take the square root of both sides of the equation

(x+\frac{b}{2a})=(+/-)\sqrt{\frac{b^{2}-4ac}{4a^{2}}}

Simplify the right side of the equation

(x+\frac{b}{2a})=(+/-)\frac{\sqrt{b^{2}-4ac}}{2a}

Subtract the quantity b over 2 times a from both sides of the equation

x=-\frac{b}{2a}(+/-)\frac{\sqrt{b^{2}-4ac}}{2a}

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

Part 7) in this problem we have  

3x^{2} +45x+24=0  

so

a=3\\b=45\\c=24

substitute in the formula

x=\frac{-(45)(+/-)\sqrt{45^{2}-4(3)(24)}} {2(3)}

x=\frac{-(45)(+/-)\sqrt{1,737}} {6}

x1=\frac{-(45)(+)\sqrt{1,737}} {6}=-0.55

x2=\frac{-(45)(-)\sqrt{1,737}} {6}=-14.45

therefore

The other solution is

-0.55

Part 8) in this problem we have

2x^{2} -8x+7=0  

so

a=2\\b=-8\\c=7

substitute in the formula

x=\frac{-(-8)(+/-)\sqrt{-8^{2}-4(2)(7)}} {2(2)}

x=\frac{8(+/-)\sqrt{8}} {4}

x=\frac{8(+/-)2\sqrt{2}} {4}

x=\frac{4(+/-)\sqrt{2}} {2}

therefore

The denominator is 2

Part 9) What are the values a, b, and c in the following quadratic equation?

−6x2 − 8x + 12

in this problem we have

-6x^{2} -8x+12=0  

so

a=-6\\b=-8\\c=12

Mathematics
Step-by-step answer
P Answered by Master
1:  x=3, x=1
2:  x= -5
3:  There are 2 real solutions.
4:  There are 2 real solutions.
5:  There are no real solutions.
6.  There is 1 real solution.
7.  x=\frac{2\pm \sqrt{10}}{3}
8.  x= -6, x = -2
9.  x = -1/6, x=1
10.  x=\frac{3\pm \sqrt{5}}{2}

Explanation:
1.  The quadratic formula is 
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
Substituting our known information we have:
x=\frac{--4\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)}
\\
\\=\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm \sqrt{4}}{2}=\frac{4\pm2}{2}
\\
\\=\frac{4+2}{2},\frac{4-2}{2}=\frac{6}{2},\frac{2}{2}=3,1
2.  Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:
x=\frac{-10\pm \sqrt{10^2-4(1)(25)}}{2(1)}=\frac{-10\pm \sqrt{100-100}}{2}
\\
\\=\frac{-10\pm \sqrt{0}}{2}=\frac{-10\pm0}{2}=\frac{-10}{2}=-5
3.  The discriminant is b²-4ac.  For this problem, that is 20²-4(-4)(25)=400--400=800.  Since this is greater than 0, there are 2 real solutions.
4.  The discriminant in this problem is 7²-4(2)(-15)=49--120=49+120=169.  This is greater than 0, so there are 2 real solutions.
5.  The discriminant in this problem is 1²-4(-2)(-28)=1-224=-223.  Since this is less than 0, there are no real solutions.
6.  If the discriminant of a quadratic is 0, then by definition there is 1 real solution.
7.  Rewriting the quadratic we have 3x²-4x-2=0.  Using the quadratic formula we have:
x=\frac{--4\pm \sqrt{(-4)^2-4(3)(-2)}}{2(3)}=\frac{4\pm \sqrt{16--24}}{6}
\\
\\=\frac{4\pm \sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}=\frac{2\pm \sqrt{10}}{3}
8.  Factoring this trinomial we want factors of 12 that sum to 8.  6*2 = 12 and 6+2=8, so those are our factors.  This gives us:
(x+6)(x+2)=0
Using the zero product property we know that either x+6=0 or x+2=0.  Solving these equations we get x= -6 or x= -2.
9.  Factoring this trinomial we want factors of 6(-1)=-6 that sum to -5.  (-6)(1)=-6 and -6+1=-5, so this is how we "split up" the x term:
6x²-6x+1x-1=0
We group together the first two and the last two terms:
(6x²-6x)+(1x-1)=0
Factor the GCF out of each group.  In the first group, that is 6x:
6x(x-1)+(1x-1)=0
In the second group, the GCF is 1:
6x(x-1)+1(x-1)=0
Both terms have a factor of (x-1), so we can factor it out:
(x-1)(6x+1)=0
Using the zero product property, we know either x-1=0 or 6x+1=0.  Solving these equations we get x=1 or x=-1/6.
10.  Substituting our information into the quadratic formula we get:
x=\frac{--3\pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}=\frac{3\pm \sqrt{9-4}}{2}
\\
\\=\frac{3\pm \sqrt{5}}{2}
Mathematics
Step-by-step answer
P Answered by Master
1:  x=3, x=1
2:  x= -5
3:  There are 2 real solutions.
4:  There are 2 real solutions.
5:  There are no real solutions.
6.  There is 1 real solution.
7.  x=\frac{2\pm \sqrt{10}}{3}
8.  x= -6, x = -2
9.  x = -1/6, x=1
10.  x=\frac{3\pm \sqrt{5}}{2}

Explanation:
1.  The quadratic formula is 
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
Substituting our known information we have:
x=\frac{--4\pm \sqrt{(-4)^2-4(1)(3)}}{2(1)}
\\
\\=\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm \sqrt{4}}{2}=\frac{4\pm2}{2}
\\
\\=\frac{4+2}{2},\frac{4-2}{2}=\frac{6}{2},\frac{2}{2}=3,1
2.  Rewriting the quadratic in standard form we have x²+10x-25=0. Substituting this into the quadratic formula gives us:
x=\frac{-10\pm \sqrt{10^2-4(1)(25)}}{2(1)}=\frac{-10\pm \sqrt{100-100}}{2}
\\
\\=\frac{-10\pm \sqrt{0}}{2}=\frac{-10\pm0}{2}=\frac{-10}{2}=-5
3.  The discriminant is b²-4ac.  For this problem, that is 20²-4(-4)(25)=400--400=800.  Since this is greater than 0, there are 2 real solutions.
4.  The discriminant in this problem is 7²-4(2)(-15)=49--120=49+120=169.  This is greater than 0, so there are 2 real solutions.
5.  The discriminant in this problem is 1²-4(-2)(-28)=1-224=-223.  Since this is less than 0, there are no real solutions.
6.  If the discriminant of a quadratic is 0, then by definition there is 1 real solution.
7.  Rewriting the quadratic we have 3x²-4x-2=0.  Using the quadratic formula we have:
x=\frac{--4\pm \sqrt{(-4)^2-4(3)(-2)}}{2(3)}=\frac{4\pm \sqrt{16--24}}{6}
\\
\\=\frac{4\pm \sqrt{40}}{6}=\frac{4\pm 2\sqrt{10}}{6}=\frac{2\pm \sqrt{10}}{3}
8.  Factoring this trinomial we want factors of 12 that sum to 8.  6*2 = 12 and 6+2=8, so those are our factors.  This gives us:
(x+6)(x+2)=0
Using the zero product property we know that either x+6=0 or x+2=0.  Solving these equations we get x= -6 or x= -2.
9.  Factoring this trinomial we want factors of 6(-1)=-6 that sum to -5.  (-6)(1)=-6 and -6+1=-5, so this is how we "split up" the x term:
6x²-6x+1x-1=0
We group together the first two and the last two terms:
(6x²-6x)+(1x-1)=0
Factor the GCF out of each group.  In the first group, that is 6x:
6x(x-1)+(1x-1)=0
In the second group, the GCF is 1:
6x(x-1)+1(x-1)=0
Both terms have a factor of (x-1), so we can factor it out:
(x-1)(6x+1)=0
Using the zero product property, we know either x-1=0 or 6x+1=0.  Solving these equations we get x=1 or x=-1/6.
10.  Substituting our information into the quadratic formula we get:
x=\frac{--3\pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}=\frac{3\pm \sqrt{9-4}}{2}
\\
\\=\frac{3\pm \sqrt{5}}{2}
Mathematics
Step-by-step answer
P Answered by PhD

C, A, D

edit - sorry the title was wrong earlier

Step-by-step explanation:

Question 1

in the quadratic formula, the equation is in the format of ax^2 + bx + c

so rearrange this equation so all values are on one side

-6x^2 = -9x + 7

-6x^2 + 9x - 7 =0

so a = -6, b = 9, c = -7

answer = C

Question 2

note that all the values are already on one side for this, so repeat the process from question 1 where the equation is in the format of ax^2 + bx + c

the equation = −6x2 − 8x + 12 = 0

so a = -6, b = -8, c = 12

answer = A

Question 3

(see picture below for steps)

following the same process above, a = 4, b = 45, and c = 24, so plug these values in the quadratic equation shown in the picture below

so you get the answers x = -10.69 and x = -0.56 after putting them in a calculator

answer = D


Answer these three questions for 100 points Question 1) Identifying the values a, b, and c is the fi
Mathematics
Step-by-step answer
P Answered by PhD

C, A, D

edit - sorry the title was wrong earlier

Step-by-step explanation:

Question 1

in the quadratic formula, the equation is in the format of ax^2 + bx + c

so rearrange this equation so all values are on one side

-6x^2 = -9x + 7

-6x^2 + 9x - 7 =0

so a = -6, b = 9, c = -7

answer = C

Question 2

note that all the values are already on one side for this, so repeat the process from question 1 where the equation is in the format of ax^2 + bx + c

the equation = −6x2 − 8x + 12 = 0

so a = -6, b = -8, c = 12

answer = A

Question 3

(see picture below for steps)

following the same process above, a = 4, b = 45, and c = 24, so plug these values in the quadratic equation shown in the picture below

so you get the answers x = -10.69 and x = -0.56 after putting them in a calculator

answer = D


Answer these three questions for 100 points Question 1) Identifying the values a, b, and c is the fi
Mathematics
Step-by-step answer
P Answered by Master

F,A,D,C,E,B

Step-by-step explanation:

I had this problem on algebra nation and got it right

Mathematics
Step-by-step answer
P Answered by Master

F,A,D,C,E,B

Step-by-step explanation:

I had this problem on algebra nation and got it right

Mathematics
Step-by-step answer
P Answered by PhD
Step 2

Step-by-step explanation:

Given the quadratic equation  6x² + 24x + 7 = 0, the following steps must be followed when finding the roots using the completing the square method.

Step 1: Subtract 7 from both sides

6x² + 24x + 7 - 7= 0 -7

6x² + 24x = -7

Step 2: divide through by the coefficient of x²

6x²/6 + 24x/6 = -7/6

x² + 4x = -7/6

Step 3: Add half of the coefficient of x (i.e 4/2 = 2) to both sides of the equation to complete the square of the equation x² + 4x.

x² + 4x + 2= -7/6 + 2

(x+2)² = -7/6 + 2

(x+2)²  = 5/6

Step 4: taking the square root of both sides

√(x+2)²  = √5/6

x+2 = ±√5/6

x = -2±√5/6

x = -2+√5/6 and -2-√5/6

Based on the conclusion above, it can be seen that Yvonne made the first error in step 2. Yvonne suppose to divide through by the coefficient of x² (i.e 6)

Mathematics
Step-by-step answer
P Answered by PhD
Step 2

Step-by-step explanation:

Given the quadratic equation  6x² + 24x + 7 = 0, the following steps must be followed when finding the roots using the completing the square method.

Step 1: Subtract 7 from both sides

6x² + 24x + 7 - 7= 0 -7

6x² + 24x = -7

Step 2: divide through by the coefficient of x²

6x²/6 + 24x/6 = -7/6

x² + 4x = -7/6

Step 3: Add half of the coefficient of x (i.e 4/2 = 2) to both sides of the equation to complete the square of the equation x² + 4x.

x² + 4x + 2= -7/6 + 2

(x+2)² = -7/6 + 2

(x+2)²  = 5/6

Step 4: taking the square root of both sides

√(x+2)²  = √5/6

x+2 = ±√5/6

x = -2±√5/6

x = -2+√5/6 and -2-√5/6

Based on the conclusion above, it can be seen that Yvonne made the first error in step 2. Yvonne suppose to divide through by the coefficient of x² (i.e 6)

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