All categories

3 years ago

What are the exact solutions of x2 − 3x − 1 = 0?

2 answers:

7 0

Hi, the quadratic formula is: $\boxed{X= \dfrac{-b \pm \sqrt{b ^{2}-4ac } }{2a}}$

Values: a = 1; b = - 3; c = -1
Solving:

$X ^{2}-3X-1=0$

$X= \dfrac{-(-3) \pm \sqrt{(-3) ^{2}-4(1)(-1) } }{2(1)}$

$X= \dfrac{3 \pm \sqrt{9+4} }{2}$

$X= \dfrac{3 \pm \sqrt{13} }{2}\ ===\ \textgreater \ \ Answer:\ \boxed{ \textbf{Option\ {C.}}}$

GOOD LUCK...!!!

kobusy [5.1K]3 years ago

3 0

Option C is the answer!

You might be interested in

What is 2345.232×1324.221 need it asap for class in 6th grade

lara [203]

3105605.46427
hope that helps

7 0

3 years ago

Read 2 more answers

True or False for each statement.

9966 [12]

1) true
2) false
3) true
4) false
5) true

7 0

3 years ago

Your family wants to purchase a new laptop with

zloy xaker [14]

Answer:

The laptop will not fit in the case.

Step-by-step explanation:

We need to find the width of the laptop to know if it fits in the case. We have:

<u>For the laptop</u>:

h: is the height of the laptop = 10 inch

d: is the widescreen of the laptop (diagonal) = 17 inch

w: is the width of the laptop =?

<u>For the laptop case</u>:

H: is the height of the laptop case = 10.5 inch

W: is the width of the laptop case = 13 inch

So, the width of the laptop can be found by using Pitagoras:

$d^{2} = w^{2} + h^{2}$

$w = \sqrt{(17 inch)^{2} - (10 inch)^{2}} = 13.7 inch$

Hence, the laptop is 13.7 inches wide.

Therefore, since the laptop is wider than the case, the laptop will not fit in the case.

I hope it helps you!

8 0

2 years ago

Distance between 0,2 and 8,8

Kitty [74]

Answer:

8.6

Step-by-step explanation:

7 0

3 years ago

14, 16, and 20 using elimination method showing work. Thanks so much

Nady [450]

14) x=0, y=3, z=-2

Solution Set (0,3,-2)

16) x=1, y=1 and z=1

Solution set = (1,1,1)

20) x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Step-by-step explanation:

14)

$x-y+2z=-7\\y+z=1\\x=2y+3z$

Rearranging and solving:

$x-y+2z=-7\,\,\,eq(1)\\y+z=1\,\,\,eq(2)\\x-2y-3z=0\,\,\,eq(3)$

Eliminate y:

Adding eq(1) and eq(2)

$x-y+2z=-7\,\,\,eq(1)\\ 0x+y+z=1\,\,\,eq(2)\\-------\\x+3z=-6\,\,\,eq(4)$

Multiply eq(2) with 2 and add with eq(3)

$0x+2y+2z=2\,\,\,eq(2)\\\\x-2y-3z=0\,\,\,eq(3)\\--------\\x-z=2\,\,\,eq(5)$

Eliminate x:

Subtract eq(4) and eq(5)

$x+3z=-6\,\,\,eq(4)\\x-z=2\,\,\,eq(5)\\-\,\,\,+\,\,\,\,\,\,-\\---------\\4z=-8\\z= -2$

So, value of z = -2

Now putting value of z in eq(2)

$y+z=1\\y+(-2)=1\\y-2=1\\y=1+2\\y=3$

So, value of y = 3

Now, putting value of z and y in eq(1)

$x-y+2z=-7\\x-(3)+2(-2)=-7\\x-3-4=-7\\x-7=-7\\x=-7+7\\x=0$

So, value of x = 0

So, x=0, y=3, z=-2

S.S(0,3,-2)

16)

$3x-y+z=3\\\x+y+2z=4\\x+2y+z=4$

Let:

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\x+2y+z=4\,\,\,eq(3)$

Eliminating y:

Adding eq(1) and (2)

$3x-y+z=3\,\,\,eq(1)\\x+y+2z=4\,\,\,eq(2)\\---------\\4x+3z=7\,\,\,eq(4)$

Multiply eq(1) by 2 and add with eq(3)

$6x-2y+2z=6\,\,\,eq(1)\\x+2y+z=4\,\,\,eq(3)\\---------\\7x+3z=10\,\,\,eq(5)$

Now eliminating z in eq(4) and eq(5) to find value of x

Subtracting eq(4) and eq(5)

$4x+3z=7\,\,\,eq(4)\\7x+3z=10\,\,\,eq(5)\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,-\\-----------\\-3x=-3\\x=-3/-3\\x=1$

So, value of x = 1

Putting value of x in eq(4) to find value of x:

$4x+3z=7\\4(1)+3z=7\\4+3z=7\\3z=7-4\\z=3/3\\z=1$

So, value of z = 1

Putting value of x and z in eq(2) to find value of y:

$x+y+2z=4\\1+y+2(1)=4\\1+y+2=4\\y+3=4\\y=4-3\\y=1$

So, x=1, y=1 and z=1

Solution set = (1,1,1)

20)

$x+4y-5z=-7\\3x+2y+2z=-7\\2x+y+5z=8$

Let:

$x+4y-5z=-7\,\,\,eq(1)\\3x+2y+2z=-7\,\,\,eq(2)\\2x+y+5z=8\,\,\,eq(3)$

Solving:

Eliminating z :

Adding eq(1) and eq(3)

$x+4y-5z=-7\,\,\,eq(1)\\2x+y+5z=8\,\,\,eq(3)\\---------\\3x+5y=1\,\,\,eq(4)$

Multiply eq(1) with 2 and eq(2) with 5 and add:

$2x+8y-10z=-14\,\,\,eq(1)\\15x+10y+10z=-35\,\,\,eq(2)\\----------\\17x+18y=-49\,\,\,eq(5)$

Eliminate y:

Multiply eq(4) with 18 and eq(5) with 5 and subtract:

$54x+90y=18\\85x+90y=-245\\-\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\\-------\\-31x=158\\x=-\frac{263}{31}$

So, value of x = -263/31

Putting value of x in eq(4)

$3x+5y=1\\3(-\frac{263}{31})+5y=1\\-\frac{789}{31}+5y=1 \\5y=1+\frac{789}{31}\\5y=\frac{820}{31}\\y=\frac{820}{31*5}\\y=\frac{164}{31}$

Now putting x = -263/31 and y=164/31 in eq(1) and finding z:

We get z=122/31

So, x = -263/31, y=164/31 ,z=122/31

Solution set (-263/31, 164/31 ,122/31)

Keywords: Solving system of Equations

Learn more about Solving system of Equations at:

#learnwithBrainly

4 0

2 years ago