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3 years ago

Using the quadratic formula to solve x^2=5-x what are the values of x?

Mathematics

2 answers:

kenny6666 [7]3 years ago

3 0

A
you have to solve x^2+x-5=0
So x=(-1 +/- sqrt(1^2+4*1*5))/2
x=(-1 +/- sqrt(21))/2

Law Incorporation [45]3 years ago

3 0

Answer:

Option a - $x=\frac{-1\pm\sqrt{21}}{2}$

Step-by-step explanation:

Given : Expression $x^2=5-x$

To find : What are the values of x?

Solution :

The quadratic formula of the quadratic equation $ax^2+bx+c=0$ is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Comparing with general equation, $x^2+x-5=0$

where, a=1 ,b=1, c=-5

Substitute in the formula,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm\sqrt{1^2-4(1)(-5)}}{2(1)}$

$x=\frac{-1\pm\sqrt{1+20}}{2}$

$x=\frac{-1\pm\sqrt{21}}{2}$

Therefore, The value of x is $x=\frac{-1\pm\sqrt{21}}{2}$

So, Option a is correct.

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a model rocket is launched from the ground with an initial velocity of 60 feet per second. the function g(t)= -16t^2 + 60t repre

Lemur [1.5K]

If we use our trusty ti's this will be a breeze
domain is the numbers we can use
obviously we can't have negative time so therefor the domain is all positive integers (0,1,2,2.3, 3/2,3,pi...)

zeroes is when you set absolute max using ti is
remember vertex form which is
max height of something in
ax^2+bx+c form is -b/2a
-16t^2+60t+0
a=-16
b=60
-60/(2 times -16)
-60/-32=30/16=15/8=1 and 7/8 so subsitute for t
g(15/8)=-16(15/8)^2+60(15/8)=225/4=56.25=max height

zeroes is when the equation equals 0
so set it to zero
0=-16t^2+60t
factor
0=(-4t)(4x-15)
set each to zero
-4x=0
x=0
4x-15=0
add 15
4x=15
divid 4
x=15/4
so the zeros are t=0 and t=15/4

domain is all the nuumbers that can be used logically for time
logically, we cannot have negative time, so all real positive
[0,∞)
(that means from 0 to infinity includng 0 so 0<span><</span>t<∞))

range is the output
output=height
we find the min height and max height
min=0
max=56.25
so range=0 to 56.25 or
[0,56.25]

max height=56.25 ft (15/8 seconds)
zeroes=0 sec and 15/4 sec
domain=all real positive numbers including zero or [0, ∞)
range=[0,56.25]

3 0

3 years ago

Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!

astra-53 [7]

(a) By DeMoivre's theorem, we have

$(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

On the LHS, expanding yields

$\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta$

Matching up real and imaginary parts, we have for (i) and (ii),

$\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$
$\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta$

(b) By the definition of the tangent function,

$\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}$
$=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}$

$=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
$=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

(c) Setting $\theta=\dfrac\pi5$ , we have $t=\tan\dfrac\pi5$ and $\tan5\left(\dfrac\pi5\right)=\tan\pi=0$ . So

$0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

At the given value of $t$ , the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.

$0=t^5-10t^3+5t\implies0=t^4-10t^2+5$

Remember, this is saying that

$0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5$

If we replace $\tan^2\dfrac\pi5$ with a variable $x$ , then the above means $\tan^2\dfrac\pi5$ is a root to the quadratic equation,

$x^2-10x+5=0$

Also, if $\theta=\dfrac{2\pi}5$ , then $t=\tan\dfrac{2\pi}5$ and $\tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0$ . So by a similar argument as above, we deduce that $\tan^2\dfrac{2\pi}5$ is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

$x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)$

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

$5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5$

But $\tanx>0$ for all $0$ , as is the case for $x=\dfrac\pi5$ and $x=\dfrac{2\pi}5$ , so we choose the positive root.

3 0

3 years ago

Help asap!!!!!!!!!! Thank u

Tema [17]

Answer:

The answer is: $m\angle 5=34^\circ$

Step-by-step explanation:

angles 1 and 5 are equal to each other since $a\parallel b$

so $m\angle 5=m\angle 1=34^\circ$

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3 years ago

What is the slope of the line that passes through (-2, 5) and (3, 9)?

Naya [18.7K]

Answer: 4/5

Step-by-step explanation:

1. you need to know the formula which is m = y2-y1/x2-x1

2. you plug in the numbers which would be 9-5/3-(-2)

3. you get your answer as 4/5

4. the answer is 4/5

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