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3 years ago

He equation 8 − 4x = 0 has real solution(s)

Mathematics

2 answers:

Rzqust [24]3 years ago

6 0

Answer:

1 real solution, x=2

Step-by-step explanation:

Let's solve for x and see how many solutions exist.

To solve for x, we must get x isolated. To do this, perform the inverse operation to both sides of the equation.

8-4x=0

4x is being subtracted from 8. The inverse, or opposite, of subtraction is addition. Add 4x to both sides.

8-4x+4x=0+4x

8=0+4x

8=4x

x is being multiplied by 4. The inverse of multiplication is division. Divide both sides by 4.

8/4=4x/4

8/4=x

2=x

x is equal to 2, therefore there is 1 real solution.

jarptica [38.1K]3 years ago

3 0

Answer:

x=-2

Step-by-step explanation:

subtract 8 from both sides you get -4x=-8

divide by -4

you get x=-2

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What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where

Setler79 [48]

Answer:

$k=-16,k=-8,k=8,k=16$

Step-by-step explanation:

We are given quadratic equations as

$x^2+kx+15$

and it can be factored as

$=(x+a)(x+b)$

now, we can multiply factor term

$(x+a)(x+b)=x^2+(a+b)x+ab$

now, we can compare

$x^2+(a+b)x+ab=x^2+kx+15$

so, we get

$k=a+b$

$ab=15$

we are given that

'a' and 'b' are integers

so, we can find all possible factors

$15=(-1\times -15),(1\times 15)$

$15=(-3\times -5),(3\times 5)$

so, we can find k

At $(-1\times -15)$ :

$k=a+b$

we can plug values

$k=-1-15$

$k=-16$

At $(1\times 15)$ :

$k=a+b$

we can plug values

$k=1+15$

$k=16$

At $(-3\times -5)$ :

$k=a+b$

we can plug values

$k=-3-5$

$k=-8$

At $(3\times 5)$ :

$k=a+b$

we can plug values

$k=3+5$

$k=8$

So, values of k are

$k=-16,k=-8,k=8,k=16$

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