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arsen [322]
3 years ago
13

Where do the equations y=2x and y=

{2} " align="absmiddle" class="latex-formula">-3 intersect
Mathematics
1 answer:
never [62]3 years ago
8 0
Set 2x=x²-3 to each other.
Subtract 2x from both sides and get the following

x²-3-2x

Rearranging the formula to make factoring easier, we get
x²-2x-3

Factor and you should get the following:
(x-3)(x+1)

set both equal to 0 and solve for x.

x-3=0⇒x=3
x+1=0⇒x=-1

Now plug both of these values back into one of your original equations to get your y-values. It does not matter which equation you plug your xs back into due to the fact they intersect at the same x, and thus will also intersect at the same y.

For times sake,
y=2x
plug in x=3 and get y=6
plug in x=-1 and get y=-2

Ergo, the equations y=2x and y=x²-3 intersect at
(3,6) and (-1,-2)
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If x = (√2 + 1)^-1/3 then the value of x^3 + 1/x^3 is​
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Step-by-step explanation:

<u>Given</u><u>:</u> x = {√(2) + 1}^(-1/3)

<u>Asked</u><u>:</u> x³+(1/x³) = ?

<u>Solution</u><u>:</u>

We have, x = {√(2) + 1}^(-1/3)

⇛x = [1/{√(2) + 1}^(1/3)]

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Cubing on both sides, then

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⇛x³ = [1/{√(2) + 1}]

Here, we see that on RHS, the denominator is √(2)+1. We know that the rationalising factor of √(a)+b = √(a)-b. Therefore, the rationalising factor of √(2)+1 = √(2) - 1. On rationalising the denominator them

⇛x³ = [1/{√(2) + 1}] * [{√(2) - 1}/{√(2) - 1}]

⇛x³ = [1{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Multiply the numerator with number outside of the bracket with numbers on the bracket.

⇛x³ = [{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Now, Comparing the denominator on RHS with (a+b)(a-b), we get

  • a = √2
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Using identity (a+b)(a-b) = a² - b², we get

⇛x³ = [{√(2) - 1}/{√(2)² - (1)²}]

⇛x³ = [{√(2) - 1}/{√(2*2) - (1*1)}]

⇛x³ = [{√(2) - 1}/(2-1)]

⇛x³ = [{√(2) - 1}/1]

Therefore, x³ = √(2) - 1 → → →Eqn(1)

Now, 1/x³ = [1/{√(2) - 1]

⇛1/x³ = [1/{√(2) - 1] * [{√(2) + 1}/{√(2) + 1}]

⇛1/x³ = [1{√(2) + 1}/{√(2) - 1}{√(2) + 1}]

⇛1/x³ = {√(2) + 1}/[{√(2) - 1}{√(2) + 1}]

⇛1/x³ = [{√(2) + 1}/{√(2)² - (1)²}]

⇛1/x³ = [{√(2) + 1}/{√(2*2) - (1*1)}]

⇛1/x³ = [{√(2) + 1}/(2-1)]

⇛1/x³ = [{√(2) + 1}/1]

Therefore, 1/x³ = √(2) + 1 → → →Eqn(2)

On adding equation (1) and equation (2), we get

x³ + (1/x³) = √(2) -1 + √(2) + 1

Cancel out -1 and 1 on RHS.

⇛x³ + (1/x³) = √(2) + √(2)

⇛x³ + (1/x³) = 2

Therefore, x³ + (1/x³) = 2

<u>Answer</u><u>:</u> Hence, the required value of x³ + (1/x³) is 2.

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