Who Has The Bulls Eye?

Algebra Level 5

$\dfrac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}-\dfrac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=8x\sqrt{x^2-3x+2}$

How many real values of $x$ satisfy the above equation?

Note/Hint : Real roots may lead to complex values.

3 1 None of these choices 4 2

2 solutions

Chaitnya Shrivastava
Feb 24, 2016

Begin by taking LCM of LHS and simplifying the numerator and denominator which leaves us with $4x*$ $\sqrt{x^2-1}$ as numerator and 1 as denominator now writing it along with the value of RHS we get $4x*$ $\sqrt{x^2-1}$ $=8x*$ $\sqrt{x^2-{3x+2}}$ Now here if we cancel out x we will miss 0 as one of its roots but we can do that keeping in our mind 0 as one of its roots. NOW we cancel out x along with other common factors from both sides and then square both sides which after rearranging leaves us with the final equation $3x^2-12x+9=0$ whose roots are 3 and 1.So we finally get 3 values of x =( 0,1&3).We should check whether they satisfy the orignal equation which all three do and 0 lea||

How $x=0$ can be a solution ? Though, plugging $x=0$ , yields both RHS and LHS equal to 0, but that doesn't mean it is a solution to the given equation because individual terms are assuming complex values at $x=0$ . Rectify me if I'm wrong.

Aditya Sky - 5 years, 3 months ago

1》Individual terms assuming complex values doesn't mean 0 is not a solution. 2》To prove 0 as a solution, in the equation $4x*$ $\sqrt{x^2-1}$ $=8x*$ $\sqrt{x^2-{3x+2}}$ bring everything on LHS i.e. $4x*$ $\sqrt{x^2-1}$ $-8x*$ $\sqrt{x^2-{3x+2}}$ $=0$ Now take $4x$ common which implies to $4x*[$ $\sqrt{x^2-1}$ $-2*$ $\sqrt{x^2-{3x+2}}$ $]= 0$ Hence either $4x=0$ i.e. $x=0$ or $\sqrt{x^2-1}$ $-2*$ $\sqrt{x^2-{3x+2}}$ $=0$ the second case again leads us to the equation $3x^2-12x+9=0$ .In this way get our three roots.Hope you are convinced 😇.

Chaitnya Shrivastava - 5 years, 3 months ago

Awawa... Why did that hint have to be there... I also disregarded $x=0$ because of that...

Manuel Kahayon - 5 years, 3 months ago

The hint was there just to help because zero must have confused many.😊

Chaitnya Shrivastava - 5 years, 3 months ago

@Chaitnya Shrivastava – Yeah. Though that hint still scared me a bit initially.

Shreyash Rai - 5 years, 3 months ago

Avi Solanki
Mar 6, 2016

U can apply componendo and dividendo and get three.solutions 1,0,3 as well.

Hello, that seems to be an interesting idea. Can you show us your working?

Rohit Udaiwal - 5 years, 3 months ago

Who Has The Bulls Eye?

2 solutions

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