Solve the following Quadratic Equation and find the value of x

[@ishanshu]

 

[Deleted member 4993]

View attachment 30337

How do you know that the function in question is a quadratic function?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[Steven G]

I suspect that the power of 7 should be 1-x

If it is, then you can see one solution by inspection
Clearly you want 7^1-x + x^1/3 = 2 (so when you subtract 2 you get 0)

You should note that y⁰ = 1 and 1^y = 1.
Now using these two facts, how to you get 2 from 7^1-x + x^1/3?

 

[BigBeachBanana]

There's clearly a solution when graphing the equation.

 

[Steven G]

There's clearly a solution when graphing the equation.
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I too graphed it and did not get any roots. I wonder which function I graphed!

 

[BigBeachBanana]

I too graphed it and did not get any roots. I wonder which function I graphed!

I did find a solution through the method of graphing. However, I'm wondering what's the required method of the OP.

 

[Deleted member 4993]

I did find a solution through the method of graphing. However, I'm wondering what's the required method of the OP.

@bbb , What solution did you get?

If you take the exponent of 7 to be (1-x) then there are two solutions → x = 1 and x = ~8

If you take the exponent of 7 to be (1+x) then there is no solution (according plot by WA).

 

[Otis]

I did find a solution through the method of graphing

You didn't graph the given equation. It has no Real solutions.

 

[BigBeachBanana]

@bbb , What solution did you get?

If you take the exponent of 7 to be (1-x) then there are two solutions → x = 1 and x = ~8

If you take the exponent of 7 to be (1+x) then there is no solution (according plot by WA).

You didn't graph the given equation. It has no Real solutions.

 

[Otis]

From wikipedia, "Unless specified otherwise, the principal root is often implied when talking about roots".

 

[BigBeachBanana]

From wikipedia, "Unless specified otherwise, the principal root is often implied when talking about roots".

Yes, if we're talking about principal based roots, we have no real solutions.