Algebra without a calculator

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Main Question or Discussion Point

How would one solve the following WITHOUT a calculator?

Sub in t as [itex]x^{\frac{1}{2}}[/itex]

[itex]4x+8=33x^{\frac{1}{2}}[/itex]

I just don't see how to do this without a calculator.


If I take t^2=x, then sub it in I need to do the quadratic formula which is too hard without a calculator/would take too long in an exam. Even if I do the formula I get t=8,2 which I would then need to square route to get x but would be the wrong answer since x = 64 and 1/16....


Any help. Is t^2=x a wrong assumption? Surely it is right since x^1/2 squared gives x?
 

Answers and Replies

  • #2
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How would one solve the following WITHOUT a calculator?

Sub in t as [itex]x^{\frac{1}{2}}[/itex]

[itex]4x+8=33x^{\frac{1}{2}}[/itex]

I just don't see how to do this without a calculator.


If I take t^2=x, then sub it in I need to do the quadratic formula which is too hard without a calculator/would take too long in an exam. Even if I do the formula I get t=8,2 which I would then need to square route to get x but would be the wrong answer since x = 64 and 1/16....


Any help. Is t^2=x a wrong assumption? Surely it is right since x^1/2 squared gives x?
Your substitution is the right one. After making the substitution, you will have a quadratic equation that can be factored.

There are two solutions for t. One solution for t is t = 8, but your other solution (t = 2) is not a solution.

Please show us your work.
 
  • #3
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Your substitution is the right one. After making the substitution, you will have a quadratic equation that can be factored.

There are two solutions for t. One solution for t is t = 8, but your other solution (t = 2) is not a solution.

Please show us your work.
Sorry for not showing my work, I just did so much and failed I thought it would be useless.

Nonetheless I tried 4t^2-33t+8=0 and then factoring that but without using the quadratic formula to give me the answer of t=8, I didn't see how else I could do it (the quadratic formula requiring a calculator since you have to square 33 to make 1089 and then get a square route...).

Then I tried factorising it into 4(t^2+2)=33t but I didn't really see where to go from there. Most of my working has just been repetitions of these.

I'm fairly sure that you can factorise it into (2t + 16)(2t + 1/2) but there was no way I would have ever gotten that without a calculator, so it is again useless.

Do you know of any methods through which to solve this?

Thanks for replying!



Quadratic formula workings; (33+sqrt(33^2-4(4)(8)))/8=8, 1/4

But then 8 and 1/4 are not the right answers (the answers are 64 and 1/16) :(
 
  • #4
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But then, how can 8 and 1/4 be the answers if in the ( )( ) equation they will have to multiply to give 8? Surely it must be 8 and 1?
 
  • #5
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But then 8 and 1/4 are not the right answers (the answers are 64 and 1/16)
You made the transformation t = x1/2. You have the roots for the polynomial in t, so convert them back to x.
 
  • #6
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You made the transformation t = x1/2. You have the roots for the polynomial in t, so convert them back to x.
The roots would be 2(sqrt2) and 1/2 however that's not the right answer (the right answer being 64 and 1/16).

That still doesn't answer how to do it without a calculator though. Got any insights?
 
  • #7
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You can try the rational roots theorem

if a rational number p/q is a root of the polynomial

[tex] a_n x^n + a_{n-1} x^{n-1} + .... + a_0 [/tex]

then p is a factor of a_0 and q a factor of a_n.

In case of 4t^2 - 33t +8 =0, you get p = 8 or 4 or 2 or 1 and q = 4 or 2 or 1.

possible roots are 8,4,2,1,(1/2),(1/4), you just try all of these. Test exercises
nearly always have rational roots.

The producut of the roots is a_0 = 8, so if you found one root, the other is easy.
 
  • #8
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4,773
Nonetheless I tried 4t^2-33t+8=0 and then factoring that but without using the quadratic formula to give me the answer of t=8, I didn't see how else I could do it (the quadratic formula requiring a calculator since you have to square 33 to make 1089 and then get a square route...).
You don't need the quadratic formula. Factoring works just fine.
Then I tried factorising it into 4(t^2+2)=33t but I didn't really see where to go from there. Most of my working has just been repetitions of these.
Breaking up the equation as you show here is a complete dead end.

I'm fairly sure that you can factorise it into (2t + 16)(2t + 1/2) but there was no way I would have ever gotten that without a calculator, so it is again useless.
Your factors here are incorrect. They should multiply to 4t2 - 33t + 8, but instead they multiply to 4t2 + 33t + 8.
Do you know of any methods through which to solve this?
Yes - factoring. You should have been exposed to lots and lots of trinomial factorization problems in your algebra class, including some where the coefficient of the squared term wasn't 1. Do you remember doing this?
Thanks for replying!



Quadratic formula workings; (33+sqrt(33^2-4(4)(8)))/8=8, 1/4

But then 8 and 1/4 are not the right answers (the answers are 64 and 1/16) :(
 
  • #9
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I talked to my friend about this and he pointed out that you can factorise it into the brackets

(4t-1)(t-8)

Giving t=8,1/4

And since t=x^1/2, you need to square 8 and 1/4 rather than square routing it like I was doing.

Sorry for wasting your time, this turned out to be a very elementary mistake.
 
  • #10
33,072
4,773
I talked to my friend about this and he pointed out that you can factorise it into the brackets

(4t-1)(t-8)

Giving t=8,1/4

And since t=x^1/2, you need to square 8 and 1/4 rather than square routing it like I was doing.

Sorry for wasting your time, this turned out to be a very elementary mistake.
Yes, that's exactly it.
 

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