Root of equation

[sugarfreeboy]

Given that p is a root of the equation 2x²=3x-4, show that 4p³=p-12

What I have tried to figure out so far:

(1) Since p is a root, then 2p²=3p-4

(2) Discriminant is B²-4AC=(-3)²-4(2)(4)=-23 (<0) so no real roots meaning the curve does not cut the x-axis (in fact it lies above the x-axis)

But how to show 4p³=p-12 ??

Any kind person can help?

Thanks in advance.

 

[BigGlenntheHeavy]

Hint: 4p³-p+12 = (2p+3)(2p²-3p+4) = 0

 

[sugarfreeboy]

Hi thanks so much for the hint

So can I conclude that to do this question, we need to work backwards to determine the unknown factor of (2p+3) using long division?

My presentation of working after your hint:

2x²=3x-4
Since p is a root of the above eqn, 2p²=3p-4.
Rewritting, 2p²-3p+4=0
Multiplying both sides of eqn by 2p+3, (2p+3)(2p²-3p+4)=0
Expanding, 4p³-6p²+8p+6p²-9p+12=0
Regrouping, 4p³=p-12 (shown)

 

[Deleted member 4993]

Hi thanks so much for the hint

So can I conclude that to do this question, we need to work backwards to determine the unknown factor of (2p+3) using long division?

My presentation of working after your hint:

2x²=3x-4
Since p is a root of the above eqn, 2p²=3p-4.
Rewritting, 2p²-3p+4=0

Only thing I would suggest, that since we do not know the existence of (2p+3), we divide (4p[sup:84netv87]3[/sup:84netv87] - p -12) by (2p[sup:84netv87]2[/sup:84netv87] - 3p +4)

Multiplying both sides of eqn by 2p+3, (2p+3)(2p²-3p+4)=0
Expanding, 4p³-6p²+8p+6p²-9p+12=0
Regrouping, 4p³=p-12 (shown)

 

[BigGlenntheHeavy]

4p³-p+12 = (2p+3)(2p²-3p+4) = 0

(2p²-3p+4) = 0/(2p+3) = 0, p does not equal -3/2.

Hence 2p² = 3p-4, which you wanted to show, see (1). QED

 

[sugarfreeboy]

4p³-p+12 = (2p+3)(2p²-3p+4) = 0

(2p²-3p+4) = 0/(2p+3) = 0, p does not equal -3/2.

Hence 2p² = 3p-4, which you wanted to show, see (1). QED

Hi, with all due respect, by writing in the first line that 4p³-p+12=0, it would defeat the purpose of proving/showing since that is what the qn wants us to show. Correct me if I am wrong. :wink:

And the reason why p can't be -3/2 is because there is no real root for the eqn rite?

 

[BigGlenntheHeavy]

Given that p is a root of the equation 2x²=3x-4, show that 4p³=p-12

$\displaystyle 2x^{2} \ = \ 3x-4$

$\displaystyle 2x^{2}-3x+4 \ = \ 0$

$\displaystyle (2x^{2}-3x+4) \ = \ 0$

$\displaystyle (2x+3)(2x^{2}-3x+4) \ = \ (2x+3)(0)$

$\displaystyle (2x+3)(2x^{2}-3x+4) \ = \ 0$

$\displaystyle 4x^{3}-x+12 \ = \ 0$

$\displaystyle 4x^{3} \ = \ x-12$

$\displaystyle 4p^{3} = p-12, \ since \ p \ is \ a \ root \ of \ 2x^{2} \ = \ 3x-4, \ p \ = \ x. \ QED$

 

[Deleted member 4993]

Given that p is a root of the equation 2x²=3x-4, show that 4p³=p-12

$\displaystyle 2x^{2} \ = \ 3x-4$

$\displaystyle 2x^{2}-3x+4 \ = \ 0$

$\displaystyle (2x^{2}-3x+4) \ = \ 0$

$\displaystyle (2x+3)(2x^{2}-3x+4) \ = \ (2x+3)(0)$<<<< Why did you decide to multiply by (2x+3)

$\displaystyle (2x+3)(2x^{2}-3x+4) \ = \ 0$

$\displaystyle 4x^{3}-x+12 \ = \ 0$

$\displaystyle 4x^{3} \ = \ x-12$

$\displaystyle 4p^{3} = p-12, \ since \ p \ is \ a \ root \ of \ 2x^{2} \ = \ 3x-4, \ p \ = \ x. \ QED$

 

[BigGlenntheHeavy]

Subhotosh Khan, what does 4p^3-p+12 factor to?

Note: Given that p is a root of the equation 2x²=3x-4, show that 4p³=p-12 could just as easily have been written as

"Given that x is a root of the equation 2x²=3x-4, show that 4x³=x-12" as p and x are nothing but dummy variables, however

they (who wrote this) obviously wanted to muddy the waters.

 

[daon]

BigGlennthehHeavy's solution is perfectly adequate and "easier".

A more straight-forward approach might be solving for p in the quadratic equation and plugging in to the cubic one. This requires tedius algebra however.

From the first we see:

$\displaystyle p =\frac{3 \pm \sqrt{23}i}{4}$

Then verify:

$\displaystyle 4(\frac{3 \pm \sqrt{23}i}{4})^3 - \frac{3 \pm \sqrt{23}i}{4} + 12 = 0$

edit:forgot coefficient

 

[BigGlenntheHeavy]

Thank you daon.

We are given that p is a root of 2x^2-3x+4. Now the roots of 2x^2-3x+4 are x=[3+i?(23)]/4 and x = [3-i?(23)]/4.

So p is one of those roots. now 2x^2-3x+4=0, so anything multipied to the equation will have more roots, but the original root of p will always be there.
For example, if we multiply 2x^2-3x+4 by a linear factor, then we will have three roots, one of them being p.

If we multiply by a cubic, then we will have five roots, but one of them will be p.

Hence it is immaterial whether p is real or complex as it will always be there.