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How to solve this equation : $$(a^2 -9d^2)(a^2-d^2)=40\quad \mathrm{where} \ a = -1/2$$

i am trying it by reducing it into quadratic equation $144(d^2)^2-40d-159=0$ and also by prime factorization but given answer is $d = +3/2$ and $-3/2$. pls help.

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closed as off-topic by Dietrich Burde, Claude Leibovici, Guy Fsone, Namaste, I am Back Dec 1 '17 at 13:24

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    $\begingroup$ I obtain the biquadratic equation $144d^4 - 40d^2 - 639=0$. Set $x=d^2$ and solve the quadratic equation in $x$. The solutions are $d=\pm 3/2$ and $d=\pm \sqrt{-71}/6$. $\endgroup$ – Dietrich Burde Dec 1 '17 at 8:56
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As Dietrich said, if you set $x=d^2$ You will obtain $144x^2-40x-639=0$ and the discriminant(delta) is equal to $\sqrt{1600+144.639} = 608$. Hence $d^2=x={40\mp608\over288}$ But $d^2 \ne {40-608\over288}$ since it is negative. So $d =\pm\sqrt{{40+608\over288}}=\pm3/2$

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    $\begingroup$ You seem to assume that $d$ is a real number. The OP has not said this (although he probably means it). $\endgroup$ – Dietrich Burde Dec 1 '17 at 9:14
  • $\begingroup$ thanks for your explanation..now my doubt is clear.. $\endgroup$ – Indu sahu Dec 1 '17 at 9:15
  • $\begingroup$ @DietrichBurde since he gave the answer, I assumed $d\in\Bbb{R}$, but you are right. $\endgroup$ – Cem Sarıer Dec 1 '17 at 9:19
  • $\begingroup$ yes..d is assumed to be a real number..it is a constraint on equation.. $\endgroup$ – Indu sahu Dec 1 '17 at 9:19
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We're given that :

$$(a^2 -9d^2)(a^2-d^2)=40\quad \mathrm{where} \ a = -1/2$$

Applying some operations :

$$a^4- a^2d^2-9a^2d^2 +9d^4=40$$

Substituting $a= -1/2$ in it :

$$\bigg(-\frac{1}{2}\bigg)^4- \bigg(-\frac{1}{2}\bigg)^2d^2 - 9\bigg(-\frac{1}{2}\bigg)^2d^2+9d^4 = 40$$

Which after some operations/calculations, become :

$$144d^4 - 40d^2 - 639 =0$$

Since we can observe that all we have is the terms $d^4,d^2$ , let's substitute $x=d^2$. Then, our equation will become :

$$144x^2 - 40x - 639 =0$$

This is now, an elementary equation to solve (using the quadratic solution method). After you've found the solutions $x_{1,2}$ to this equation, substitute them back to $x=d^2$ to find the $4$ solutions of $d$ for the initial equation.

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  • $\begingroup$ thank u so much for your guidance....this is very helpful for me...thanks again. $\endgroup$ – Indu sahu Dec 1 '17 at 9:14
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For a different approach, there is almost an invitation to multiply through by $16$ and factorise the left-hand side using the difference of two squares. Changing the sign of both factors and putting $c=2d$ gives $$(3c+1)(c+1)(c-1)(3c-1)=640=2^7\times 5$$

If all the factors on the left-hand side are positive, there are not many to choose - it is clear that $640\gt 9c^2$ for $c\ge 2$ and that all factors have to be integers, with $c$ odd (otherwise all factors on LHS are odd). So $c\lt 9$, and is odd, and the factorisation on the right quickly excludes most values, leaving $c=3$.


To exclude values - $c=7: 7-1=6$ would bring in a factor $3$ as would $c=5: 5+1=6$ and $c=1$ makes the left-hand side zero, leaving just $c=3$


We have a polynomial in $c^2$ so that $c=-3$ is a negative solution. The left-hand side is clearly increasing with $c\gt 1$ and is symmetric, and it is clear that the left-hand side is nowhere near as large as $640$ for $-1\le c \le 1$, so these are the only real solutions.

There are some vague statements above which would need to be made precise for a proof.

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  • $\begingroup$ $d$ is not an integer, and nor is $c$. $\endgroup$ – zwim Dec 1 '17 at 10:52
  • $\begingroup$ Actually 40 can be factorised as to get answers in brackets. Its a little complicated method but answer is coming correct $\endgroup$ – Atharva Shetty Dec 1 '17 at 11:05
  • $\begingroup$ @zwim What do you think $c$ is? If $c=\frac pq$ in lowest terms you can only possibly cancel one factor three in the denominator and that in just two of the four terms, and you have an integer in the right-hand side. $c=\pm 3$ does give an integer solution corresponding to $d=\\frac c2 = pm \frac 32$. $c$ could be the square root of an integer, but then $c^2-1$ and $9c^2-1$ would be integers having only the factors $2$ and $5$ and it is possible to explore that case. Possibly I phrased this badly, but it is easy to find an integer solution $c$ this way, or to tell if there isn't one. $\endgroup$ – Mark Bennet Dec 1 '17 at 13:32

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