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One of the roots of the equation $$8x^2 - 6x - a - 3$$ are the square of the other. Which means if $β\ and \ ⍺$ are the roots then $β = ⍺^2$. Then we have to find a.

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Hint:

$$\alpha+\alpha^2=\dfrac68$$

$$\alpha\cdot\alpha^2=-\dfrac{a+3}8$$

So, $\left(\dfrac34\right)^3=(\alpha+\alpha^2)^3=\alpha^3+(\alpha^3)^2+3\alpha^3\cdot\dfrac34$

Replace the value of $\alpha^3$ to form a quadratic equation $a^3$

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Firsty, we can get the two roots, $\alpha$ and $\beta$, in terms of constant and $a$, which is done via the formula of quadratic equation. Then, by $\beta = \alpha^2$ you solved it.

Edit: How to solve quadratic equation - https://www.purplemath.com/modules/quadform.htm

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    $\begingroup$ i think OP needs help solving the actual equation $\endgroup$ – Saketh Malyala Jun 6 at 6:52
  • $\begingroup$ @SakethMalyala Maybe you are right. I am in a hurry and may need another person to answer it... I have edited to add a link :) $\endgroup$ – fzyzcjy Jun 6 at 6:53
  • $\begingroup$ I need to find the value of "a" in the equation I mentioned in my question @SakethMalyala $\endgroup$ – weegee Jun 6 at 6:53
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$\alpha+\alpha^2=\dfrac{6}{8}$ and $ \alpha^3=\dfrac{-a-3}{8}$. Then from the first equation you solve for $\alpha$, plug that in the second to get $a$

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  • $\begingroup$ What should be the next step? $\endgroup$ – lab bhattacharjee Jun 6 at 7:01
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You have $8x^2-6x-a-3=0$.

This is equivalent to $\displaystyle x^2 - \frac{3}{4}x - \frac{1}{8}(a+3)=0$.

We then have that $\alpha+\alpha^2=+\frac{3}{4}$, by Vieta's formula.

This is an easy equation to solve using Quadratic equation, yielding $\displaystyle \alpha=-\frac{3}{2}$ or $\displaystyle \alpha=\frac{1}{2}$.

Then, also by Vieta's formula, we have $\alpha\cdot\alpha^2=-\frac{1}{8}(a+3)$.

This yields a value of $a=24$ for $\alpha=-\frac{3}{2}$ or $a=-4$ for $\alpha=\frac{1}{2}$

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By Vieta: $\alpha+ \beta=3/4$ and $\alpha \beta= - \frac{a+3}{8}.$

Can you proceed ?

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  • $\begingroup$ What should be the next step? $\endgroup$ – lab bhattacharjee Jun 6 at 7:01
  • $\begingroup$ Next step: use $\beta^2= \alpha.$ This gives $\alpha=-3/2$ or $\alpha =1/2$. This yields $a=24$ or $a=-4.$ $\endgroup$ – Fred Jun 6 at 7:08
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We know that for a quadratic equation of the form $ax^2+bx+c = 0$, the two roots $x = \alpha$ and $x = \beta$ must satisfy the following equations:

$$\alpha+\beta = -\frac{b}{a} \tag{1}$$

$$\alpha\beta = \frac{c}{a} \tag{2}$$

Your equation is $8x^2-6x-a-3 = 0$, so $a = 8$, $b = -6$, and $c = -a-3$. Using the fact that $\beta = a^2$, you get the following equations:

$$\alpha+\alpha^2 = \frac{3}{4}$$

$$\alpha^3 = \frac{-a-3}{8}$$

Solving the first equation (by completing the square, factoring, or using the Quadratic Formula) yields $\alpha = -\frac{1}{2}\pm 1$. Plugging this in the second equation gives

$$-a-3 = 8\alpha^3 \iff a = -3-8\alpha^3 = \begin{cases} -3-8(0.5)^3 = -4 \\ -3-8(-1.5)^3 = 24 \end{cases}$$

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  • $\begingroup$ Thanks for the explanation. I got the answer $\endgroup$ – weegee Jun 6 at 7:16

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