# if one root of the equation is the square of the other, then the values of a is

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.

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 $$a^3$$

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.

• i think OP needs help solving the actual equation – Saketh Malyala Jun 6 '19 at 6:52
• @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 :) – fzyzcjy Jun 6 '19 at 6:53
• I need to find the value of "a" in the equation I mentioned in my question @SakethMalyala – weegee Jun 6 '19 at 6:53

$$\alpha+\alpha^2=\dfrac{6}{8}$$ and $$\alpha^3=\dfrac{-a-3}{8}$$. Then from the first equation you solve for $$\alpha$$, plug that in the second to get $$a$$

• What should be the next step? – lab bhattacharjee Jun 6 '19 at 7:01

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 $$\alpha=\frac{1}{2}$$

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

Can you proceed ?

• What should be the next step? – lab bhattacharjee Jun 6 '19 at 7:01
• Next step: use $\beta^2= \alpha.$ This gives $\alpha=-3/2$ or $\alpha =1/2$. This yields $a=24$ or $a=-4.$ – Fred Jun 6 '19 at 7:08

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

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

• Thanks for the explanation. I got the answer – weegee Jun 6 '19 at 7:16