# Value of $a$ for which one root of the Quadratic equation is square of another

I need some urgent help! If one root of the equation $8x^2-6x-a-3=0$ is the square of the other root, find the value of $a$. (Let the roots be $\alpha$ and $\beta$) The answer for a is -4 and 24.

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You can't demand such things urgently. – user9413 May 19 '11 at 8:37
@Chandru: Haha, I guess I am too impatient for the solution. Been thinking about this for a whole day! – Sophia May 19 '11 at 8:39
Ok. Then please add what you have done. I have added an hint for your question. Does that help. – user9413 May 19 '11 at 8:41

Put $\beta =\alpha ^{2}$. Factor the polynomial $8x^{2}-6x-a-3$ as

$$8x^{2}-6x-a-3=8(x-\alpha )(x-\beta )=8(x-\alpha )(x-\alpha ^{2}).$$

Expand the RHS to get

$$8x^{2}-6x-a-3=8x^{2}+\left( -8\alpha ^{2}-8\alpha \right) x+8\alpha ^{3}.$$

Equate coefficients and solve for $\alpha$ and $a$.

$$8\alpha ^{3}=-a-3,\qquad (1)$$

$$-8\alpha ^{2}-8\alpha =-6.\qquad (2)$$

The two solutions of $(2)$ are $\alpha =-3/2,\alpha =1/2$. For $\alpha =-3/2$, $(1)$ becomes $8\left( -3/2\right) ^{3}=-a-3$, whose solution is $a=24$; and for $\alpha =1/2$, $(1)$ becomes $8\left( 1/2\right) ^{3}=-a-3$, whose solution is $a=-4$.

Added: Polynomials can be factored in 1st degree terms $(x-x_i)$, where $x_i$'s are the roots of the polynomial. The cofficient of the product of all $(x-x_i)$ is the coefficient of the leading term.

So the quadratic polynomial $Ax^2+Bx+C$ is factored in two terms. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\qquad (\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\ast)$ are equal.

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@Americo Tavares: I am sorry but can you explain the process of factorising the polynomial? – Sophia May 19 '11 at 9:03
Is it because the coefficient of x^2 must be 1? – Sophia May 19 '11 at 9:05
@Sophia: The quadratic polynomial $Ax^2+Bx+C$ is factored in terms of its roots. Let $x_1,x_2$ be the roots. Then $$Ax^2+Bx+C=A(x-x_1)(x-x_2).\qquad (\ast)$$ Observe that $Ax^2+Bx+C=0$ if and only if $A(x-x_1)(x-x_2)=0$. The coefficient $A$ must be in front of $(x-x_1)(x-x_2)$ so that both sides of $(\ast)$ are equal. – Américo Tavares May 19 '11 at 9:10
@Sopia: Is now clear? I have added this explanation to the answer. – Américo Tavares May 19 '11 at 9:12
@Americo Tavares: But by putting A in front of the roots of the factorised equation will throw it out of balance, no? – Sophia May 19 '11 at 9:22

Let $\alpha,\beta$ be the $2$ roots. Given $\beta = \alpha^{2}$.

• For a quadratic equation $ax^{2}+bx+c$ which one root is $$\alpha = \frac{-b +\sqrt{D}}{2a}, \qquad \beta = \frac{-b-\sqrt{D}}{2a}$$

$D$ denotes the discriminant whose value is $b^{2}-4ac$. Now use the relation $\alpha =\beta^{2}$ and then solve for $a$.

Ok. So let $\alpha$ be on root and $\beta$ be the other root of the equation $8x^{2}-6x-(a+3)=0$. Now we take $$\alpha = \frac{6 + \sqrt{36+32(a+3)}}{16} , \qquad \beta = \frac{6 -\sqrt{36+32(a+3)}}{16}$$

Now $$\beta^{2} = \frac{1}{96} \cdot \Bigl[ 36 - 2 \times 6 \times \sqrt{36+32(a+3)} + 36+32(a+3)\Bigr]$$ Since this should be equal to $\alpha$ we have $$\frac{1}{256} \cdot \Bigl[168 - 12\sqrt{36+32(a+3)} + 32a\Bigr] = \frac{1}{16} \cdot \Bigl[6 + \sqrt{36+32(a+3)}\Bigr]$$ $$\Longrightarrow 168 -12 \cdot \sqrt{36 +32(a+3)} + 32a = 96 + 16\sqrt{36+32(a+3)}$$

$$\Longrightarrow 18+8a =7 \cdot \sqrt{36 + 32(a+3)}$$

$$9 +4a = 7 \sqrt{8a +33}$$ Again squaring both sides we get $$\Longrightarrow 16a^{2} + 72a +81 = 392a + 1617$$ from this we get $$\Longrightarrow 16a^{2} -320a -1536=0$$ $$\Longrightarrow 4a^{2}-80a -384 =0$$ $$\Longrightarrow a^{2}-20a -96=0$$

Hopefully you can finish off from here.

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@Sophia: This should be fine. – user9413 May 19 '11 at 10:04
@Americo: Nice answer. Good to see so many ways of answering a problem. +1 :) – user9413 May 22 '11 at 11:50

Here try this:

The two roots of $8x^2−6x−a−3=0$ can be found by the quadratic formula:

$x_1=\frac{8-\sqrt{8a+33}}{8}$

$x_2=\frac{8+\sqrt{8a+33}}{8}$

Since the problem requires one to be the square of the other, this means that the larger root, $x_2$, must be the square of the smaller, $x_1$. So do that:

$x_1^2=x_2$

$\left(\frac{8-\sqrt{8a+33}}{8}\right)^2= \frac{8+\sqrt{8a+33}}{8}$

Can you square and simplify that? If so, you'll then obtain another quadratic equation--this time with the variable $a$. Solve it using the quadratic formula. You'll get the two solutions for $a$, as you claimed. The pairs of roots of the original equation are $\{\frac{1}{2},\frac{1}{4}\}$ and $\{\frac{-3}{2}, \frac{9}{4}\}$.

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You should not assume that the root with the plus sign gives the square. Were it $\frac{1}{4}\pm \frac{\sqrt{129}-9}{16}$, the version with the minus sign is the square of the one with plus. But the approach is a good one. – Ross Millikan May 19 '11 at 13:02