Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer
@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.

share|cite|improve this answer
@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:



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:


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

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 $\{\frac{-3}{2}, \frac{9}{4}\}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.