Roots of a quadratic

Algebra Level 3

Let α \alpha and β \beta be the roots of the equation x 2 + x − 3 = 0 x^2 + x - 3 = 0 such that α < β \alpha < \beta . Find the value of the expression 4 β 2 − α 3 4 \beta^2 - \alpha^3 .


The answer is 19.

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Md Zuhair by Md Zuhair , 20, India

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3 solutions

Tapas Mazumdar
Mar 5, 2017

By Vieta's formula, we have

α + β = − 1 \alpha + \beta = -1

Note that

4 β 2 − α 3 = 4 ( − 1 − α ) 2 − α 3 = − α 3 + 4 α 2 + 8 α + 4 = − α ( α 2 + α − 3 ) 0 + 5 ( α 2 + α ) ⏟ = 3 + 4 = 5 ( 3 ) + 4 = 19 \begin{aligned} 4 \beta^2 - \alpha^3 &= 4 {(-1-\alpha)}^2 - \alpha^3 \\ &= -\alpha^3 + 4\alpha^2 + 8 \alpha +4 \\ &= -\alpha \cancel{(\alpha^2 + \alpha -3)}^{0} + 5 \underbrace{(\alpha^2 + \alpha)}_{= 3} + 4 \\ &= 5(3) + 4 \\ &= 19 \end{aligned}

15 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice solution buddy! B)

Sahil Silare - 4 years, 3 months ago

x 2 − x \cancel{x^2-x} , just trying latex, dont worry

Md Zuhair - 4 years, 1 month ago

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That command is to be used like this

\require{cancel} \cancel{x}

Which will give

x \cancel{x}

'\require{cancel}' command is important here.

Moreover

\require{cancel} \cancel{x}^1

will give

x 1 \cancel{x}^1

and

\require{cancel} \cancel{x}_1

will give

x 1 \cancel{x}_1

Tapas Mazumdar - 4 years, 1 month ago

BRUTE FORCE SOLUTION

By the quadratic formula, the roots of x 2 + x − 3 x^2+x-3 , which will be denoted by α \alpha and β \beta , where α < β \alpha<\beta , are

α = − 1 − 13 2 β = − 1 + 13 2 . \begin{aligned} \alpha &= \frac{-1-\sqrt{13}}{2} \\ \beta &= \frac{-1+\sqrt{13}}{2}. \end{aligned}

So,

4 β 2 − α 3 = 4 × ( − 1 + 13 ) 2 4 − ( − 1 − 13 ) 3 8 = ( − 1 + 13 ) 2 − ( − 1 − 13 ) 3 8 . \begin{aligned} 4\beta^2 - \alpha^3 &= 4 \times \frac{(-1+\sqrt{13})^2}{4} - \frac{(-1-\sqrt{13})^3}{8} \\ &= (-1+\sqrt{13})^2 - \frac{(-1-\sqrt{13})^3}{8}. \end{aligned}

Observe that

( − 1 + 13 ) 2 = 14 − 2 13 ( − 1 − 13 ) 3 = ( 14 + 2 13 ) ( − 1 − 13 ) = − 40 − 16 13 = − 8 ( 5 + 2 13 ) . \begin{aligned} (-1+\sqrt{13})^2 &= 14 - 2\sqrt{13} \\ (-1-\sqrt{13})^3 &= (14+2\sqrt{13})(-1-\sqrt{13}) \\ &= -40-16\sqrt{13} = -8(5+2\sqrt{13}). \end{aligned}

As a result,

4 β 2 − α 3 = 14 − 2 13 − − 8 ( 5 + 2 13 ) 8 = 14 − 2 13 + 5 + 2 13 = 19 , \begin{aligned} 4\beta^2 - \alpha^3 &= 14 - 2\sqrt{13} - \frac{-8(5+2\sqrt{13})}{8} \\ &= 14 - 2\sqrt{13} + 5 + 2\sqrt{13} \\ &= 19, \end{aligned}

as required.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Its quire tough to do it like this

Md Zuhair - 4 years, 1 month ago

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Brute force solutions may not be the best way of arriving to solutions to many people, but it is a foolproof way of proving things. If you take issue with it, that's your issue, not mine's.

A Former Brilliant Member - 4 years, 1 month ago
Skye Rzym
Apr 12, 2017

From Vieta's Theorem, we get

β + α = − 1 \beta + \alpha = -1

β α = − 3 \beta\alpha = -3

And then, because α < β \alpha < \beta , so

β − α = Discriminant \beta - \alpha = \sqrt{\text{Discriminant}}

β − α = ( − 1 ) 2 − 4.1. ( − 3 ) \beta - \alpha = \sqrt{(-1)^{2}-4.1.(-3)}

β − α = 1 + 4.3 = 13 \beta - \alpha = \sqrt{1+4.3}=\sqrt{13}

Now, let

x = 4 β 2 − α 3 x = 4\beta^{2} - \alpha^{3} , and

y = 4 α 2 − β 3 y = 4\alpha^{2} - \beta^{3}

Now

(1)...

x + y = 4 β 2 − α 3 + 4 α 2 − β 3 x+y = 4\beta^{2} - \alpha^{3} + 4\alpha^{2} - \beta^{3} x + y = 4 ( β 2 + α 2 ) − ( α 3 + β 3 ) x+y = 4(\beta^{2} + \alpha^{2}) - (\alpha^{3} + \beta^{3}) x + y = 4 ( ( β + α ) 2 − 2 β α ) − ( α + β ) ( α 2 − β α + β 2 ) x+y = 4((\beta + \alpha)^{2} - 2\beta\alpha) - (\alpha + \beta)(\alpha^{2} - \beta\alpha + \beta^{2}) x + y = 4 ( ( − 1 ) 2 − 2 ( − 3 ) ) − ( − 1 ) ( ( α + β ) 2 − 3 α β ) x+y = 4((-1)^{2} - 2(-3)) - (-1)((\alpha + \beta)^{2} - 3\alpha\beta) x + y = 4 ( 1 + 2.3 ) + ( ( − 1 ) 2 − 3 ( − 3 ) ) x+y = 4(1 + 2.3) + ((-1)^{2} - 3(-3)) x + y = 4.7 + ( 1 + 3.3 ) = 28 + 10 = 38 x+y = 4.7 + (1 + 3.3) = 28 + 10 = 38

And then

(2)...

x − y = 4 β 2 − α 3 − 4 α 2 + β 3 x-y = 4\beta^{2} - \alpha^{3} - 4\alpha^{2} + \beta^{3} x − y = 4 ( β 2 − α 2 ) + ( β 3 − α 3 ) x-y = 4(\beta^{2} - \alpha^{2}) + (\beta^{3} - \alpha^{3}) x − y = 4 ( β + α ) ( β − α ) + ( β − α ) ( α 2 + β α + β 2 ) x-y = 4(\beta + \alpha)(\beta - \alpha) + (\beta - \alpha)(\alpha^{2} + \beta\alpha + \beta^{2}) x − y = 4 ( ( − 1 ) ( 13 ) + ( 13 ) ( ( α + β ) 2 − α β ) x-y = 4((-1)(\sqrt{13}) + (\sqrt{13})((\alpha + \beta)^{2} - \alpha\beta) x − y = − 4 13 + 13 ( ( − 1 ) 2 − ( − 3 ) ) x-y = -4\sqrt{13} + \sqrt{13}((-1)^{2} - (-3)) x − y = − 4 13 + 13 ( 1 + 3 ) = − 4 13 + 4 13 = 0 x-y = -4\sqrt{13} + \sqrt{13}(1 + 3) = -4\sqrt{13} + 4\sqrt{13} = 0

Now, we find the value of x x .

(1)+(2)

x + y + x − y = 38 + 0 x+y+x-y = 38+0 2 x = 38 2x = 38 x = 19 x = 19 4 β 2 − α 3 = 19 4\beta^{2} - \alpha^{3} = \boxed{19}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Let suppose for a general quadratic a x 2 + b x + c = 0 ax^2+bx+c=0 whose roots are α \alpha and β \beta with α < β \alpha < \beta , so we have

β = − b + b 2 − 4 a c 2 a α = − b − b 2 − 4 a c 2 a ⟹ β − α = 2 ( b 2 − 4 a c 2 a ) = Discriminant a \beta = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \\ \alpha = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \\ \implies \beta - \alpha = 2 \left( \dfrac{\sqrt{b^2-4ac}}{2a} \right) = \dfrac{\sqrt{\text{Discriminant}}}{a}

But as in the third step of your solution, you've simply stated β − α = Discriminant \beta - \alpha = \sqrt{\text{Discriminant}} which may give a false information to someone reading your solution who doesn't have a good understanding of quadratics yet.

Tapas Mazumdar - 4 years, 1 month ago

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