Problem 1: Symmetrical Properties of Roots

Algebra Level 2

If α \alpha and β \beta are the roots of the equation 2 x 2 − 4 x − 3 = 0 2x^2-4x-3=0 , where α > β \alpha>\beta , find the value of α 2 + β 2 \alpha^2+\beta^2 .


The answer is 7.

View wiki
Victor Loh by Victor Loh , 19, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Daniel Lim
Jul 22, 2014

α + β = 4 2 \alpha+\beta=\dfrac{4}{2} and α β = − 3 2 \alpha\beta=-\dfrac{3}{2}

Since ( α + β ) 2 = α 2 + β 2 + 2 α β (\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta , to find α + β \alpha+\beta , just subtract 2 α β 2\alpha\beta from ( α + β ) 2 (\alpha+\beta)^2

So the answer to this is 2 2 − ( − 3 ) = 7 2^2-(-3)=\boxed{7}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Almost right. Because the problem mentioned that α > β \alpha > \beta , you should prove that the roots of this equation are real, else the condition can't be fulfilled.

Proof of real roots:

Since, roots are real, therefore in the quadratic formula:

x = − b ± b 2 − 4 a c 2 a x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}

The part in the square root, (the discriminant of the equation Δ \Delta ) must be greater than or equal to 0 0 .

Comparing, 2 x 2 − 4 x − 3 = 0 2x^{2}-4x-3=0 with a x 2 + b x + c = 0 ax^{2}+bx+c=0 , we obtain:

a = 2 , b = − 4 , c = − 3 \color{#D61F06}{a}=2~,~\color{#20A900}{b}=-4~,~\color{#3D99F6}{c}=-3

Hence,

Δ = b 2 − 4 a c = ( − 4 ) 2 − 4 ( 2 ) ( − 3 ) = 40 ≥ 0 ■ \begin{aligned} \Delta & = b^{2} - 4ac \\ & = {(-4)}^{2} - 4(2)(-3) \\ & = 40 \ge 0 ~~~~~~ \blacksquare \end{aligned}

Thus completes the proof that the roots are real.

Also, when roots are real, then without loss of generality, we have,

α = − b + b 2 − 4 a c 2 a , β = − b − b 2 − 4 a c 2 a \color{#D61F06}{\alpha} = \dfrac{-b + \sqrt{b^{2}-4ac}}{2a} \quad , \quad \color{#3D99F6}{\beta} = \dfrac{-b - \sqrt{b^{2}-4ac}}{2a}

and by plugging in values of a , b a,b and c c we can find value of α \alpha and β \beta which are not necessary for this problem.

Tapas Mazumdar - 4 years, 8 months ago
Megha Pardhi
Jul 23, 2014

solution solution

1 Helpful 0 Interesting 0 Brilliant 0 Confused

your photo is unavailable

Daniel Lim - 6 years, 10 months ago

α 2 + β 2 = ( α + β ) 2 − 2 α β \color{#69047E}{\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta} From Vieta,s formulas,we know that α + β = 4 2 = 2 , α β = − 3 2 \color{#20A900}{\alpha+\beta=\frac{4}{2}=2,\alpha\beta=-\frac{3}{2}} .Replacing these values,we get 2 2 − 2 ( − 3 2 ) = 4 − ( − 3 ) = 4 + 3 = 7 \color{#3D99F6}{2^2-2(\frac{-3}{2})=4-(-3)=4+3=\boxed{7}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Almost right. Because the problem mentioned that α > β \alpha > \beta , you should prove that the roots of this equation are real, else the condition can't be fulfilled.

Proof of real roots:

Since, roots are real, therefore in the quadratic formula:

x = − b ± b 2 − 4 a c 2 a x=\dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}

The part in the square root, (the discriminant of the equation Δ \Delta ) must be greater than or equal to 0 0 .

Comparing, 2 x 2 − 4 x − 3 = 0 2x^{2}-4x-3=0 with a x 2 + b x + c = 0 ax^{2}+bx+c=0 , we obtain:

a = 2 , b = − 4 , c = − 3 \color{#D61F06}{a}=2~,~\color{#20A900}{b}=-4~,~\color{#3D99F6}{c}=-3

Hence,

Δ = b 2 − 4 a c = ( − 4 ) 2 − 4 ( 2 ) ( − 3 ) = 40 ≥ 0 ■ \begin{aligned} \Delta & = b^{2} - 4ac \\ & = {(-4)}^{2} - 4(2)(-3) \\ & = 40 \ge 0 ~~~~~~ \blacksquare \end{aligned}

Thus completes the proof that the roots are real.

Also, when roots are real, then without loss of generality, we have,

α = − b + b 2 − 4 a c 2 a , β = − b − b 2 − 4 a c 2 a \color{#D61F06}{\alpha} = \dfrac{-b + \sqrt{b^{2}-4ac}}{2a} \quad , \quad \color{#3D99F6}{\beta} = \dfrac{-b - \sqrt{b^{2}-4ac}}{2a}

and by plugging in values of a , b a,b and c c we can find value of α \alpha and β \beta which are not necessary for this problem.

Tapas Mazumdar - 4 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...