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July 10th, 2017, 01:41 AM  #1 
Newbie Joined: Jul 2017 From: Australia Posts: 3 Thanks: 0  On the Roots of a Quadratic Equation
Hello, I am having some trouble with this question: The quadratic equation 2x^2+4x+1=0 has roots α and β. Find the quadratic equation with roots α^2 and β^2. Help with solving the question would be very much appreciated! 
July 10th, 2017, 02:29 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,074 Thanks: 695 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Some hints for you: 1. Factorise the lefthand side of $\displaystyle 2x^2 + 4x +1 = 0$ 2. You can use the factorised result to obtain values for the roots. 3. Those roots are $\displaystyle \alpha$ and $\displaystyle \beta$ 4. Try doing the reverse of steps 13 using new roots, where the new roots are calculated by taking the square of the old roots. Give us a shout with how you get one... if you show your working here it's easier for us to help you too 
July 10th, 2017, 03:45 AM  #3 
Senior Member Joined: Feb 2010 Posts: 632 Thanks: 103 
It is possible to get the answer without actually finding $\displaystyle \alpha$ and $\displaystyle \beta$. Google "elementary symmetric functions".

July 10th, 2017, 04:23 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1422 
As 2x² + 4x + 1 = 0 has roots $\alpha$ and $\beta$, 2x²  4x + 1 = 0 has roots $\alpha$ and $\beta$. Hence ±$\alpha$ and ±$\beta$ are the roots of (2x² + 4x + 1)(2x²  4x + 1) = 0, i.e. 4x$^4$  12x² + 1 = 0. It follows that $\alpha$² and $\beta$² are the roots of 4x²  12x + 1 = 0. 
July 10th, 2017, 12:23 PM  #5  
Newbie Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 23 Thanks: 1  a nicer solution. Quote:
Last edited by skipjack; July 10th, 2017 at 01:26 PM.  
July 10th, 2017, 02:55 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1422 
For x = $\alpha$ or $\beta$, 2(x)²  4(x) + 1 = 2x² + 4x + 1 = 0, so 2x²  4x + 1 = 0 is satisfied by x = $\alpha$ and x = $\beta$. If either quadratic expression is zero, their product is zero, so ±$\alpha$ and ±$\beta$ are roots of (2x² + 4x + 1)(2x²  4x + 1) = 0. That can be written as 4(x²)²  12x² + 1 = 0, which is a quadratic in x², so $\alpha$² and $\beta$² are roots of 4x²  12x + 1 = 0. It's often useful to start on a problem by considering what simple observations you can make, and whether they can help you find a way to tackle the problem. Initially, I saw that $\alpha$² and $\beta$² are the same as ($\alpha$)² and ($\beta$)², so if they are zeros of a quadratic in x, x = ±$\alpha$ and x = ±$\beta$ are zeros of a corresponding quadratic in x². Now, one can easily see how to find that quadratic in x² by considering its zeros in the order $\alpha$, $\beta$, $\alpha$, $\beta$. In effect, I had found a way to reduce the problem to the much easier problem of finding a quadratic equation with roots $\alpha$ and $\beta$. 
July 11th, 2017, 04:33 PM  #7  
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 175  Quote:
α + β = 4/2 = 2 αβ = 1/2 α² + β² = (α + β)² 2αβ = (2)² 2*(1/2) = 4  1 = 3 α²β² = (αβ)² = (1/2)² = 1/4 The quadratic equation with roots α² and β² is: x²  3x + 1/4 = 0$\iff$4x²  12x + 1 = 0 Last edited by skipjack; July 11th, 2017 at 04:54 PM.  

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