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 July 10th, 2017, 12:41 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, 01:29 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,021 Thanks: 666 Math Focus: Physics, mathematical modelling, numerical and computational solutions Some hints for you: 1. Factorise the left-hand 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 1-3 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 Thanks from topsquark and CapitalSwine
 July 10th, 2017, 02:45 AM #3 Senior Member     Joined: Feb 2010 Posts: 621 Thanks: 98 It is possible to get the answer without actually finding $\displaystyle \alpha$ and $\displaystyle \beta$. Google "elementary symmetric functions".
 July 10th, 2017, 03:23 AM #4 Global Moderator   Joined: Dec 2006 Posts: 17,533 Thanks: 1322 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. Thanks from topsquark, Shariq Faraz and CapitalSwine
July 10th, 2017, 11:23 AM   #5
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a nicer solution.

Quote:
 Originally Posted by skipjack 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.
Could please share your thought process regarding how you made such indirect approach to the required quadratic by presuming it to be the product of the given quadratic and the other replacing x by -x?

Last edited by skipjack; July 10th, 2017 at 12:26 PM.

 July 10th, 2017, 01:55 PM #6 Global Moderator   Joined: Dec 2006 Posts: 17,533 Thanks: 1322 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 -$\beta$. Thanks from Shariq Faraz
July 11th, 2017, 03:33 PM   #7
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Quote:
 Originally Posted by CapitalSwine The quadratic equation 2x^2+4x+1=0 has roots α and β. Find the quadratic equation with roots α^2 and β^2.
Vieta's formulas:

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 03:54 PM.

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