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 April 21st, 2013, 09:39 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Sum & Product of the Roots of a Quadratic Equation My answer seems relatively close to that of the text book. Can anyone see where I've gone wrong, or does the books answer contain a typo? Many thanks. Q. If $\alpha$ & $\beta$ are the roots of the equation $2x^2-x-4=0$, form the equation whose roots are $\alpha+5$ & $\beta+5$. Attempt: $a=2,$ $b=-1,$ $c=-4$ Sum of Roots: $\alpha+5+\beta+5=\alpha+\beta+10=\frac{-b}{a}+10$...(where $\alpha+\beta=\frac{-b}{a})$ $=\frac{21}{2}$ Product of Roots: $(\alpha+5)(\beta+5)=\alpha\beta+5\alpha+5\beta+25= \frac{c}{a}+5(\alpha+\beta)+25$...(where $\alpha\beta=\frac{c}{a})$ $=\frac{-4}{2}+5(\frac{21}{2})+25=\frac{151}{2}$ $x^2-\frac{21}{2}(x)+\frac{151}{2}=2x^2-21x+151$ $Ans:$ (From text book $2x^2-21x+51$
 April 21st, 2013, 10:39 AM #2 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Sum & Product of the Roots of a Quadratic Equation the answer is (2(x-5)^2-(x-5)-4) = 2x^2-21x+51 = 0. 2x^2-21x+51 isn't an equation so the answer from the textbook is wrong.
April 21st, 2013, 11:55 AM   #3
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Re: Sum & Product of the Roots of a Quadratic Equation

Hello, bilano99!

Quote:
 $\text{If }\alpha\text{ and }\beta\text{ are the roots of the equation: }\:2x^2\,-\,x\,-\,4\:=\:0 \;\;\;\text{form the equation whose roots are: }\,\alpha\,+\,5\,\text{ and }\,\beta\,+\,5$

$\text{W\!e have: }\:x^2\,-\,\frac{1}{2}x\,-\,2 \:=\:0$

$\text{Hence: }\:\alpha\,+\,\beta \,=\,\frac{1}{2}\,,\;\;\alpha\beta \:=\:-2\;\;[1]$

$\text{W\!e have the equation: }\:\big[x\,-\,(\alpha\,+\,5)\big]\,\big[x\,-\,(\beta\,+\,5)\big] \:=\:0$

[color=beige]. . . . . . [/color]$x^2\,-\,(\alpha\,+\,5)x\,-\,(\beta\,+\,5)x \,+\,(\alpha\,+\,5)(\beta\,+\,5) \:=\:0$

[color=beige]. . . . . . [/color]$x^2 \,-\,(\alpha\,+\,\beta\,+\,10)x\,+\,(\alpha\beta \,+\,5\alpha\,+\,5\beta\,+\,25) \:=\:0$

[color=beige]. . . . . [/color]$x^2\,-\,\big[(\alpha\,+\,\beta)\,+\,10\big]x \,+\,\big[\alpha\beta \,+\,5(\alpha\,+\,\beta)\,+\,25\big] \:=\:0$

$\text{Substitute [1]:}\;\;\;x^2\,-\,\left[\frac{1}{2}\,+\,10\right]x\,+\,\left[-2\,+\,5\left(\frac{1}{2}\right)\,+\,25\right] \:=\:0$

[color=beige]. . . . . . . . . . . . . . . . . [/color]$x^2\,-\,\frac{21}{2}x\,+\,\frac{51}{2} \:=\:0$

[color=beige]. . . . . . . . . . . . . . . . [/color]$2x ^2\,-\,21x \,+\,51 \;=\:0$

 April 21st, 2013, 12:22 PM #4 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Sum & Product of the Roots of a Quadratic Equation Thank you very much.
 April 21st, 2013, 01:22 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond Re: Sum & Product of the Roots of a Quadratic Equation $2x^2\,-\,x\,-\,4\,=\,0\,\Rightarrow\,x\,=\,\frac{1\,\pm\,\sqrt{ 33}}{4}$ $$$x\,-\,\frac{21\,+\,\sqrt{33}}{4}$$$$x\,-\,\frac{21\,-\,\sqrt{33}}{4}$$\,=\,0$ $x^2\,-\,\frac{21}{2}x\,+\,\frac{51}{2}\,=\,0$ $2x^2\,-\,21x\,+\,51\,=\,0$
April 21st, 2013, 06:08 PM   #6
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Re: Sum & Product of the Roots of a Quadratic Equation

Quote:
 Originally Posted by bilano99 . . . $=\frac{-4}{2}+5(\frac{21}{2})+25=\frac{151}{2}$
That should be $\,=\,\frac{-4}{2}\,+\,5$$\frac{1}{2}$$\,+\,25\,=\,\frac{51}{2} \,.$

The required equation is 2(x - 5)² - (x - 5) - 4 = 0, i.e., 2x² - 21x + 51 = 0.

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