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avatar+180 

One root of x^2+12x+k=0 is twice the other root. Find k.

 Dec 12, 2019
 #1
avatar+19820 
0

This might work for ya

(x+4)(x+8)     roots  -4   and -8       k = 32

 Dec 12, 2019
 #2
avatar+180 
+1

I see, I did this using Vieta's Formula and got the same answer. Thank you!!!

 Dec 12, 2019
 #3
avatar+107089 
+1

I can do it a third way, although to be honest I do not know what Vieta's formula is

 

Sum of the roots in any polynomial is  = -b/a

 

In this case that is  -12/1 = -12

 

let one root be T the other is twice the size so   

T+2T=-12

3T=-12

T= -4

 

So the roots are  -4 and -8  so k is 32

 Dec 12, 2019
 #4
avatar+23905 
+1

One root of \(x^2+12x+k=0\) is twice the other root.

Find \(k\).

 

Let the roots are \(x_1\), \(x_2\).

 

\(\begin{array}{|rcll|} \hline x^2+12x+k=0 &=& (x-x_1)(x-x_2)= 0 \quad | \quad x_2 = 2x_1 \\\\ &=& (x-x_1)\Big(x-(2x_1)\Big) \\ &=& (x-x_1)(x-2x_1) \\ &=& x^2-2x_1x-x_1x+2x^2_1 \\ x^2+12x+k=0 &=& x^2-3x_1x+2x^2_1 \\ && \text{compare} \\ && \boxed{12=-3x_1\ \text{ or }\ x_1=-4 \\ k= 2x^2_1\ \text{ or }\ k=2*(-4)^2=32 } \\ \hline \end{array} \)

 

\(\boxed{k=32}\)

 

laugh

 Dec 12, 2019

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