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# Algebra Question

0
94
4
+180

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

Dec 12, 2019

#1
+19820
0

This might work for ya

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

Dec 12, 2019
#2
+180
+1

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

Dec 12, 2019
#3
+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
+23905
+1

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

Find $$k$$.

Let the roots are $$x_1$$, $$x_2$$.

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

$$\boxed{k=32}$$

Dec 12, 2019