+0

# Algebra Question

0
281
1
+4

Wanted to check if essay writer solved my problem X(squared) - 6x - 39 = 0 right:

x2 - 6x - 39 = 0

D = b2 - 4ac = (-6)2 - 4·1·(-39) = 36 + 156 = 192

x1 = 6 - √1922·1 = 3 - 4√3 ≈ -3.9282032302755088

x2 = 6 + √1922·1 = 3 + 4√3 ≈ 9.928203230275509

Is it correct?

Feb 22, 2018

#1
+22010
0

Wanted to check if essay writer solved my problem X(squared) - 6x - 39 = 0 right:

It is correct!

We proof: $$x_1 = 3-4\sqrt{3}:$$

$$\large{ \begin{array}{|rcll|} \hline x^2-6x-39 &=& 0 \quad & | \quad x_1 = 3 - 4\sqrt{3} \\ (3 - 4\sqrt{3})^2-6(3 - 4\sqrt{3})-39 & \overset{?}{=} & 0 \\ 9 - 24\sqrt{3} + 16\cdot 3 - 18 + 24\sqrt{3} - 39 & \overset{?}{=} & 0 \\ 9 + 16\cdot 3 - 18 - 39 & \overset{?}{=} & 0 \\ 9 + 48 - 18 - 39 & \overset{?}{=} & 0 \\ 57-57 & \overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array} }$$

We proof $$x_2 = 3+4\sqrt{3}:$$

$$\large{ \begin{array}{|rcll|} \hline x^2-6x-39 &=& 0 \quad & | \quad x_2 = 3 + 4\sqrt{3} \\ (3 + 4\sqrt{3})^2-6(3 + 4\sqrt{3})-39 & \overset{?}{=} & 0 \\ 9 + 24\sqrt{3} + 16\cdot 3 - 18 - 24\sqrt{3} - 39 & \overset{?}{=} & 0 \\ 9 + 16\cdot 3 - 18 - 39 & \overset{?}{=} & 0 \\ 9 + 48 - 18 - 39 & \overset{?}{=} & 0 \\ 57-57 & \overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array} }$$

Feb 22, 2018