+0

# Algebra Question

0
64
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?

florachavezz  Feb 22, 2018
Sort:

#1
+19066
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 $$x_2 = 3+4\sqrt{3}:$$

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

heureka  Feb 22, 2018

### 5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details