Find all real solutions of the equation. $x ^ { 2 } - \sqrt { 29 } x + 1 = 0$
A) $x = 12 , x = - 17 / 2$
B) $x = \sqrt { 29 } / 2 , x = \sqrt { 17 } / 2$
C) $x = \sqrt { 19 } / 2 , x = - \sqrt { 17 } / 2$
D) $x = \frac { \sqrt { 29 } - 5 } { 2 } , x = \frac { \sqrt { 29 } + 5 } { 2 }$
E) none of these

Question

Find all real solutions of the equation. $x ^ { 2 } - \sqrt { 29 } x + 1 = 0$ 
A) $x = 12 , x = - 17 / 2$
B) $x = \sqrt { 29 } / 2 , x = \sqrt { 17 } / 2$
C) $x = \sqrt { 19 } / 2 , x = - \sqrt { 17 } / 2$
D) $x = \frac { \sqrt { 29 } - 5 } { 2 } , x = \frac { \sqrt { 29 } + 5 } { 2 }$
E) none of these

Quizplus · Accepted Answer

We can use the quadratic formula, which states that if $ax^2+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, we have $a=1, b=-\sqrt{29}, c=1$, so
$$x=\frac{\sqrt{29}\pm\sqrt{29^2-4}}{2}=\frac{\sqrt{29}\pm\sqrt{109}}{2}.$$
Thus the solutions are $x=\frac{\sqrt{29}-\sqrt{109}}{2}$ and $x=\frac{\sqrt{29}+\sqrt{109}}{2}$.
We can simplify this using the factored form of 109: $109=7\cdot 13$. Then, we have
$$\frac{\sqrt{29}-\sqrt{109}}{2}=\frac{\sqrt{29}-\sqrt{7}\sqrt{13}}{2}=\frac{\sqrt{29}-\sqrt{91}}{2}=\frac{\sqrt{29}-5}{2}$$
and
$$\frac{\sqrt{29}+\sqrt{109}}{2}=\frac{\sqrt{29}+\sqrt{7}\sqrt{13}}{2}=\frac{\sqrt{29}+\sqrt{91}}{2}=\frac{\sqrt{29}+5}{2}.$$
Therefore, the solutions are $x=\frac{\sqrt{29}-5}{2}$ and $x=\frac{\sqrt{29}+5}{2}$, which corresponds to choice D.

Find All Real Solutions of the Equation x2−29x+1=0x ^ { 2 } - \sqrt { 29 } x + 1 = 0x2−29​x+1=0

Find All Real Solutions of the Equation $x ^ { 2 } - \sqrt { 29 } x + 1 = 0$