+26367 How would I solve this equation algebraically?
\(\large{ \sqrt{5x^2+11}} = x + 5\)
\(\begin{array}{|rcll|} \hline \sqrt{5x^2+11} &=& x + 5 \quad & | \quad \text{square both sides} \\ 5x^2+11 &=& (x + 5)^2 \\ 5x^2+11 &=& x^2+10x+25 \\ 4x^2-10x-14 &=& 0 \quad & | \quad : 2 \\ 2x^2-5x-7 &=& 0 \\ \\ x &=& \dfrac{ 5\pm\sqrt{25-4\cdot 2 \cdot (-7) } } {2\cdot 2 } \\\\ x &=& \dfrac{ 5\pm\sqrt{25+56 } } { 4 } \\\\ x &=& \dfrac{ 5\pm\sqrt{81} } { 4 } \\\\ x &=& \dfrac{ 5\pm 9 } { 4 } \\\\ x_1 &=& \dfrac{ 5 + 9 } { 4 } \\\\ x_1 &=& \dfrac{ 14 } { 4 } \\\\ \mathbf{ x_1 } &\mathbf{ =}& \mathbf{ \dfrac{ 7 } { 2 }} \\\\ x_2 &=& \dfrac{ 5 - 9 } { 4 } \\\\ x_2 &=& -\dfrac{ 4 } { 4 } \\\\ \mathbf{ x_2} &\mathbf{ =}& \mathbf{ -1} \\ \hline \end{array}\)
+997 \({ \sqrt{5x^2+11}} = x + 5\)... you have this equation.
Sqaure both sides to give you \(5x^2 + 11 = (x+5)^2\)
Solving this, we get \(5x^2 + 11 = x^2 + 10x + 25\).
This is equal to \(4x^2 - 10x - 14 = 0\)
Factorizing this equation, we get \((2x-7)(2x+2) = 0\). (You can test that out to see if it matches the previous equation of \(4x^2 - 10x - 14 = 0\).)
Since \((2x-7)(2x+2)\) has to equal zero, at least one of the brackets need to equal zero for the equation to equal zero. Therefore, we can end up with two solutions.
\(x\) is either \(\boxed{\frac{7}{2}}\), or \(x\) is \(\boxed{-1}\).
