what is the solution of $sqrt{x^{2}+49}=x + 5$?\n$x=\frac{12}{5}$\n$x=-\frac{12}{5}$\n$x=-6$ or $x=-3$\nno…

what is the solution of $sqrt{x^{2}+49}=x + 5$?\n$x=\frac{12}{5}$\n$x=-\frac{12}{5}$\n$x=-6$ or $x=-3$\nno solution

what is the solution of $sqrt{x^{2}+49}=x + 5$?\n$x=\frac{12}{5}$\n$x=-\frac{12}{5}$\n$x=-6$ or $x=-3$\nno solution

Answer

Answer:

A. $x = \frac{12}{5}$

Explanation:

Step1: Square both sides

$(\sqrt{x^{2}+49})^2=(x + 5)^2$ $x^{2}+49=x^{2}+10x + 25$

Step2: Simplify the equation

Subtract $x^{2}$ from both sides: $49=10x + 25$

Step3: Solve for x

Subtract 25 from both sides: $10x=49 - 25$ $10x=24$ Divide both sides by 10: $x=\frac{24}{10}=\frac{12}{5}$

Step4: Check the solution

Substitute $x = \frac{12}{5}$ into the original equation: Left - hand side: $\sqrt{(\frac{12}{5})^{2}+49}=\sqrt{\frac{144}{25}+49}=\sqrt{\frac{144 + 1225}{25}}=\sqrt{\frac{1369}{25}}=\frac{37}{5}$ Right - hand side: $\frac{12}{5}+5=\frac{12 + 25}{5}=\frac{37}{5}$ The solution $x=\frac{12}{5}$ is valid.