solve the equation $x^{2}+6x = 25$ for $x$ by completing the square. round decimal answers to the nearest…

solve the equation $x^{2}+6x = 25$ for $x$ by completing the square. round decimal answers to the nearest tenth. enter the answers in the boxes.\n$x^{2}+6x = 25$\n$x^{2}+6x+square=25+square$\n$(x + square)^{2}=square$\n$x+square=pmsquare$\n$x=square$ and $x=-square$

solve the equation $x^{2}+6x = 25$ for $x$ by completing the square. round decimal answers to the nearest tenth. enter the answers in the boxes.\n$x^{2}+6x = 25$\n$x^{2}+6x+square=25+square$\n$(x + square)^{2}=square$\n$x+square=pmsquare$\n$x=square$ and $x=-square$

Answer

Explanation:

Step1: Find the value to add to complete the square

For the quadratic equation $x^{2}+bx$, we add $(\frac{b}{2})^{2}$. Here $b = 6$, so $(\frac{6}{2})^{2}=9$. $x^{2}+6x + 9=25 + 9$

Step2: Rewrite the left - hand side as a perfect square

Using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a=x$ and $b = 3$, we get $(x + 3)^{2}=34$

Step3: Take the square root of both sides

$x+3=\pm\sqrt{34}$

Step4: Solve for x

$x=-3\pm\sqrt{34}$ $x=-3+\sqrt{34}\approx - 3+5.8 = 2.8$ and $x=-3-\sqrt{34}\approx-3 - 5.8=-8.8$

Answer:

$x^{2}+6x+9 = 25+9$; $(x + 3)^{2}=34$; $x+3=\pm\sqrt{34}$; $x = 2.8$ and $x=-8.8$