Answer :
Answer:
[tex]\textsf{1.}\quad x=-2\;\;\text{or}\;\;x=-12[/tex]
[tex]\textsf{2.}\quad x=8\;\;\text{or}\;\;x=2[/tex]
[tex]\textsf{3.}\quad x=-0.4\;\;\text{or}\;\;x=-2[/tex]
[tex]\textsf{4.}\quad x=1\;\;\text{or}\;\;x=\dfrac{1}{5}[/tex]
Step-by-step explanation:
To solve a quadratic equation by the method of completing the square when the leading coefficient is one:
- Move the variable terms to the left side and the constants to the right side of the equation.
- Add the square of half the coefficient of the x-term to both sides of the equation.
- The left side of the equation is now a perfect square trinomial. Factor this.
- Square root both sides of the equation.
- Solve for x by isolating the variable.
[tex]\dotfill[/tex]
Question 1
Correct solution:
[tex]\begin{aligned}x^2+14x&=-24\\\\x^2+14x+\left(\dfrac{14}{2}\right)^2&=-24+\left(\dfrac{14}{2}\right)^2\\\\x^2+14x+7^2&=-24+7^2\\\\x^2+14x+49&=-24+49\\\\x^2+14x+49&=25\\\\(x+7)^2&=25\\\\\sqrt{(x+7)^2}&=\sqrt{25}\\\\x+7&=\pm5\\\\x+7&=5 \implies x=-2\\x+7&=-5 \implies x=-12\end{aligned}[/tex]
Therefore, the corrected worked solution is:
[tex]\boxed{\begin{aligned}x^2+14x&=-24\\x^2+14x+49&=25\\(x+7)^2&=25\\\\x+7=5\;\;&\text{or}\;\;x+7=-5\\\phantom{d}\:\:x=-2\;\;&\text{or}\;\;x=-12\end{aligned}}[/tex]
[tex]\dotfill[/tex]
Question 2
Correct solution:
[tex]\begin{aligned}x^2-10x+16&=0\\\\x^2-10x&=-16\\\\x^2-10x+\left(\dfrac{-10}{2}\right)^2&=-16+\left(\dfrac{-10}{2}\right)^2\\\\x^2-10x+(-5)^2&=-16+(-5)^2\\\\x^2-10x+25&=-16+25\\\\x^2-10x+25&=9\\\\(x-5)^2&=9\\\\\sqrt{(x-5)^2}&=\sqrt{9}\\\\x-5&=\pm3\\\\x-5&=3 \implies x=8\\x-5&=-3 \implies x=2\end{alilgned}[/tex]
Therefore, the corrected worked solution is:
[tex]\boxed{\begin{aligned}x^2-10x+16&=0\\x^2-10x+25&=9\\(x-5)^2&=3\\\\x-5=3\;\;&\text{or}\;\;x-5=-3\\\phantom{d}\:\:x=8\;\;&\text{or}\;\;x= 2\end{aligned}}[/tex]
[tex]\dotfill[/tex]
Question 3
Correct solution:
[tex]\begin{aligned}x^2+2.4x&=-0.8\\\\x^2+2.4x+\left(\dfrac{2.4}{2}\right)^2&=-0.8+\left(\dfrac{2.4}{2}\right)^2\\\\x^2+2.4x+1.2^2&=-0.8+1.2^2\\\\x^2+2.4x+1.44&=-0.8+1.44\\\\x^2+2.4x+1.44&=0.64\\\\(x+1.2)^2&=0.64\\\\\sqrt{(x+1.2)^2}&=\sqrt{0.64}\\\\x+1.2&=\pm 0.8\\\\x+1.2&=0.8 \implies x=-0.4\\x+1.2&=-0.8\implies x=-2\end{aligned}[/tex]
Therefore, the corrected worked solution is:
[tex]\boxed{\begin{aligned}x^2+2.4x&=-0.8\\x^2+2.4x+1.44&=0.64\\(x+1.2)^2&=0.64\\x+1.2&=\pm 0.8\\\\x+1.2=0.8\;\;&\text{or}\;\;x+1.2=-0.8\\\phantom{d}\:\:x=-0.4\;\;&\text{or}\;\;x= -2\end{aligned}}[/tex]
[tex]\dotfill[/tex]
Question 4
Correct solution:
[tex]\begin{aligned}x^2-\dfrac{6}{5}x+\dfrac{1}{5}&=0\\\\x^2-\dfrac{6}{5}x&=-\dfrac{1}{5}\\\\x^2-\dfrac{6}{5}x+\left(\dfrac{-\frac65}{2}\right)^2&=-\dfrac{1}{5}+\left(\dfrac{-\frac65}{2}\right)^2\\\\x^2-\dfrac{6}{5}x+\left(-\dfrac{3}{5}\right)^2&=-\dfrac{1}{5}+\left(-\dfrac{3}{5}\right)^2\\\\x^2-\dfrac{6}{5}x+\dfrac{9}{25}&=-\dfrac{1}{5}+\dfrac{9}{25}\\\\x^2-\dfrac{6}{5}x+\dfrac{9}{25}&=\dfrac{4}{25}\\\\\left(x-\dfrac{3}{5}\right)^2&=\dfrac{4}{25}\end{aligned}[/tex]
[tex]\begin{aligned}\sqrt{\left(x-\dfrac{3}{5}\right)^2}&=\sqrt{\dfrac{4}{25}}\\\\x-\dfrac{3}{5}&=\pm\dfrac{2}{5}\\\\x-\frac{3}{5}&=\frac{2}{5} \implies 1\\\\x-\frac{3}{5}&=-\frac{2}{5} \implies \frac{1}{5}\end{aligned}[/tex]
Therefore, the corrected worked solution is:
[tex]\boxed{\begin{aligned}x^2-\dfrac{6}{5}x+\dfrac{1}{5}&=0\\\\x^2-\dfrac{6}{5}x+\dfrac{9}{25}&=\dfrac{4}{25}\\\\\left(x-\dfrac{3}{5}\right)^2&=\dfrac{4}{25}\\\\x-\frac{3}{5}=\frac{2}{5}\;\;&\text{or}\;\;x-\frac{3}{5}=-\frac{2}{5}\\\phantom{d}\:\:x=1\;\;&\text{or}\;\;x= \frac{1}{5}\end{aligned}}[/tex]
