The answer to 9k2 – 6k = 11 is k = 11/12. Though, I'm not sure by the square part.
Here's the decimal form: k=0.91667
Steps:
Multiply the numbers. (9*2)
Add similar elements. (18k - 6k = 11)
Divide both sides by 12.
Simplify.
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Hope this helped!
Answer:
The solutions of the given equation are [tex]k=\frac{1}{3}+\frac{2}{\sqrt{3}}[/tex] and [tex]k=\frac{1}{3}-\frac{2}{\sqrt{3}}[/tex].
Explanation:
The given equation is
[tex]9k^2-6k=11[/tex]
Divide both sides by 9.
[tex]k^2-\frac{2}{3}k=\frac{11}{9}[/tex]
If an expression is [tex]x^2+bx[/tex] then we need to add [tex](\frac{b}{2})^2[/tex] in the expression to make it perfect square.
In the above equation [tex]b=\frac{2}{3}[/tex], so add [tex](\frac{1}{3})^2[/tex] on both the sides.
[tex]k^2-\frac{2}{3}k+(\frac{1}{3})^2=\frac{11}{9}+(\frac{1}{3})^2[/tex]
[tex](k-\frac{1}{3})^2=\frac{11}{9}+\frac{1}{9}[/tex]
[tex](k-\frac{1}{3})^2=\frac{12}{9}[/tex]
[tex](k-\frac{1}{3})^2=\frac{4}{3}[/tex]
Taking square root both the sides.
[tex]k-\frac{1}{3}=\pm \sqrt{\frac{4}{3}}[/tex]
Add 1/3 on both the sides.
[tex]k=\frac{1}{3}\pm \frac{2}{\sqrt{3}}[/tex]
Therefore the solutions of the given equation are [tex]k=\frac{1}{3}+\frac{2}{\sqrt{3}}[/tex] and [tex]k=\frac{1}{3}-\frac{2}{\sqrt{3}}[/tex].
