Solution:

\(3y^2-5y+2=0\)

\(\frac{3y^2}{3}-\frac{5y}{3}+\frac{2}{3}=\frac{0}{3}\)

\(\Rightarrow y^2-\frac{5y}{3}+\frac{2}{3}=0\)

\(\Rightarrow y^2-\frac{5y}{3}=-\frac{2}{3}\)

Now we find \(\frac{1}{2} \text{of } -\frac{5}{3}\) and square it.

i.e \(\frac{1}{2}(-\frac{5}{3})=(-\frac{5}{6})^2\)

add \((-\frac{5}{6})^2 \text{ to} -\frac{5}{3}y \text{ and} -\frac{2}{3}.\)

\(\Rightarrow y^2-\frac{5}{3}y+(-\frac{5}{6})^2=-\frac{2}{3}+(-\frac{5}{6})^2\)

\(\Rightarrow (y-\frac{5}{6})^2=-\frac{2}{3}+\frac{25}{36}\)

\(\Rightarrow (y-\frac{5}{6})^2=\frac{1}{36}\)

Finding square root of both sides

\(\Rightarrow \sqrt{(y-\frac{5}{6})^2}=\sqrt{\frac{1}{36}}\)

\(\Rightarrow y-\frac{5}{6}=\pm\frac{1}{6}\)

\(\Rightarrow y=\pm\frac{1}{6}+\frac{5}{6}\)

Either \(y=\frac{1}{6}+\frac{5}{6}\) or \(y=-\frac{1}{6}+\frac{5}{6}\)

For \( y=\frac{1}{6}+\frac{5}{6}\)

\(y=1\)

And for \(y=-\frac{1}{6}+\frac{5}{6}\)

\(y=-\frac{4}{6}=\frac{2}{3}=0.67\) to two decimal places

Hence \(y=1.00 \text{ and } 0.67\) to two decimal places.

By: Abdul Karim

Solution:

\(\frac{1}{x}+\frac{1}{x+3}=\frac{1}{2}\)

\(\frac{(x+3)+x}{x(x+3)}=\frac{1}{2}\)

\(\frac{2x+3}{x(x+3)}=\frac{1}{2}\) [Cross multiply]

\(\Rightarrow 2(2x+3)=1\times x(x+3)\)

\(\Rightarrow 4x+6=x^2+3x\)

\(\Rightarrow 0=x^2+3x-4x-6\)

\(\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\)

\(\Rightarrow x(x-3)+2(x-3)=0\)

\(\Rightarrow (x-3)(x+2)=0\)

Either \(x-3=0 \text{ or } x+2=0\)

For \(x-3=0 \Rightarrow x=3\)

And for \(x+2=0 \Rightarrow x=-2\)

Hence \(x=3 \text{ or} -2.\)

By: Abdul Karim

Solution:

When \(k\) is added to \(y^2-12y\) means \(y^2-12y+k.\) The expression becomes \((y+p)^2\) implies \(y^2-12y+k=(y+p)^2\)

\(\Rightarrow y^2-12y+k=(y+p)(y+p)\)

\(\Rightarrow y^2-12y+k=y^2+2yp+p^2\)

Comparing the L.H.S and R.H.S of the equation

\(\Rightarrow [y^2=y^2, -12y=2yp, k=p^2]\)

\(\Rightarrow -12=2p ……(1)\)

\(\Rightarrow k=p^2 …….(2)\)

From eqn (1), \(-12=2p\)

\(\Rightarrow \frac{-12}{2}=\frac{2p}{2}\)

\(\Rightarrow p=-6\)

Put \(p=-6\) into eqn (2), \(k=p^2\)

\(\Rightarrow k=(-6)^2\)

\(\Rightarrow k=36\)

By: Abdul Karim