Theory Questions – Surds

Question 1

Question: If \(\scriptsize x = \normalsize \frac{1}{2} \left( \scriptsize1 \: – \: \sqrt{2}  \right)\)

Evaluate \(\left( \scriptsize 2x^2 \: – \: 2x \right )\)

Solution

⇒ \(\scriptsize x = \normalsize \frac{1}{2} \left( \scriptsize1 \: – \: \sqrt{2}  \right)\)

Substitute the value of x into \(\scriptsize 2x^2 \: – \: 2x\)

 = \(\scriptsize 2\left [ \normalsize \frac{1}{2} \left(\scriptsize1 \: – \: \sqrt{2} \right)  \right ]^2 \scriptsize \: – \: 2\left [ \normalsize \frac{1}{2} \left(\scriptsize 1 \: – \: \sqrt{2} \right)  \right ]\)

= \(\scriptsize 2\left [ \normalsize \frac{1}{4} \left(\scriptsize1 \: – \: \sqrt{2} \right)^2  \right ]\scriptsize  \: – \:  \left( 1 \: – \: \sqrt{2} \right)\)

Expanding

⇒ \(\left( \scriptsize 1 \: – \: \sqrt{2} \right)^2 \\ \scriptsize = (1 \: – \: \sqrt{2})(1 \: – \: \sqrt{2}) \\ \scriptsize =  1 \: – \: \sqrt{2} \: – \: \sqrt{2} \: + \: 2 \\ \scriptsize = 1 \: – \: 2\sqrt{2} \: + \: 2\)

Hence, \(\scriptsize 2x^2 \: – \: 2x\)

= \(\scriptsize 2 \left [ \normalsize \frac{1}{4} \left( \scriptsize 1 \: – \: 2\sqrt{2} \: + \: 2 \right) \right] \scriptsize  \: – \:  1 \: + \: \sqrt{2}\)

= \(\frac{1}{2} \left ( \scriptsize 1  \: – \: 2\sqrt{2} \: + \: 2 \right) \scriptsize  \: – \:  1 \: + \: \sqrt{2}\)

= \(\frac{1 \: – \: 2\sqrt{2}\: + \: 2 \: – \: 2 \: + \: 2 \sqrt{2}}{2}\)

= \(\frac{1}{2}\)

Question 2

Question: Using the method of completing the square, solve \(\scriptsize 4x^2 \: -\: 4\sqrt{3x} \: + \: 3 = 0\)

Leaving the answer in surd form.

Solution

Using completing the square method

⇒ \(\scriptsize 4x^2 \: -\: 4\sqrt{3x} \: + \: 3 = 0\)

Move constant to the R.H.S

⇒ \(\scriptsize 4x^2 \: -\: 4\sqrt{3x} = -3\)

Divide through by 4

⇒ \(\scriptsize x^2 \: – \: \sqrt{3x} = \normalsize \frac{-3}{4}\)

Add the square of half of the coefficient of x to both sides

⇒ \(\scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-\sqrt{3}}{2} \right)^2 = \normalsize \frac{-3}{4} \: + \: \left ( \normalsize \frac{-\sqrt{3}}{2} \right)^2\)

⇒ \(\scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-\sqrt{3}}{2} \right)^2 = \normalsize \frac{-3}{4} \: + \: \frac{3}{4}\)

⇒ \(\scriptsize x^2 \: – \: \sqrt{3x} \: + \: \left ( \normalsize \frac{-\sqrt{3}}{2} \right)^2 =0\)

Combine the squares on the L.H.S

⇒ \(\left(\scriptsize x \: – \:   \normalsize \frac{\sqrt{3}}{2}\right) \scriptsize = 0\)

⇒ \(\scriptsize x  \: – \: \normalsize \frac{\sqrt{3}}{2} \scriptsize = 0\)

⇒ \(\scriptsize x = \normalsize \frac{\sqrt {3}}{2}\)