Question 9: If x = \\frac{1}{8-\sqrt{60}}), what is the value of x^{3} - 5x^{2} + 8x - 4?

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In all these questions, before plugging the value of x in the expression that has to be evaluated, rationalize x and get x expressed as a number with a rational denominator.

You are likely to find a clue after completing this step that will make things easier to solve.

**Step 1**: Rationalise \\frac{1}{8 - \sqrt{60}})

\\frac{1}{8 - \sqrt{60}}\\) x \\frac{8 + \sqrt{60}}{8 + \sqrt{60}})

= \\frac{8 + \sqrt{60}}{4}) = 2 + \\frac{\sqrt{15}}{2})

Or x - 2 = \\frac{\sqrt{15}}{2})

**Step 2**: Rewrite x^{3} - 5x^{2} + 8x - 4

Because x = 2 + \\frac{\sqrt{15}}{2}), try and express (x^{3} - 5x^{2} + 8x - 4) in terms of (x - 2)^{3}, (x - 2)^{2} and (x - 2).

(x - 2)^{3} = x^{3} - 6x^{2} + 12x - 8 -------- (1)

(x - 2)^{2} = x^{2} - 4x + 4 -------- (2)

Add or subtract the two equations to check whether you get the given expression

(1) + (2) = x^{3} - 5x^{2} + 8x - 4

So, x^{3} - 5x^{2} + 8x - 4 = (x - 2)^{3} + (x - 2)^{2}

= \{(\frac{\sqrt{15}}{2})^3}) + \{(\frac{\sqrt{15}}{2})^2})

= \\frac{15\sqrt{15}}{8}) + \\frac{15}{4})

= \\frac{15{(\sqrt{15} + 2)}}{8})

Class 9 Maths

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