Question

Find the roots of equation, if they exists, by applying the quadratic formula :

$x {}^{2} - ( \sqrt{3} + 1)x + \sqrt{3} = 0$

Answer 1

Quadratic Equations

A quadratic equation is one of the form ax2+ bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.

Factoring

This approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.

Example 1.

2x2 - 5x - 12 = 0.

(2x + 3)(x - 4) = 0.

2x + 3 = 0 or x - 4 = 0.

x = -3/2, or x = 4.

Square Root Principle

If x2 = k, then x = ± sqrt(k).

Example 2.

x2 - 9 = 0.

x2 = 9.

x = 3, or x = -3.

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Quadratic Formula

The solutions for the equation ax2 + bx + c = 0 are

- b +-√ b2- 4ac/2a
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A/Q

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The given equation is x2−(3‾√+1)x+3‾√=0.

Comparing it with ax2+bx+c=0, we get

a = 1, b = −(3‾√+1) and c = 3‾√

∴ Discriminant, D = b2−4ac=[−(3‾√+1)]2−4×1×3‾√=3+1+23‾√−43‾√=3−23‾√+1=(3‾√−1)2>0

So, the given equation has real roots.

Now, D‾‾√=(3‾√−1)2‾‾‾‾‾‾‾‾‾‾√=3‾√−1

∴ α=−b+D√2a=−[−(3√+1)]+(3√−1)2×1=3√+1+3√−12=23√2=3‾√β=−b−D√2a=−[−(3√+1)]−(3√−1)2×1=3√+1−3√+12=22=1

Hence, 3‾√ and 1 are the roots of the given equation.

Answer 2

Given Equation is x^2 - (√3 + 1)x + √3 = 0.

Here, a = 1, b = -(√3 + 1) = -√3 - 1, c = √3.

(i)

$=> \frac{-b+ \sqrt{b^2 - 4ac}}{2a}$

$=> \frac{-(-1-\sqrt{3})+\sqrt{(1+\sqrt{3})^2 - 4\sqrt{3}}}{2}$

$=>\frac{1+\sqrt{3} +\sqrt{3} -1}{2}$

$=>\frac{2 \sqrt{3}}{2}$

$=>\sqrt{3}$

(ii)

$=>x=\frac{-b- \sqrt{b^2 - 4ac}}{2a}$

$=> \frac{-( \sqrt{3}-1)-\sqrt{(1+\sqrt{3})^2 - 4\sqrt{3}}}{2}$

$=> \frac{-( \sqrt{3} - 1)-\sqrt{3} + 1}{2}$

$=> \frac{1 + 1}{2}$

$=> 1$

Therefore, the roots are:

$=> x = \boxed{\sqrt{3},1}$

Hope it helps!