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### Solved Examples

**Question 1:**Simplify $\frac{45x^{12}}{30x^6}$

**Solution:**

Given $\frac{45x^{12}}{30x^6}$

Step 1:

Factorized the numerator and denominator

45 x

30 x

Step 2:

$\frac{45x^{12}}{30x^6}$ = $\frac{3 * 3 * 5 * x^{12}}{ 2 * 3 * 5 * x^6}$

= $\frac{3 * x^{12}}{ 2 * x^6}$

= $\frac{3}{2}$$ x^{12 - 6}$

[ $\frac{x^m}{x^n}$ = x

= $\frac{3}{2}$$ x^6$

=> $\frac{45x^{12}}{30x^6}$ = $\frac{3}{2}$$ x^6$

Step 1:

Factorized the numerator and denominator

45 x

^{12}= 3 * 3 * 5 * x^{12}30 x

^{6}= 2 * 3 * 5 * x^{6}Step 2:

$\frac{45x^{12}}{30x^6}$ = $\frac{3 * 3 * 5 * x^{12}}{ 2 * 3 * 5 * x^6}$

= $\frac{3 * x^{12}}{ 2 * x^6}$

= $\frac{3}{2}$$ x^{12 - 6}$

[ $\frac{x^m}{x^n}$ = x

^{m - n}]= $\frac{3}{2}$$ x^6$

=> $\frac{45x^{12}}{30x^6}$ = $\frac{3}{2}$$ x^6$

**Question 2:**Solve the quadratic equation by using formula

2x

^{2}+ 5x - 3 = 0

**Solution:**

Given quadratic equation 2x

Comparing with equation ax

a = 2, b = 5 and c = - 3

b

= 25 + 24

= 49

and $\sqrt{b^2 - 4ac}$ = $\sqrt{49}$

= 7

=> $\sqrt{b^2 - 4ac}$ = 7

therefore

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

= $\frac{- 5 \pm 7}{2 * 2}$

= $\frac{- 5 \pm 7}{4}$

=> x = $\frac{- 5 + 7}{4}$ or x = $\frac{- 5 - 7}{4}$

=> x = $\frac{1}{2}$, - 3.

^{2}+ 5x - 3 = 0Comparing with equation ax

^{2}+ bx + c = 0a = 2, b = 5 and c = - 3

b

^{2}- 4ac = 5^{2}- 4 * 2 * - 3= 25 + 24

= 49

and $\sqrt{b^2 - 4ac}$ = $\sqrt{49}$

= 7

=> $\sqrt{b^2 - 4ac}$ = 7

therefore

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

= $\frac{- 5 \pm 7}{2 * 2}$

= $\frac{- 5 \pm 7}{4}$

=> x = $\frac{- 5 + 7}{4}$ or x = $\frac{- 5 - 7}{4}$

=> x = $\frac{1}{2}$, - 3.

**Question 3:**Simplify $\frac{x^3 - 2^3}{x^2 + 2x - 8}$

**Solution:**

Given, $\frac{x^3 - 2^3}{x^2 + 2x - 8}$

Step 1:

x

= (x - 2)(x

[x

and

x

= x(x + 4) - 2(x + 4)

= (x + 4)(x - 2)

Step 2:

$\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{ (x - 2)(x^2 + 2x + 4)}{ (x + 4)(x - 2) }$

= $\frac{x^2 + 2x + 4}{x + 4}$

=> $\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{x^2 + 2x + 4}{x + 4}$

Step 1:

x

^{3}- 2^{3}= (x - 2)(x^{2}+ 2x + 2^{2})= (x - 2)(x

^{2}+ 2x + 4)[x

^{3}- y^{3}= (x - y)(x^{2}+ xy + y^{2})]and

x

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

= (x + 4)(x - 2)

Step 2:

$\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{ (x - 2)(x^2 + 2x + 4)}{ (x + 4)(x - 2) }$

= $\frac{x^2 + 2x + 4}{x + 4}$

=> $\frac{x^3 - 2^3}{x^2 + 2x - 8}$ = $\frac{x^2 + 2x + 4}{x + 4}$