Q1. $(2\text{x} - 3) \ (\text{x} + 2) = 0$
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Converting into standard form:
$\Rightarrow (2\text{x} - 3) \ (\text{x} + 2) = 0$
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 3\text{x} - 6 = 0$
$\Rightarrow 2\text{x}^{2} + \text{x} - 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 3\text{x} - 6 = 0$
$\Rightarrow 2\text{x}(\text{x} + 2) - 3(\text{x} + 2) = 0$
$\Rightarrow (\text{x} + 2) \ (2\text{x} - 3) = 0$
$\therefore \text{x} + 2 = 0$ or $2\text{x} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 2$ or $\text{x} = \dfrac{3}{2}$ Ans
$\Rightarrow (2\text{x} - 3) \ (\text{x} + 2) = 0$
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 3\text{x} - 6 = 0$
$\Rightarrow 2\text{x}^{2} + \text{x} - 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 3\text{x} - 6 = 0$
$\Rightarrow 2\text{x}(\text{x} + 2) - 3(\text{x} + 2) = 0$
$\Rightarrow (\text{x} + 2) \ (2\text{x} - 3) = 0$
$\therefore \text{x} + 2 = 0$ or $2\text{x} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 2$ or $\text{x} = \dfrac{3}{2}$ Ans
Q2. $\text{x}(\text{x} - 1) = 42$
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Converting into standard form:
$\Rightarrow \text{x}(\text{x} - 1) = 42$
$\Rightarrow \text{x}^{2} - \text{x} = 42$
$\Rightarrow \text{x}^{2} - \text{x} - 42 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 7\text{x} + 6\text{x} - 42 = 0$
$\Rightarrow \text{x}(\text{x} - 7) + 6(\text{x} - 7) = 0$
$\Rightarrow (\text{x} - 7) \ (\text{x} + 6) = 0$
$\therefore \text{x} - 7 = 0$ or $\text{x} + 6 = 0$
As such, we have:
$\Rightarrow \text{x} = 7$ or $\text{x} = - \ 6$ Ans
$\Rightarrow \text{x}(\text{x} - 1) = 42$
$\Rightarrow \text{x}^{2} - \text{x} = 42$
$\Rightarrow \text{x}^{2} - \text{x} - 42 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 7\text{x} + 6\text{x} - 42 = 0$
$\Rightarrow \text{x}(\text{x} - 7) + 6(\text{x} - 7) = 0$
$\Rightarrow (\text{x} - 7) \ (\text{x} + 6) = 0$
$\therefore \text{x} - 7 = 0$ or $\text{x} + 6 = 0$
As such, we have:
$\Rightarrow \text{x} = 7$ or $\text{x} = - \ 6$ Ans
Q3. $8\text{x}^{2} + \text{x} = 6 - \text{x}$
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Converting into standard form:
$\Rightarrow 8\text{x}^{2} + \text{x} = 6 - \text{x}$
$\Rightarrow 8\text{x}^{2} + \text{x} - (6 - \text{x}) = 0$
$\Rightarrow 8\text{x}^{2} + \text{x} - 6 + \text{x} = 0$
$\Rightarrow 8\text{x}^{2} + 2\text{x} - 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 8\text{x}^{2} + 8\text{x} - 6\text{x} - 6 = 0$
$\Rightarrow 8\text{x}(\text{x} + 1) - 6(\text{x} + 1) = 0$
$\Rightarrow (\text{x} + 1) \ (8\text{x} - 6) = 0$
$\therefore \text{x} + 1 = 0$ or $8\text{x} - 6 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 1$ or $\text{x} = \dfrac{6}{8} = \dfrac{3}{4}$ Ans
$\Rightarrow 8\text{x}^{2} + \text{x} = 6 - \text{x}$
$\Rightarrow 8\text{x}^{2} + \text{x} - (6 - \text{x}) = 0$
$\Rightarrow 8\text{x}^{2} + \text{x} - 6 + \text{x} = 0$
$\Rightarrow 8\text{x}^{2} + 2\text{x} - 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 8\text{x}^{2} + 8\text{x} - 6\text{x} - 6 = 0$
$\Rightarrow 8\text{x}(\text{x} + 1) - 6(\text{x} + 1) = 0$
$\Rightarrow (\text{x} + 1) \ (8\text{x} - 6) = 0$
$\therefore \text{x} + 1 = 0$ or $8\text{x} - 6 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 1$ or $\text{x} = \dfrac{6}{8} = \dfrac{3}{4}$ Ans
Q4. $\dfrac{5\text{x} - 1}{3} = \dfrac{6}{\text{x}}$
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Converting into standard form:
$\Rightarrow \dfrac{5\text{x} - 1}{3} = \dfrac{6}{\text{x}}$
$\Rightarrow \text{x}(5\text{x} - 1) = 6 \times 3$
$\Rightarrow 5\text{x}^{2} - \text{x} = 18$
$\Rightarrow 5\text{x}^{2} - \text{x} - 18 = 0$
Now factorizing the left hand side:
$\Rightarrow 5\text{x}^{2} - 10\text{x} + 9\text{x} - 18 = 0$
$\Rightarrow 5\text{x}(\text{x} - 2) + 9(\text{x} - 2) = 0$
$\Rightarrow (\text{x} - 2) \ (5\text{x} + 9) = 0$
$\therefore \text{x} - 2 = 0$ or $5\text{x} + 9 = 0$
As such, we have:
$\Rightarrow \text{x} = 2$ or $\text{x} = \dfrac{9}{5}$ Ans
$\Rightarrow \dfrac{5\text{x} - 1}{3} = \dfrac{6}{\text{x}}$
$\Rightarrow \text{x}(5\text{x} - 1) = 6 \times 3$
$\Rightarrow 5\text{x}^{2} - \text{x} = 18$
$\Rightarrow 5\text{x}^{2} - \text{x} - 18 = 0$
Now factorizing the left hand side:
$\Rightarrow 5\text{x}^{2} - 10\text{x} + 9\text{x} - 18 = 0$
$\Rightarrow 5\text{x}(\text{x} - 2) + 9(\text{x} - 2) = 0$
$\Rightarrow (\text{x} - 2) \ (5\text{x} + 9) = 0$
$\therefore \text{x} - 2 = 0$ or $5\text{x} + 9 = 0$
As such, we have:
$\Rightarrow \text{x} = 2$ or $\text{x} = \dfrac{9}{5}$ Ans
Q5. $\text{x}^{2} - 64 = 0$
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Converting into standard form:
$\Rightarrow \text{x}^{2} - 64 = 0$
$\Rightarrow (\text{x} - 8) \ (\text{x} + 8) = 0$
$\therefore \text{x} - 8 = 0$ or $\text{x} + 8 = 0$
As such, we have:
$\Rightarrow \text{x} = 8$ or $\text{x} = - \ 8$ Ans
Q6. $2\text{x}^{2} - 3\text{x} + 1 = 0$
Solution 6:
Factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} - 3\text{x} + 1 = 0$
$\Rightarrow 2\text{x}^{2} - 2\text{x} - \text{x} + 1 = 0$
$\Rightarrow 2\text{x}(\text{x} - 1) - 1(\text{x} - 1) = 0$
$\Rightarrow (\text{x} - 1) \ (2\text{x} - 1) = 0$
$\therefore \text{x} - 1 = 0$ or $2\text{x} - 1 = 0$
As such, we have:
$\Rightarrow \text{x} = 1$ or $\text{x} = \dfrac{1}{2}$ Ans
$\Rightarrow \text{x}^{2} - 64 = 0$
$\Rightarrow (\text{x} - 8) \ (\text{x} + 8) = 0$
$\therefore \text{x} - 8 = 0$ or $\text{x} + 8 = 0$
As such, we have:
$\Rightarrow \text{x} = 8$ or $\text{x} = - \ 8$ Ans
Q6. $2\text{x}^{2} - 3\text{x} + 1 = 0$
Solution 6:
Factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} - 3\text{x} + 1 = 0$
$\Rightarrow 2\text{x}^{2} - 2\text{x} - \text{x} + 1 = 0$
$\Rightarrow 2\text{x}(\text{x} - 1) - 1(\text{x} - 1) = 0$
$\Rightarrow (\text{x} - 1) \ (2\text{x} - 1) = 0$
$\therefore \text{x} - 1 = 0$ or $2\text{x} - 1 = 0$
As such, we have:
$\Rightarrow \text{x} = 1$ or $\text{x} = \dfrac{1}{2}$ Ans
Q7. $\text{x} - \dfrac{10}{\text{x} - 3} = 0$
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Converting into standard form:
$\Rightarrow \text{x} - \dfrac{10}{\text{x} - 3} = 0$
$\Rightarrow \dfrac{\text{x}(\text{x} - 3) - 10}{\text{x} - 3} = 0$
$\Rightarrow \dfrac{\text{x}^{2} - 3\text{x} - 10}{\text{x} - 3} = 0$
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0(\text{x} - 3)$
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0$
$\Rightarrow \text{x}^{2} - 5\text{x} + 2\text{x} - 10 = 0$
$\Rightarrow \text{x}(\text{x} - 5) + 2(\text{x} - 5) = 0$
$\Rightarrow (\text{x} - 5) \ (\text{x} + 2) = 0$
$\therefore \text{x} - 5 = 0$ or $\text{x} + 2 = 0$
As such, we have:
$\Rightarrow \text{x} = 5$ or $\text{x} = - \ 2$ Ans
$\Rightarrow \text{x} - \dfrac{10}{\text{x} - 3} = 0$
$\Rightarrow \dfrac{\text{x}(\text{x} - 3) - 10}{\text{x} - 3} = 0$
$\Rightarrow \dfrac{\text{x}^{2} - 3\text{x} - 10}{\text{x} - 3} = 0$
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0(\text{x} - 3)$
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 3\text{x} - 10 = 0$
$\Rightarrow \text{x}^{2} - 5\text{x} + 2\text{x} - 10 = 0$
$\Rightarrow \text{x}(\text{x} - 5) + 2(\text{x} - 5) = 0$
$\Rightarrow (\text{x} - 5) \ (\text{x} + 2) = 0$
$\therefore \text{x} - 5 = 0$ or $\text{x} + 2 = 0$
As such, we have:
$\Rightarrow \text{x} = 5$ or $\text{x} = - \ 2$ Ans
Q8. $2\text{y}^{2} = 12 - 5\text{y}$
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Converting into standard form:
$\Rightarrow 2\text{y}^{2} = 12 - 5\text{y}$
$\Rightarrow 2\text{y}^{2} - (12 - 5\text{y}) = 0$
$\Rightarrow 2\text{y}^{2} - 12 + 5\text{y} = 0$
$\Rightarrow 2\text{y}^{2} + 5\text{y} - 12 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{y}^{2} + 5\text{y} - 12 = 0$
$\Rightarrow 2\text{y}^{2} + 8\text{y} - 3\text{y} - 12 = 0$
$\Rightarrow 2\text{y}(\text{y} + 4) - 3(\text{y} + 4) = 0$
$\Rightarrow (\text{y} + 4) \ (2\text{y} - 3) = 0$
$\therefore \text{y} + 4 = 0$ or $2\text{y} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 4$ or $\text{x} = \dfrac{3}{2}$ Ans
$\Rightarrow 2\text{y}^{2} = 12 - 5\text{y}$
$\Rightarrow 2\text{y}^{2} - (12 - 5\text{y}) = 0$
$\Rightarrow 2\text{y}^{2} - 12 + 5\text{y} = 0$
$\Rightarrow 2\text{y}^{2} + 5\text{y} - 12 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{y}^{2} + 5\text{y} - 12 = 0$
$\Rightarrow 2\text{y}^{2} + 8\text{y} - 3\text{y} - 12 = 0$
$\Rightarrow 2\text{y}(\text{y} + 4) - 3(\text{y} + 4) = 0$
$\Rightarrow (\text{y} + 4) \ (2\text{y} - 3) = 0$
$\therefore \text{y} + 4 = 0$ or $2\text{y} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 4$ or $\text{x} = \dfrac{3}{2}$ Ans
Q9. $\text{x}(\text{x} - 2) = 9 - 2\text{x}$
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Converting into standard form:
$\Rightarrow \text{x}(\text{x} - 2) = 9 - 2\text{x}$
$\Rightarrow \text{x}^{2} - 2\text{x} = 9 - 2\text{x}$
$\Rightarrow \text{x}^{2} - 2\text{x} - (9 - 2\text{x}) = 0$
$\Rightarrow \text{x}^{2} - 2\text{x} - 9 + 2\text{x} = 0$
$\Rightarrow \text{x}^{2} - 9 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 9 = 0$
$\Rightarrow (\text{x} - 3) \ (\text{x} + 3) = 0$
$\therefore \text{x} - 3 = 0$ or $\text{x} + 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 3$ or $\text{x} = - \ 3$ Ans
$\Rightarrow \text{x}(\text{x} - 2) = 9 - 2\text{x}$
$\Rightarrow \text{x}^{2} - 2\text{x} = 9 - 2\text{x}$
$\Rightarrow \text{x}^{2} - 2\text{x} - (9 - 2\text{x}) = 0$
$\Rightarrow \text{x}^{2} - 2\text{x} - 9 + 2\text{x} = 0$
$\Rightarrow \text{x}^{2} - 9 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 9 = 0$
$\Rightarrow (\text{x} - 3) \ (\text{x} + 3) = 0$
$\therefore \text{x} - 3 = 0$ or $\text{x} + 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 3$ or $\text{x} = - \ 3$ Ans
Q10. $9\text{x}^{2} = 25$
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Converting into standard form:
$\Rightarrow 9\text{x}^{2} = 25$
$\Rightarrow 9\text{x}^{2} - 25 = 0$
Now factorizing the left hand side:
$\Rightarrow (3\text{x})^{2} - (5)^{2} = 0$
$\Rightarrow (3\text{x} - 5)^{2} = 0$
$\Rightarrow (3\text{x} - 5) \ (3\text{x} + 5) = 0$
$\therefore 3\text{x} - 5 = 0$ or $3\text{x} + 5 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{5}{3}$ or $\text{x} = - \ \dfrac{5}{3}$ Ans
$\Rightarrow 9\text{x}^{2} = 25$
$\Rightarrow 9\text{x}^{2} - 25 = 0$
Now factorizing the left hand side:
$\Rightarrow (3\text{x})^{2} - (5)^{2} = 0$
$\Rightarrow (3\text{x} - 5)^{2} = 0$
$\Rightarrow (3\text{x} - 5) \ (3\text{x} + 5) = 0$
$\therefore 3\text{x} - 5 = 0$ or $3\text{x} + 5 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{5}{3}$ or $\text{x} = - \ \dfrac{5}{3}$ Ans
Q11. $\text{a}^{2} + \text{a} = 90$
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Converting into standard form:
$\Rightarrow \text{a}^{2} + \text{a} = 90$
$\Rightarrow \text{a}^{2} + \text{a} - 90 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{a}^{2} + 10\text{a} - 9\text{a} - 90 = 0$
$\Rightarrow \text{a}(\text{a} + 10) - 9(\text{a} + 10) = 0$
$\Rightarrow (\text{a} + 10) \ (\text{a} - 9) = 0$
$\therefore \text{a} + 10 = 0$ or $\text{a} - 9 = 0$
As such, we have:
$\Rightarrow \text{a} = - \ 10$ or $\text{a} = 9$ Ans
$\Rightarrow \text{a}^{2} + \text{a} = 90$
$\Rightarrow \text{a}^{2} + \text{a} - 90 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{a}^{2} + 10\text{a} - 9\text{a} - 90 = 0$
$\Rightarrow \text{a}(\text{a} + 10) - 9(\text{a} + 10) = 0$
$\Rightarrow (\text{a} + 10) \ (\text{a} - 9) = 0$
$\therefore \text{a} + 10 = 0$ or $\text{a} - 9 = 0$
As such, we have:
$\Rightarrow \text{a} = - \ 10$ or $\text{a} = 9$ Ans
Q12. $\dfrac{15}{\text{x}^{2}} = 1 - \dfrac{2}{\text{x}}$
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Converting into standard form:
$\Rightarrow \dfrac{15}{\text{x}^{2}} = 1 - \dfrac{2}{\text{x}}$
$\Rightarrow \dfrac{15}{\text{x}^{2}} = \dfrac{\text{x} - 2}{\text{x}}$
$\Rightarrow 15 = \text{x}^{2}(\dfrac{\text{x} - 2}{\text{x}})$
$\Rightarrow 15 = \text{x}(\text{x} - 2)$
$\Rightarrow \text{x}^{2} - 2\text{x} = 15$
$\Rightarrow \text{x}^{2} - 2\text{x} - 15 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 5\text{x} + 3\text{x} - 15 = 0$
$\Rightarrow \text{x}(\text{x} - 5) + 3(\text{x} - 5) = 0$
$\Rightarrow (\text{x} - 5) \ (\text{x} + 3) = 0$
$\therefore \text{x} - 5 = 0$ or $\text{x} + 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 5$ or $\text{x} = - \ 3$ Ans
$\Rightarrow \dfrac{15}{\text{x}^{2}} = 1 - \dfrac{2}{\text{x}}$
$\Rightarrow \dfrac{15}{\text{x}^{2}} = \dfrac{\text{x} - 2}{\text{x}}$
$\Rightarrow 15 = \text{x}^{2}(\dfrac{\text{x} - 2}{\text{x}})$
$\Rightarrow 15 = \text{x}(\text{x} - 2)$
$\Rightarrow \text{x}^{2} - 2\text{x} = 15$
$\Rightarrow \text{x}^{2} - 2\text{x} - 15 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 5\text{x} + 3\text{x} - 15 = 0$
$\Rightarrow \text{x}(\text{x} - 5) + 3(\text{x} - 5) = 0$
$\Rightarrow (\text{x} - 5) \ (\text{x} + 3) = 0$
$\therefore \text{x} - 5 = 0$ or $\text{x} + 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 5$ or $\text{x} = - \ 3$ Ans
Q13. $6\text{x}^{2} = \text{x} + 1$
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Converting into standard form:
$\Rightarrow 6\text{x}^{2} = \text{x} + 1$
$\Rightarrow 6\text{x}^{2} - (\text{x} + 1) = 0$
$\Rightarrow 6\text{x}^{2} - \text{x} - 1 = 0$
Now factorizing the left hand side:
$\Rightarrow 6\text{x}^{2} - 3\text{x} + 2\text{x} - 1 = 0$
$\Rightarrow 3\text{x}(2\text{x} - 1) + 1(2\text{x} - 1) = 0$
$\Rightarrow (2\text{x} - 1) \ (3\text{x} + 1) = 0$
$\therefore 2\text{x} - 1 = 0$ or $3\text{x} + 1 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{1}{2}$ or $\text{x} = - \ \dfrac{1}{3}$ Ans
$\Rightarrow 6\text{x}^{2} = \text{x} + 1$
$\Rightarrow 6\text{x}^{2} - (\text{x} + 1) = 0$
$\Rightarrow 6\text{x}^{2} - \text{x} - 1 = 0$
Now factorizing the left hand side:
$\Rightarrow 6\text{x}^{2} - 3\text{x} + 2\text{x} - 1 = 0$
$\Rightarrow 3\text{x}(2\text{x} - 1) + 1(2\text{x} - 1) = 0$
$\Rightarrow (2\text{x} - 1) \ (3\text{x} + 1) = 0$
$\therefore 2\text{x} - 1 = 0$ or $3\text{x} + 1 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{1}{2}$ or $\text{x} = - \ \dfrac{1}{3}$ Ans
Q14. $(2\text{a} + 1) \ (\text{a} + 3) + 3 = 0$
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Converting into standard form:
$\Rightarrow (2\text{a} + 1) \ (\text{a} + 3) + 3 = 0$
$\Rightarrow 2\text{a}^{2} + 6\text{a} + \text{a} + 3 + 3 = 0$
$\Rightarrow 2\text{a}^{2} + 7\text{a} + 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{a}^{2} + 4\text{a} + 3\text{a} + 6 = 0$
$\Rightarrow 2\text{a}(\text{a} + 2) + 3(\text{a} + 2) = 0$
$\Rightarrow (\text{a} + 2) \ (2\text{a} + 3) = 0$
$\therefore \text{a} + 2 = 0$ or $2\text{a} + 3 = 0$
As such, we have:
$\Rightarrow \text{a} = - \ 2$ or $\text{a} = - \ \dfrac{3}{2}$ Ans
$\Rightarrow (2\text{a} + 1) \ (\text{a} + 3) + 3 = 0$
$\Rightarrow 2\text{a}^{2} + 6\text{a} + \text{a} + 3 + 3 = 0$
$\Rightarrow 2\text{a}^{2} + 7\text{a} + 6 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{a}^{2} + 4\text{a} + 3\text{a} + 6 = 0$
$\Rightarrow 2\text{a}(\text{a} + 2) + 3(\text{a} + 2) = 0$
$\Rightarrow (\text{a} + 2) \ (2\text{a} + 3) = 0$
$\therefore \text{a} + 2 = 0$ or $2\text{a} + 3 = 0$
As such, we have:
$\Rightarrow \text{a} = - \ 2$ or $\text{a} = - \ \dfrac{3}{2}$ Ans
Q15. $(3\text{x} + 1) \ (2\text{x} + 3) = 3$
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Converting into standard form:
$\Rightarrow (3\text{x} + 1) \ (2\text{x} + 3) = 3$
$\Rightarrow 6\text{x}^{2} + 9\text{x} + 2\text{x} + 3 - 3 = 0$
$\Rightarrow 6\text{x}^{2} + 11\text{x} = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}(6\text{x} + 11) = 0$
$\therefore \text{x} = 0$ or $6\text{x} + 11 = 0$
As such, we have:
$\Rightarrow \text{x} = 0$ or $\text{x} = - \ \dfrac{11}{6}$ Ans
$\Rightarrow (3\text{x} + 1) \ (2\text{x} + 3) = 3$
$\Rightarrow 6\text{x}^{2} + 9\text{x} + 2\text{x} + 3 - 3 = 0$
$\Rightarrow 6\text{x}^{2} + 11\text{x} = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}(6\text{x} + 11) = 0$
$\therefore \text{x} = 0$ or $6\text{x} + 11 = 0$
As such, we have:
$\Rightarrow \text{x} = 0$ or $\text{x} = - \ \dfrac{11}{6}$ Ans
Q16. $(\text{x} - 3)^{2} = 25$
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Converting into standard form:
$\Rightarrow (\text{x} - 3)^{2} = 25$
$\Rightarrow (\text{x} - 3) \ (\text{x} - 3) = 25$
$\Rightarrow \text{x}^{2} - 3\text{x} - 3\text{x} + 9 = 25$
$\Rightarrow \text{x}^{2} - 3\text{x} - 3\text{x} + 9 - 25 = 0$
$\Rightarrow \text{x}^{2} - 6\text{x} - 16 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 8\text{x} + 2\text{x} - 16 = 0$
$\Rightarrow \text{x}(\text{x} - 8) + 2(\text{x} - 8) = 0$
$\Rightarrow (\text{x} - 8) \ (\text{x} + 2) = 0$
$\therefore \text{x} - 8 = 0$ or $\text{x} + 2 = 0$
As such, we have:
$\Rightarrow \text{x} = 8$ or $\text{x} = - 2$ Ans
$\Rightarrow (\text{x} - 3)^{2} = 25$
$\Rightarrow (\text{x} - 3) \ (\text{x} - 3) = 25$
$\Rightarrow \text{x}^{2} - 3\text{x} - 3\text{x} + 9 = 25$
$\Rightarrow \text{x}^{2} - 3\text{x} - 3\text{x} + 9 - 25 = 0$
$\Rightarrow \text{x}^{2} - 6\text{x} - 16 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 8\text{x} + 2\text{x} - 16 = 0$
$\Rightarrow \text{x}(\text{x} - 8) + 2(\text{x} - 8) = 0$
$\Rightarrow (\text{x} - 8) \ (\text{x} + 2) = 0$
$\therefore \text{x} - 8 = 0$ or $\text{x} + 2 = 0$
As such, we have:
$\Rightarrow \text{x} = 8$ or $\text{x} = - 2$ Ans
Q17. $(\text{x} - 1)^{2} + (\text{x} + 3)^{2} = 26$
Show Answer
Converting into standard form:
$\Rightarrow (\text{x} - 1)^{2} + (\text{x} + 3)^{2} = 26$
$\Rightarrow (\text{x} - 1) \ (\text{x} - 1) + (\text{x} + 3) \ (\text{x} + 3) = 26$
$\Rightarrow \text{x}^{2} - \text{x} - \text{x} + 1 + \text{x}^{2} + 3\text{x} + 3\text{x} + 9 = 26$
$\Rightarrow \text{x}^{2} - 2\text{x} + 1 + \text{x}^{2} + 6\text{x} + 9 = 26$
$\Rightarrow 2\text{x}^{2} + 4\text{x} + 10 = 26$
$\Rightarrow 2\text{x}^{2} + 4\text{x} + 10 - 26 = 0$
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 16 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} + 8\text{x} - 4\text{x} - 16 = 0$
$\Rightarrow 2\text{x}(\text{x} + 4) - 4(\text{x} + 4) = 0$
$\Rightarrow (\text{x} + 4) \ (2\text{x} - 4) = 0$
$\therefore \text{x} + 4 = 0$ or $2\text{x} - 4 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 4$ or $\text{x} = \dfrac{4}{2} = 2$ Ans
$\Rightarrow (\text{x} - 1)^{2} + (\text{x} + 3)^{2} = 26$
$\Rightarrow (\text{x} - 1) \ (\text{x} - 1) + (\text{x} + 3) \ (\text{x} + 3) = 26$
$\Rightarrow \text{x}^{2} - \text{x} - \text{x} + 1 + \text{x}^{2} + 3\text{x} + 3\text{x} + 9 = 26$
$\Rightarrow \text{x}^{2} - 2\text{x} + 1 + \text{x}^{2} + 6\text{x} + 9 = 26$
$\Rightarrow 2\text{x}^{2} + 4\text{x} + 10 = 26$
$\Rightarrow 2\text{x}^{2} + 4\text{x} + 10 - 26 = 0$
$\Rightarrow 2\text{x}^{2} + 4\text{x} - 16 = 0$
Now factorizing the left hand side:
$\Rightarrow 2\text{x}^{2} + 8\text{x} - 4\text{x} - 16 = 0$
$\Rightarrow 2\text{x}(\text{x} + 4) - 4(\text{x} + 4) = 0$
$\Rightarrow (\text{x} + 4) \ (2\text{x} - 4) = 0$
$\therefore \text{x} + 4 = 0$ or $2\text{x} - 4 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 4$ or $\text{x} = \dfrac{4}{2} = 2$ Ans
Q18. $\dfrac{\text{x}}{3} + \dfrac{9}{\text{x}} = 4$
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Converting into standard form:
$\Rightarrow \dfrac{\text{x}}{3} + \dfrac{9}{\text{x}} = 4$
$\Rightarrow \dfrac{\text{x}(\text{x}) + 3(9)}{3\text{x}} = 4$
$\Rightarrow \dfrac{\text{x}^{2} + 27}{3\text{x}} = 4$
$\Rightarrow \text{x}^{2} + 27 = 3\text{x}(4)$
$\Rightarrow \text{x}^{2} + 27 = 12\text{x}$
$\Rightarrow \text{x}^{2} - 12\text{x} + 27 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 9\text{x} - 3\text{x} + 27 = 0$
$\Rightarrow \text{x}(\text{x} - 9) - 3(\text{x} - 9) = 0$
$\Rightarrow (\text{x} - 9) \ (\text{x} - 3) = 0$
$\therefore \text{x} - 9 = 0$ or $\text{x} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 9$ or $\text{x} = 3$ Ans
$\Rightarrow \dfrac{\text{x}}{3} + \dfrac{9}{\text{x}} = 4$
$\Rightarrow \dfrac{\text{x}(\text{x}) + 3(9)}{3\text{x}} = 4$
$\Rightarrow \dfrac{\text{x}^{2} + 27}{3\text{x}} = 4$
$\Rightarrow \text{x}^{2} + 27 = 3\text{x}(4)$
$\Rightarrow \text{x}^{2} + 27 = 12\text{x}$
$\Rightarrow \text{x}^{2} - 12\text{x} + 27 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} - 9\text{x} - 3\text{x} + 27 = 0$
$\Rightarrow \text{x}(\text{x} - 9) - 3(\text{x} - 9) = 0$
$\Rightarrow (\text{x} - 9) \ (\text{x} - 3) = 0$
$\therefore \text{x} - 9 = 0$ or $\text{x} - 3 = 0$
As such, we have:
$\Rightarrow \text{x} = 9$ or $\text{x} = 3$ Ans
Q19. $\dfrac{\text{x}}{2} + 1 = \dfrac{12}{\text{x}}$
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Converting into standard form:
$\Rightarrow \dfrac{\text{x}}{2} + 1 = \dfrac{12}{\text{x}}$
$\Rightarrow \dfrac{\text{x} + 2}{2} = \dfrac{12}{\text{x}}$
$\Rightarrow \text{x}(\text{x} + 2) = 2(12)$
$\Rightarrow \text{x}^{2} + 2\text{x} = 24$
$\Rightarrow \text{x}^{2} + 2\text{x} - 24 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} + 6\text{x} - 4\text{x} - 24 = 0$
$\Rightarrow \text{x}(\text{x} + 6) - 4(\text{x} + 6) = 0$
$\Rightarrow (\text{x} + 6) \ (\text{x} - 4) = 0$
$\therefore \text{x} + 6 = 0$ or $\text{x} - 4 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 6$ or $\text{x} = 4$ Ans
$\Rightarrow \dfrac{\text{x}}{2} + 1 = \dfrac{12}{\text{x}}$
$\Rightarrow \dfrac{\text{x} + 2}{2} = \dfrac{12}{\text{x}}$
$\Rightarrow \text{x}(\text{x} + 2) = 2(12)$
$\Rightarrow \text{x}^{2} + 2\text{x} = 24$
$\Rightarrow \text{x}^{2} + 2\text{x} - 24 = 0$
Now factorizing the left hand side:
$\Rightarrow \text{x}^{2} + 6\text{x} - 4\text{x} - 24 = 0$
$\Rightarrow \text{x}(\text{x} + 6) - 4(\text{x} + 6) = 0$
$\Rightarrow (\text{x} + 6) \ (\text{x} - 4) = 0$
$\therefore \text{x} + 6 = 0$ or $\text{x} - 4 = 0$
As such, we have:
$\Rightarrow \text{x} = - \ 6$ or $\text{x} = 4$ Ans
Q20. $\dfrac{\text{x} + 8}{3\text{x} - 5} = \dfrac{\text{x} - 8}{\text{x} + 5}$
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Converting into standard form:
$\Rightarrow \dfrac{\text{x} + 8}{3\text{x} - 5} = \dfrac{\text{x} - 8}{\text{x} + 5}$
$\Rightarrow (\text{x} + 5) \ (\text{x} + 8) = (3\text{x} - 5) \ (\text{x} - 8)$
$\Rightarrow \text{x}^{2} + 8\text{x} + 5\text{x} + 40 = 3\text{x}^{2} - 24\text{x} - 5\text{x} + 40$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 = 3\text{x}^{2} - 29\text{x} + 40$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 - (3\text{x}^{2} - 29\text{x} + 40) = 0$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 - 3\text{x}^{2} + 29\text{x} - 40 = 0$
$\Rightarrow - \ 2\text{x}^{2} + 42\text{x} = 0$
Now factorizing the left hand side:
$\Rightarrow - \ 2\text{x}(\text{x} - 21) = 0$
$\therefore - \ 2\text{x} = 0$ or $\text{x} - 21 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{0}{2} = 0$ or $\text{x} = 21$ Ans
$\Rightarrow \dfrac{\text{x} + 8}{3\text{x} - 5} = \dfrac{\text{x} - 8}{\text{x} + 5}$
$\Rightarrow (\text{x} + 5) \ (\text{x} + 8) = (3\text{x} - 5) \ (\text{x} - 8)$
$\Rightarrow \text{x}^{2} + 8\text{x} + 5\text{x} + 40 = 3\text{x}^{2} - 24\text{x} - 5\text{x} + 40$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 = 3\text{x}^{2} - 29\text{x} + 40$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 - (3\text{x}^{2} - 29\text{x} + 40) = 0$
$\Rightarrow \text{x}^{2} + 13\text{x} + 40 - 3\text{x}^{2} + 29\text{x} - 40 = 0$
$\Rightarrow - \ 2\text{x}^{2} + 42\text{x} = 0$
Now factorizing the left hand side:
$\Rightarrow - \ 2\text{x}(\text{x} - 21) = 0$
$\therefore - \ 2\text{x} = 0$ or $\text{x} - 21 = 0$
As such, we have:
$\Rightarrow \text{x} = \dfrac{0}{2} = 0$ or $\text{x} = 21$ Ans