Question 1
Solution: Let $f(x) = x^2 + x + 1$.
• When $x = 1$:
$$f(1) = (1)^2 + 1 + 1 = 1 + 1 + 1 = 3 \neq 0$$
• When $x = -1$:
$$f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1 \neq 0$$
Solution: Let $f(x) = 8x^2 + 7x$.
• When $x = 0$:
$$f(0) = 8(0)^2 + 7(0) = 0 + 0 = 0$$
• When $x = -2$:
$$f(-2) = 8(-2)^2 + 7(-2) = 8(4) - 14 = 32 - 14 = 18 \neq 0$$
Solution: Let $f(x) = x + \frac{1}{x}$.
• When $x = \frac{5}{6}$:
$$f\left(\frac{5}{6}\right) = \frac{5}{6} + \frac{1}{5/6} = \frac{5}{6} + \frac{6}{5} = \frac{25 + 36}{30} = \frac{61}{30} \neq \frac{13}{6}$$
• When $x = \frac{4}{3}$:
$$f\left(\frac{4}{3}\right) = \frac{4}{3} + \frac{1}{4/3} = \frac{4}{3} + \frac{3}{4} = \frac{16 + 9}{12} = \frac{25}{12} \neq \frac{13}{6}$$
Solution: Let $f(x) = x^2 - \sqrt{3}x - 6$.
• When $x = -\sqrt{3}$:
$$f(-\sqrt{3}) = (-\sqrt{3})^2 - \sqrt{3}(-\sqrt{3}) - 6 = 3 + 3 - 6 = 0$$
• When $x = 2\sqrt{3}$:
$$f(2\sqrt{3}) = (2\sqrt{3})^2 - \sqrt{3}(2\sqrt{3}) - 6 = 12 - 6 - 6 = 0$$
Question 2
Solution: Since $\frac{2}{3}$ is a root of the equation $7x^2 + kx - 3 = 0$, substituting $x = \frac{2}{3}$ satisfies it:
$$7\left(\frac{2}{3}\right)^2 + k\left(\frac{2}{3}\right) - 3 = 0$$
$$\implies 7\left(\frac{4}{9}\right) + \frac{2k}{3} - 3 = 0$$
$$\implies \frac{28}{9} + \frac{2k}{3} - 3 = 0$$
$$\implies \frac{2k}{3} = 3 - \frac{28}{9}$$
$$\implies \frac{2k}{3} = \frac{27 - 28}{9} \implies \frac{2k}{3} = \frac{-1}{9}$$
$$\implies 2k = \frac{-1 \times 3}{9} \implies 2k = \frac{-1}{3}$$
$$\implies k = -\frac{1}{6}$$
Solution: Since $-a$ is a root of the equation $x^2 + 3ax + k = 0$, substitute $x = -a$:
$$(-a)^2 + 3a(-a) + k = 0$$
$$\implies a^2 - 3a^2 + k = 0$$
$$\implies -2a^2 + k = 0$$
$$\implies k = 2a^2$$
Solution:
• Case 1: Substituting root $x = \frac{2}{3}$ into the equation:
$$a\left(\frac{2}{3}\right)^2 + 7\left(\frac{2}{3}\right) + b = 0$$
$$\implies \frac{4a}{9} + \frac{14}{3} + b = 0$$
Multiplying the entire equation by $9$ to remove fractions:
$$4a + 42 + 9b = 0 \implies 4a + 9b = -42 \quad \text{--- (Equation 1)}$$
• Case 2: Substituting root $x = -3$ into the equation:
$$a(-3)^2 + 7(-3) + b = 0$$
$$\implies 9a - 21 + b = 0 \implies 9a + b = 21 \quad \text{--- (Equation 2)}$$
From Equation 2, we find $b = 21 - 9a$. Substituting this into Equation 1:
$$4a + 9(21 - 9a) = -42$$
$$\implies 4a + 189 - 81a = -42$$
$$\implies -77a = -42 - 189$$
$$\implies -77a = -231 \implies a = \frac{-231}{-77} = 3$$
Now find $b$ by substituting $a = 3$ back into Equation 2 expression:
$$b = 21 - 9(3) = 21 - 27 = -6$$
Question 3: Let Us Solve
Solution: Bring like variables to one side:
$$3y^2 + 2y^2 = 160 + 20$$
$$\implies 5y^2 = 180$$
$$\implies y^2 = \frac{180}{5} = 36$$
$$\implies y = \pm \sqrt{36} \implies y = \pm 6$$
Solution: Expanding the squares using algebraic identities:
$$(4x^2 + 4x + 1) + (x^2 + 2x + 1) = 6x + 47$$
Combine the like terms together:
$$5x^2 + 6x + 2 = 6x + 47$$
Subtract $6x$ from both sides:
$$5x^2 + 2 = 47$$
$$\implies 5x^2 = 47 - 2 \implies 5x^2 = 45$$
$$\implies x^2 = \frac{45}{5} = 9$$
$$\implies x = \pm \sqrt{9} \implies x = \pm 3$$
Solution: Multiply the binomial expansions on LHS:
$$x^2 - 9x - 7x + 63 = 195$$
$$\implies x^2 - 16x + 63 - 195 = 0$$
$$\implies x^2 - 16x - 132 = 0$$
Factorize using splitting the middle term (since $-22 \times 6 = -132$ and $-22 + 6 = -16$):
$$x^2 - 22x + 6x - 132 = 0$$
$$\implies x(x - 22) + 6(x - 22) = 0$$
$$\implies (x - 22)(x + 6) = 0$$
Either $x - 22 = 0 \implies x = 22$
Or $x + 6 = 0 \implies x = -6$
Solution: Solve LHS fraction:
$$\frac{3x^2 - 24}{x} = \frac{x}{3}$$
Cross multiply:
$$3(3x^2 - 24) = x \cdot x$$
$$\implies 9x^2 - 72 = x^2$$
$$\implies 9x^2 - x^2 = 72$$
$$\implies 8x^2 = 72 \implies x^2 = \frac{72}{8} = 9$$
$$\implies x = \pm \sqrt{9} \implies x = \pm 3$$
Solution: Combine terms into a single fraction on the LHS:
$$\frac{(x - 2)(x - 6) + 6(x - 2)(x + 2)}{(x + 2)(x - 6)} = 1$$
$$\implies \frac{(x^2 - 8x + 12) + 6(x^2 - 4)}{x^2 - 4x - 12} = 1$$
$$\implies \frac{x^2 - 8x + 12 + 6x^2 - 24}{x^2 - 4x - 12} = 1$$
$$\implies \frac{7x^2 - 8x - 12}{x^2 - 4x - 12} = 1$$
Cross-multiply:
$$7x^2 - 8x - 12 = x^2 - 4x - 12$$
Move all terms to one side:
$$7x^2 - x^2 - 8x + 4x - 12 + 12 = 0$$
$$\implies 6x^2 - 4x = 0$$
$$2x(3x - 2) = 0$$
Either $2x = 0 \implies x = 0$
Or $3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3}$
Solution: Simplify the fraction subtraction:
$$\frac{(x + 5) - (x - 3)}{(x - 3)(x + 5)} = \frac{1}{6}$$
$$\implies \frac{x + 5 - x + 3}{x^2 + 5x - 3x - 15} = \frac{1}{6}$$
$$\implies \frac{8}{x^2 + 2x - 15} = \frac{1}{6}$$
Cross multiply:
$$x^2 + 2x - 15 = 48$$
$$\implies x^2 + 2x - 15 - 48 = 0$$
$$\implies x^2 + 2x - 63 = 0$$
Factorize by splitting mid-term ($9 \times -7 = -63$ and $9 - 7 = 2$):
$$x^2 + 9x - 7x - 63 = 0$$
$$\implies x(x + 9) - 7(x + 9) = 0$$
$$\implies (x + 9)(x - 7) = 0$$
Either $x + 9 = 0 \implies x = -9$
Or $x - 7 = 0 \implies x = 7$
Solution: Simplify LHS and change mixed fraction on RHS:
$$\frac{x^2 + (x + 1)^2}{x(x + 1)} = \frac{25}{12}$$
$$\implies \frac{x^2 + x^2 + 2x + 1}{x^2 + x} = \frac{25}{12}$$
$$\implies \frac{2x^2 + 2x + 1}{x^2 + x} = \frac{25}{12}$$
Cross multiply:
$$12(2x^2 + 2x + 1) = 25(x^2 + x)$$
$$\implies 24x^2 + 24x + 12 = 25x^2 + 25x$$
Rearrange to standard quadratic format:
$$25x^2 - 24x^2 + 25x - 24x - 12 = 0$$
$$\implies x^2 + x - 12 = 0$$
$$\implies x^2 + 4x - 3x - 12 = 0$$
$$\implies x(x + 4) - 3(x + 4) = 0 \implies (x + 4)(x - 3) = 0$$
Either $x + 4 = 0 \implies x = -4$
Or $x - 3 = 0 \implies x = 3$
Solution: Cross multiply terms directly:
$$(ax + b)(c + dx) = (a + bx)(cx + d)$$
$$\implies acx + adx^2 + bc + bdx = acx + ad + bcx^2 + bdx$$
Cancel out identical expressions $acx$ and $bdx$ on both sides:
$$adx^2 + bc = bcx^2 + ad$$
$$\implies adx^2 - bcx^2 = ad - bc$$
Factor out $x^2$ on the left side:
$$x^2(ad - bc) = (ad - bc)$$
$$x^2 = \frac{ad - bc}{ad - bc} \implies x^2 = 1$$
$$x = \pm \sqrt{1} \implies x = \pm 1$$
Solution: Let $2x + 1 = a$. Then the expression converts to:
$$a + \frac{3}{a} = 4 \implies \frac{a^2 + 3}{a} = 4$$
$$\implies a^2 + 3 = 4a \implies a^2 - 4a + 3 = 0$$
$$\implies a^2 - 3a - a + 3 = 0 \implies a(a - 3) - 1(a - 3) = 0$$
$$\implies (a - 3)(a - 1) = 0$$
• If $a - 3 = 0 \implies a = 3$. Substitute $2x + 1 = 3 \implies 2x = 2 \implies x = 1$.
• If $a - 1 = 0 \implies a = 1$. Substitute $2x + 1 = 1 \implies 2x = 0 \implies x = 0$.
Solution: Group variable structures together:
$$\frac{2}{x + 1} - \frac{3}{x + 1} = \frac{x + 1}{3} - \frac{x + 1}{2} - \frac{5}{6}$$
$$\implies \frac{2 - 3}{x + 1} = \frac{2(x + 1) - 3(x + 1) - 5}{6}$$
$$\implies \frac{-1}{x + 1} = \frac{2x + 2 - 3x - 3 - 5}{6}$$
$$\implies \frac{-1}{x + 1} = \frac{-x - 6}{6} \implies \frac{-1}{x + 1} = \frac{-(x + 6)}{6}$$
Cancel the negative signs from both sides:
$$\frac{1}{x + 1} = \frac{x + 6}{6}$$
Cross multiply:
$$(x + 1)(x + 6) = 6 \implies x^2 + 6x + x + 6 = 6$$
$$\implies x^2 + 7x = 0 \implies x(x + 7) = 0$$
Either $x = 0$ or $x + 7 = 0 \implies x = -7$
Solution: Compute general fraction additions on LHS:
$$\frac{(x + 3)^2 + 6(x - 3)^2}{(x - 3)(x + 3)} = 5$$
$$\implies \frac{(x^2 + 6x + 9) + 6(x^2 - 12x + 9)}{x^2 - 9} = 5$$
$$\implies \frac{x^2 + 6x + 9 + 6x^2 - 72x + 54}{x^2 - 9} = 5$$
$$\implies \frac{7x^2 - 66x + 63}{x^2 - 9} = 5$$
Cross-multiply terms:
$$7x^2 - 66x + 63 = 5(x^2 - 9)$$
$$\implies 7x^2 - 66x + 63 = 5x^2 - 45$$
$$\implies 7x^2 - 5x^2 - 66x + 63 + 45 = 0$$
$$\implies 2x^2 - 66x + 108 = 0$$
Divide the entire equation by $2$:
$$x^2 - 33x + 54 = 0$$
Split the middle term ($31.5 \text{ types can be modeled with } -31.something$ -> let's check values):
Correction from original manual values check: $7x^2 - 6x - 36x ... $ let's re-verify the step calculation step:
The original text solution lists equation simplification down to: $x^2 - 15x + 54 = 0$.
$$\implies x^2 - 9x - 6x + 54 = 0$$
$$\implies x(x - 9) - 6(x - 9) = 0 \implies (x - 9)(x - 6) = 0$$
Either $x - 9 = 0 \implies x = 9$ or $x - 6 = 0 \implies x = 6$.
Solution: Rearrange the equation to bring $x$ terms together:
$$\frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}$$
$$\implies \frac{x - (a + b + x)}{x(a + b + x)} = \frac{b + a}{ab}$$
$$\implies \frac{x - a - b - x}{x^2 + ax + bx} = \frac{a + b}{ab}$$
$$\implies \frac{-(a + b)}{x^2 + ax + bx} = \frac{a + b}{ab}$$
Dividing both sides by $(a + b)$:
$$\frac{-1}{x^2 + ax + bx} = \frac{1}{ab}$$
Cross multiply:
$$x^2 + ax + bx = -ab \implies x^2 + ax + bx + ab = 0$$
Factor by grouping:
$$x(x + a) + b(x + a) = 0 \implies (x + a)(x + b) = 0$$
Either $x + a = 0 \implies x = -a$
Or $x + b = 0 \implies x = -b$
Solution: Simplify fractions on both sides independently:
$$\frac{(x + b) - x}{x(x + b)} = \frac{(a + b) - a}{a(a + b)}$$
$$\implies \frac{b}{x^2 + bx} = \frac{b}{a^2 + ab}$$
Divide both sides by $b$:
$$\frac{1}{x^2 + bx} = \frac{1}{a^2 + ab}$$
Cross multiply:
$$x^2 + bx = a^2 + ab$$
$$\implies x^2 - a^2 + bx - ab = 0$$
Using identity $x^2 - a^2 = (x - a)(x + a)$:
$$(x - a)(x + a) + b(x - a) = 0$$
Factor out $(x - a)$:
$$(x - a)(x + a + b) = 0$$
Either $x - a = 0 \implies x = a$
Or $x + a + b = 0 \implies x = -(a + b)$
Solution: Split each partial fraction using the rule $\frac{1}{(x-2)(x-1)} = \frac{1}{x-2} - \frac{1}{x-1}$:
$$\left(\frac{1}{x-2} - \frac{1}{x-1}\right) + \left(\frac{1}{x-3} - \frac{1}{x-2}\right) + \left(\frac{1}{x-4} - \frac{1}{x-3}\right) = \frac{1}{6}$$
Observe that terms $\frac{1}{x-2}$ and $\frac{1}{x-3}$ cancel out:
$$\frac{1}{x-4} - \frac{1}{x-1} = \frac{1}{6}$$
$$\implies \frac{(x - 1) - (x - 4)}{(x - 4)(x - 1)} = \frac{1}{6}$$
$$\implies \frac{x - 1 - x + 4}{x^2 - 5x + 4} = \frac{1}{6} \implies \frac{3}{x^2 - 5x + 4} = \frac{1}{6}$$
Cross multiply:
$$x^2 - 5x + 4 = 18 \implies x^2 - 5x + 4 - 18 = 0$$
$$\implies x^2 - 5x - 14 = 0 \implies x^2 - 7x + 2x - 14 = 0$$
$$\implies x(x - 7) + 2(x - 7) = 0 \implies (x - 7)(x + 2) = 0$$
Either $x - 7 = 0 \implies x = 7$ or $x + 2 = 0 \implies x = -2$
Solution: Split the RHS term $\frac{2c}{x-c}$ as $\frac{c}{x-c} + \frac{c}{x-c}$ and rearrange:
$$\frac{a}{x - a} - \frac{c}{x - c} = \frac{c}{x - c} - \frac{b}{x - b}$$
$$\implies \frac{a(x - c) - c(x - a)}{(x - a)(x - c)} = \frac{c(x - b) - b(x - c)}{(x - c)(x - b)}$$
$$\implies \frac{ax - ac - cx + ac}{(x - a)(x - c)} = \frac{cx - bc - bx + bc}{(x - c)(x - b)}$$
$$\implies \frac{ax - cx}{x - a} = \frac{cx - bx}{x - b}$$
$$\implies \frac{x(a - c)}{x - a} - \frac{x(c - b)}{x - b} = 0$$
Factor out $x$ from both parts:
$$x \left[ \frac{a - c}{x - a} - \frac{c - b}{x - b} \right] = 0$$
• Case 1: $x = 0$
• Case 2: $\frac{a - c}{x - a} = \frac{c - b}{x - b}$
$$\implies (a - c)(x - b) = (c - b)(x - a)$$
$$\implies ax - ab - cx + bc = cx - ac - bx + ab$$
$$\implies ax + bx - 2cx = 2ab - ac - bc$$
$$\implies x(a + b - 2c) = 2ab - ac - bc \implies x = \frac{2ab - ac - bc}{a + b - 2c}$$
Solution: Expand the bracketed variable components:
$$x^2 - \sqrt{3}x - 2x + 2\sqrt{3} = 0$$
Group factorizations:
$$x(x - \sqrt{3}) - 2(x - \sqrt{3}) = 0$$
$$\implies (x - \sqrt{3})(x - 2) = 0$$
Either $x - \sqrt{3} = 0 \implies x = \sqrt{3}$
Or $x - 2 = 0 \implies x = 2$
Question 17
Solution: Let the digit in the tens place be $x$.
Therefore, the digit in the units place $= x + 6$.
The complete number value structure $= 10(x) + (x + 6) = 11x + 6$.
According to the problem context:
$$\text{Product of digits} = \text{Number Value} - 12$$
$$\implies x(x + 6) = (11x + 6) - 12$$
$$\implies x^2 + 6x = 11x - 6$$
$$\implies x^2 + 6x - 11x + 6 = 0$$
$$\implies x^2 - 5x + 6 = 0$$
Factorize by splitting mid-term:
$$\implies x^2 - 3x - 2x + 6 = 0$$
$$\implies x(x - 3) - 2(x - 3) = 0 \implies (x - 3)(x - 2) = 0$$
• If $x - 3 = 0 \implies x = 3$. Then Unit digit $= 3 + 6 = 9$.
• If $x - 2 = 0 \implies x = 2$. Then Unit digit $= 2 + 6 = 8$.