Mix Examples - Quadratic Equations Questions in English

(A) To verify if $x = -2\sqrt{2}$ is a solution,substitute the value of $x$ into the quadratic equation $x^{2} + \sqrt{2}x - 4 = 0$.
$LHS$ $= (-2\sqrt{2})^{2} + \sqrt{2}(-2\sqrt{2}) - 4$
$= (4 \times 2) + (-2 \times 2) - 4$
$= 8 - 4 - 4$
$= 0$
Since $LHS$ $=$ $RHS$,the given value $x = -2\sqrt{2}$ is a solution of the quadratic equation.

(B) To verify if $x = \frac{1}{2}$ is a solution,substitute $x = \frac{1}{2}$ into the quadratic equation $2x^2 - 5x + 3 = 0$.
$LHS$ $= 2(\frac{1}{2})^2 - 5(\frac{1}{2}) + 3$
$= 2(\frac{1}{4}) - \frac{5}{2} + 3$
$= \frac{1}{2} - \frac{5}{2} + 3$
$= -\frac{4}{2} + 3$
$= -2 + 3 = 1$
Since $LHS$ $\neq$ $RHS$ (where $RHS$ $= 0$),$x = \frac{1}{2}$ is not a solution of the given quadratic equation.

To verify whether $x = \frac{-2}{m+n}$ is a solution, we substitute it into the quadratic equation.
Given equation:
$(m+n)^2 x^2 + (m+n)x - 2 = 0$
Substitute $x = \frac{-2}{m+n}$:
$(m+n)^2 \left(\frac{-2}{m+n}\right)^2 + (m+n)\left(\frac{-2}{m+n}\right) - 2 = 0$
Simplifying:
$(m+n)^2 \cdot \frac{4}{(m+n)^2} + (m+n)\cdot \frac{-2}{m+n} - 2 = 0$
$4 - 2 - 2 = 0$
$0 = 0$
Since both sides are equal, $x = \frac{-2}{m+n}$ is a solution of the given quadratic equation.

(B) To verify if $x=2$ is a solution,substitute $x=2$ into the given quadratic equation: $(x-2)(x+3)+1=0$.
Substituting $x=2$:
$(2-2)(2+3)+1 = (0)(5)+1 = 0+1 = 1$.
Since the result is $1$ and not $0$,the left-hand side is not equal to the right-hand side.
Therefore,$x=2$ is not a solution of the given quadratic equation.

(A) To verify if $x = -\frac{1}{2}$ is a solution,substitute the value into the left-hand side $(LHS)$ of the equation.
$LHS$ = $\frac{1}{3 - 2(-\frac{1}{2})} + \frac{1}{5 + 2(-\frac{1}{2})}$
$LHS$ = $\frac{1}{3 - (-1)} + \frac{1}{5 + (-1)}$
$LHS$ = $\frac{1}{3 + 1} + \frac{1}{5 - 1}$
$LHS$ = $\frac{1}{4} + \frac{1}{4}$
$LHS$ = $\frac{2}{4} = \frac{1}{2}$
Since $LHS$ = $RHS$ (which is $\frac{1}{2}$),the given value $x = -\frac{1}{2}$ is a solution of the quadratic equation.

(A) To verify if $x = \frac{\sqrt{3}}{4}$ is a solution,substitute the value of $x$ into the quadratic equation $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$.
$LHS$ = $4 \sqrt{3} \left( \frac{\sqrt{3}}{4} \right)^{2} + 5 \left( \frac{\sqrt{3}}{4} \right) - 2 \sqrt{3}$
$LHS$ = $4 \sqrt{3} \left( \frac{3}{16} \right) + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $\frac{3 \sqrt{3}}{4} + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $\frac{8 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $2 \sqrt{3} - 2 \sqrt{3} = 0$
Since $LHS$ = $RHS$,the given value $x = \frac{\sqrt{3}}{4}$ is a solution of the quadratic equation.

(A) To verify if $x = 2$ is a solution,substitute $x = 2$ into the left-hand side $(LHS)$ of the equation:
$LHS$ = $\frac{1}{2+4} - \frac{1}{2-7}$
$LHS$ = $\frac{1}{6} - \frac{1}{-5}$
$LHS$ = $\frac{1}{6} + \frac{1}{5}$
$LHS$ = $\frac{5+6}{30} = \frac{11}{30}$
Since $LHS$ = $RHS$ $(\frac{11}{30})$,the value $x = 2$ is a solution of the given quadratic equation.

(A) Given equation: $x + \frac{1}{x} = 3 \frac{1}{3}$.
Convert the mixed fraction to an improper fraction: $3 \frac{1}{3} = \frac{3 \times 3 + 1}{3} = \frac{10}{3}$.
So,the equation is $x + \frac{1}{x} = \frac{10}{3}$.
Substitute $x = \frac{1}{3}$ into the left-hand side $(LHS)$ of the equation:
$LHS$ $= \frac{1}{3} + \frac{1}{1/3} = \frac{1}{3} + 3$.
$LHS$ $= \frac{1 + 9}{3} = \frac{10}{3}$.
Since $LHS$ $= \frac{10}{3}$ and $RHS$ $= \frac{10}{3}$,$LHS$ $=$ $RHS$.
Therefore,$x = \frac{1}{3}$ is a solution of the given quadratic equation.

(A) Given the quadratic equation is $kx^{2} + 2x - 3 = 0$.
Since $x = 2$ is a root of the equation,it must satisfy the equation.
Substitute $x = 2$ into the equation:
$k(2)^{2} + 2(2) - 3 = 0$
$k(4) + 4 - 3 = 0$
$4k + 1 = 0$
$4k = -1$
$k = -\frac{1}{4}$

(B) Given the quadratic equation $3x^{2} + 2kx - 3 = 0$.
Since $x = -\frac{1}{2}$ is a root of the equation,it must satisfy the equation.
Substitute $x = -\frac{1}{2}$ into the equation:
$3(-\frac{1}{2})^{2} + 2k(-\frac{1}{2}) - 3 = 0$
$3(\frac{1}{4}) - k - 3 = 0$
$\frac{3}{4} - k - 3 = 0$
$-k = 3 - \frac{3}{4}$
$-k = \frac{12 - 3}{4}$
$-k = \frac{9}{4}$
$k = -\frac{9}{4}$

(N/A) Given the quadratic equation is $x^{2} - \sqrt{p}x + q = 0$.
Since $x = -\sqrt{p}$ is a root of the equation,it must satisfy the equation.
Substitute $x = -\sqrt{p}$ into the equation:
$(-\sqrt{p})^{2} - \sqrt{p}(-\sqrt{p}) + q = 0$
$p - (-\sqrt{p} \cdot \sqrt{p}) + q = 0$
$p - (-p) + q = 0$
$p + p + q = 0$
$2p + q = 0$
Hence,it is proved.

(N/A) Given the quadratic equation is $px^{2} + qx + r = 0$.
Since $x = -2$ is a root of this equation,it must satisfy the equation.
Substituting $x = -2$ into the equation,we get:
$p(-2)^{2} + q(-2) + r = 0$
$p(4) - 2q + r = 0$
$4p - 2q + r = 0$
Hence,it is proved that $4p - 2q + r = 0$.

(A) Let $(x+2) = y$. Substituting this into the equation,we get:
$y^{2} - 14y + 45 = 0$
To factorize,we look for two numbers whose product is $45$ and sum is $-14$. These numbers are $-9$ and $-5$.
$y^{2} - 9y - 5y + 45 = 0$
$y(y - 9) - 5(y - 9) = 0$
$(y - 9)(y - 5) = 0$
So,$y = 9$ or $y = 5$.
Now,substitute $y = x + 2$ back:
Case $1$: $x + 2 = 9 \implies x = 7$
Case $2$: $x + 2 = 5 \implies x = 3$
Thus,the solutions are $x = 3, 7$.

(B) Let $y = 2x + 1$. The equation becomes $6y^2 = y + 5$.
Rearranging the terms,we get $6y^2 - y - 5 = 0$.
To factorize,we look for two numbers that multiply to $(6 \times -5) = -30$ and add to $-1$.
These numbers are $-6$ and $5$.
So,$6y^2 - 6y + 5y - 5 = 0$.
$6y(y - 1) + 5(y - 1) = 0$.
$(6y + 5)(y - 1) = 0$.
This gives $y = 1$ or $y = -\frac{5}{6}$.
Substituting $y = 2x + 1$ back:
Case $1$: $2x + 1 = 1 \implies 2x = 0 \implies x = 0$.
Case $2$: $2x + 1 = -\frac{5}{6} \implies 2x = -\frac{5}{6} - 1 = -\frac{11}{6} \implies x = -\frac{11}{12}$.
Thus,the solutions are $x = 0$ and $x = -\frac{11}{12}$.

(C) Given equation: $x + 3 + \frac{2}{x} = 0$.
Multiply the entire equation by $x$ (where $x \neq 0$):
$x^2 + 3x + 2 = 0$.
Now,factorize the quadratic equation by splitting the middle term:
$x^2 + 2x + x + 2 = 0$.
$x(x + 2) + 1(x + 2) = 0$.
$(x + 1)(x + 2) = 0$.
Setting each factor to zero:
$x + 1 = 0 \implies x = -1$.
$x + 2 = 0 \implies x = -2$.
Thus,the solutions are $x = -1, -2$.

(A) Given equation: $(x+1)^{2} + x^{2} = 221$
Expand the square: $(x^{2} + 2x + 1) + x^{2} = 221$
Combine like terms: $2x^{2} + 2x + 1 = 221$
Subtract $221$ from both sides: $2x^{2} + 2x - 220 = 0$
Divide the entire equation by $2$: $x^{2} + x - 110 = 0$
Factorize the quadratic equation: $x^{2} + 11x - 10x - 110 = 0$
Group the terms: $x(x + 11) - 10(x + 11) = 0$
$(x - 10)(x + 11) = 0$
Therefore,$x - 10 = 0$ or $x + 11 = 0$
$x = 10$ or $x = -11$
Thus,the solutions are $10, -11$.

(A) Given the equation: $x - \frac{20}{x} = 1$
Multiply the entire equation by $x$ to eliminate the fraction: $x^2 - 20 = x$
Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 - x - 20 = 0$
To factorize,find two numbers that multiply to $-20$ and add to $-1$: These numbers are $-5$ and $4$.
Rewrite the middle term: $x^2 - 5x + 4x - 20 = 0$
Group the terms: $x(x - 5) + 4(x - 5) = 0$
Factor out the common binomial: $(x - 5)(x + 4) = 0$
Set each factor to zero: $x - 5 = 0$ or $x + 4 = 0$
Thus,the solutions are $x = 5$ or $x = -4$.

(B) Given equation: $x^{2} + (x + 5)^{2} = 125$
Expand the term $(x + 5)^{2}$ using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$:
$x^{2} + (x^{2} + 10x + 25) = 125$
Combine like terms:
$2x^{2} + 10x + 25 = 125$
Subtract $125$ from both sides to set the equation to zero:
$2x^{2} + 10x - 100 = 0$
Divide the entire equation by $2$ to simplify:
$x^{2} + 5x - 50 = 0$
Factorize the quadratic equation by finding two numbers that multiply to $-50$ and add to $5$:
These numbers are $10$ and $-5$:
$x^{2} + 10x - 5x - 50 = 0$
$x(x + 10) - 5(x + 10) = 0$
$(x - 5)(x + 10) = 0$
Setting each factor to zero:
$x - 5 = 0 \implies x = 5$
$x + 10 = 0 \implies x = -10$
Thus,the solutions are $5, -10$.

(D) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}$.
Let $y = \frac{x}{x+1}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.
Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies to $2y^2 - 5y + 2 = 0$.
Factoring the quadratic: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.
So,$(2y - 1)(y - 2) = 0$,giving $y = 1/2$ or $y = 2$.
Case $1$: $\frac{x}{x+1} = \frac{1}{2} \implies 2x = x + 1 \implies x = 1$.
Case $2$: $\frac{x}{x+1} = 2 \implies x = 2x + 2 \implies x = -2$.
The solution set is $\{1, -2\}$.

(D) Given equation: $\frac{x+2}{x} + \frac{x}{x+2} = \frac{10}{3}$
Taking $LCM$: $\frac{(x+2)^2 + x^2}{x(x+2)} = \frac{10}{3}$
Expand the numerator: $\frac{x^2 + 4x + 4 + x^2}{x^2 + 2x} = \frac{10}{3}$
$\frac{2x^2 + 4x + 4}{x^2 + 2x} = \frac{10}{3}$
Cross-multiply: $3(2x^2 + 4x + 4) = 10(x^2 + 2x)$
$6x^2 + 12x + 12 = 10x^2 + 20x$
Rearrange terms: $4x^2 + 8x - 12 = 0$
Divide by $4$: $x^2 + 2x - 3 = 0$
Factorize: $x^2 + 3x - x - 3 = 0$
$x(x+3) - 1(x+3) = 0$
$(x-1)(x+3) = 0$
Thus,$x = 1$ or $x = -3$.
The solution set is $\{1, -3\}$.

(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$
Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$
$\frac{x^2 + x + x + 2}{x^2 + 2x} = \frac{5}{4}$
$\frac{x^2 + 2x + 2}{x^2 + 2x} = \frac{5}{4}$
Cross-multiplying: $4(x^2 + 2x + 2) = 5(x^2 + 2x)$
$4x^2 + 8x + 8 = 5x^2 + 10x$
Rearranging terms to form a quadratic equation: $x^2 + 2x - 8 = 0$
Factorizing the quadratic equation: $x^2 + 4x - 2x - 8 = 0$
$x(x+4) - 2(x+4) = 0$
$(x-2)(x+4) = 0$
Thus,$x = 2$ or $x = -4$.
The solution set is $\{2, -4\}$.

(B) Given equation: $x + 2 - \frac{6}{x + 2} = 1$
Let $y = x + 2$. Then the equation becomes: $y - \frac{6}{y} = 1$
Multiply by $y$: $y^2 - 6 = y$
Rearrange to standard form: $y^2 - y - 6 = 0$
Factorize: $y^2 - 3y + 2y - 6 = 0$
$y(y - 3) + 2(y - 3) = 0$
$(y - 3)(y + 2) = 0$
So,$y = 3$ or $y = -2$.
Substitute $y = x + 2$ back:
Case $1$: $x + 2 = 3 \implies x = 1$
Case $2$: $x + 2 = -2 \implies x = -4$
Thus,the solution set is $\{1, -4\}$.

(C) Given equation: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$
Let $y = \frac{x}{x-1}$,then the equation becomes $y + \frac{1}{y} = \frac{25}{12}$.
Multiplying by $12y$,we get $12y^2 + 12 = 25y$,which rearranges to $12y^2 - 25y + 12 = 0$.
Factoring the quadratic: $12y^2 - 16y - 9y + 12 = 0 \implies 4y(3y - 4) - 3(3y - 4) = 0$.
So,$(4y - 3)(3y - 4) = 0$,giving $y = \frac{3}{4}$ or $y = \frac{4}{3}$.
Case $1$: $\frac{x}{x-1} = \frac{3}{4} \implies 4x = 3x - 3 \implies x = -3$.
Case $2$: $\frac{x}{x-1} = \frac{4}{3} \implies 3x = 4x - 4 \implies x = 4$.
Thus,the solution set is $\{4, -3\}$.

(D) Given equation: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$
Taking $LCM$ on the left side: $\frac{9(x-3) - 2(x-1)}{(x-1)(x-3)} = \frac{5}{x+1}$
$\frac{9x - 27 - 2x + 2}{x^2 - 4x + 3} = \frac{5}{x+1}$
$\frac{7x - 25}{x^2 - 4x + 3} = \frac{5}{x+1}$
Cross-multiplying: $(7x - 25)(x + 1) = 5(x^2 - 4x + 3)$
$7x^2 + 7x - 25x - 25 = 5x^2 - 20x + 15$
$7x^2 - 18x - 25 = 5x^2 - 20x + 15$
$2x^2 + 2x - 40 = 0$
Dividing by $2$: $x^2 + x - 20 = 0$
Factorizing: $x^2 + 5x - 4x - 20 = 0$
$x(x + 5) - 4(x + 5) = 0$
$(x - 4)(x + 5) = 0$
Thus,$x = 4$ or $x = -5$.
The solution set is $\{-5, 4\}$.

(A) Let $y = \frac{x+1}{x-3}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.
Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies to $2y^2 - 5y + 2 = 0$.
Factoring the quadratic equation: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.
This gives $(2y - 1)(y - 2) = 0$,so $y = \frac{1}{2}$ or $y = 2$.
Case $1$: $\frac{x+1}{x-3} = \frac{1}{2} \implies 2x + 2 = x - 3 \implies x = -5$.
Case $2$: $\frac{x+1}{x-3} = 2 \implies x + 1 = 2x - 6 \implies x = 7$.
Thus,the solution set is $\{-5, 7\}$.

(B) Given equation: $\frac{8}{x+5} = \frac{3}{x-1} + \frac{1}{x-5}$
First,simplify the right side: $\frac{8}{x+5} = \frac{3(x-5) + 1(x-1)}{(x-1)(x-5)}$
$\frac{8}{x+5} = \frac{3x - 15 + x - 1}{x^2 - 6x + 5}$
$\frac{8}{x+5} = \frac{4x - 16}{x^2 - 6x + 5}$
Cross-multiply: $8(x^2 - 6x + 5) = (x+5)(4x - 16)$
$8x^2 - 48x + 40 = 4x^2 - 16x + 20x - 80$
$8x^2 - 48x + 40 = 4x^2 + 4x - 80$
$4x^2 - 52x + 120 = 0$
Divide by $4$: $x^2 - 13x + 30 = 0$
Factorize: $x^2 - 10x - 3x + 30 = 0$
$x(x - 10) - 3(x - 10) = 0$
$(x - 3)(x - 10) = 0$
Thus,$x = 3$ or $x = 10$.
The solution set is $\{3, 10\}$.

(C) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{41}{20}$.
Let $y = \frac{x}{x+1}$,then the equation becomes $y + \frac{1}{y} = \frac{41}{20}$.
Multiplying by $20y$,we get $20y^2 + 20 = 41y$,which rearranges to $20y^2 - 41y + 20 = 0$.
Factorizing the quadratic: $20y^2 - 16y - 25y + 20 = 0 \Rightarrow 4y(5y - 4) - 5(5y - 4) = 0$.
So,$(4y - 5)(5y - 4) = 0$,giving $y = 5/4$ or $y = 4/5$.
Case $1$: $\frac{x}{x+1} = \frac{5}{4} \Rightarrow 4x = 5x + 5 \Rightarrow x = -5$.
Case $2$: $\frac{x}{x+1} = \frac{4}{5} \Rightarrow 5x = 4x + 4 \Rightarrow x = 4$.
Thus,the solution set is $\{4, -5\}$.

(D) Let $y = \frac{x}{1-x}$. Then the equation becomes $y + \frac{1}{y} = \frac{13}{6}$.
Multiplying by $6y$,we get $6y^2 + 6 = 13y$,which simplifies to $6y^2 - 13y + 6 = 0$.
Factoring the quadratic equation: $6y^2 - 9y - 4y + 6 = 0 \implies 3y(2y - 3) - 2(2y - 3) = 0$.
This gives $(3y - 2)(2y - 3) = 0$,so $y = \frac{2}{3}$ or $y = \frac{3}{2}$.
Case $1$: $\frac{x}{1-x} = \frac{2}{3} \implies 3x = 2 - 2x \implies 5x = 2 \implies x = \frac{2}{5}$.
Case $2$: $\frac{x}{1-x} = \frac{3}{2} \implies 2x = 3 - 3x \implies 5x = 3 \implies x = \frac{3}{5}$.
Thus,the solution set is $\{\frac{2}{5}, \frac{3}{5}\}$.

(A) Given equation: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$
Taking the common denominator: $\frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3$
Expanding the numerators and denominators: $\frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3$
Simplifying the numerator: $\frac{2x^2 + 4}{x^2 + x - 2} = 3$
Cross-multiplying: $2x^2 + 4 = 3(x^2 + x - 2)$
$2x^2 + 4 = 3x^2 + 3x - 6$
Rearranging terms to form a quadratic equation: $x^2 + 3x - 10 = 0$
Factoring the quadratic equation: $x^2 + 5x - 2x - 10 = 0$
$x(x + 5) - 2(x + 5) = 0$
$(x - 2)(x + 5) = 0$
Thus,$x = 2$ or $x = -5$.
The solution set is $\{-5, 2\}$.

(B) Given equation: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{x}$
Taking $LCM$ on the left side: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{x}$
Simplify the numerator: $(x^2 + 2x + 1) - (x^2 - 2x + 1) = 4x$
Simplify the denominator: $(x-1)(x+1) = x^2 - 1$
So,$\frac{4x}{x^2 - 1} = \frac{5}{x}$
Cross-multiply: $4x^2 = 5(x^2 - 1)$
$4x^2 = 5x^2 - 5$
$x^2 = 5$
$x = \pm\sqrt{5}$
Thus,the solution set is $\{-\sqrt{5}, \sqrt{5}\}$.

(C) Given equation: $x^{2}+3x-5=0$.
To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{3}{2})^{2} = \frac{9}{4}$.
$x^{2}+3x+\frac{9}{4}-5-\frac{9}{4}=0$
$(x+\frac{3}{2})^{2} - \frac{20+9}{4} = 0$
$(x+\frac{3}{2})^{2} - \frac{29}{4} = 0$
$(x+\frac{3}{2})^{2} = (\frac{\sqrt{29}}{2})^{2}$
Taking the square root on both sides:
$x+\frac{3}{2} = \pm \frac{\sqrt{29}}{2}$
$x = -\frac{3}{2} \pm \frac{\sqrt{29}}{2}$
$x = \frac{-3 \pm \sqrt{29}}{2}$
Thus,the solutions are $\frac{-3-\sqrt{29}}{2}$ and $\frac{-3+\sqrt{29}}{2}$.

(D) Given equation: $2x^2 - 7x + 3 = 0$
Divide the entire equation by $2$ to make the coefficient of $x^2$ equal to $1$:
$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{7}{2})^2 = (\frac{7}{4})^2 = \frac{49}{16}$:
$x^2 - \frac{7}{2}x + \frac{49}{16} - \frac{49}{16} + \frac{3}{2} = 0$
$(x - \frac{7}{4})^2 - \frac{49}{16} + \frac{24}{16} = 0$
$(x - \frac{7}{4})^2 - \frac{25}{16} = 0$
$(x - \frac{7}{4})^2 = (\frac{5}{4})^2$
Taking the square root on both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$
Thus,the solutions are $x = 3$ or $x = \frac{1}{2}$.

(D) Given equation: $4 x^{2}+3 x+5=0$
Divide the entire equation by $4$ to make the coefficient of $x^2$ equal to $1$:
$x^{2} + \frac{3}{4}x + \frac{5}{4} = 0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{3}{4})^2 = (\frac{3}{8})^2 = \frac{9}{64}$:
$x^{2} + \frac{3}{4}x + \frac{9}{64} - \frac{9}{64} + \frac{5}{4} = 0$
$(x + \frac{3}{8})^2 + \frac{-9 + 80}{64} = 0$
$(x + \frac{3}{8})^2 + \frac{71}{64} = 0$
$(x + \frac{3}{8})^2 = -\frac{71}{64}$
Since the square of any real number cannot be negative,there is no real value of $x$ that satisfies this equation. Thus,the equation has no real roots.

(B) Given quadratic equation is $x^{2}+10x+7=0$.
To complete the square,we add and subtract $(\frac{10}{2})^{2} = 25$:
$x^{2}+10x+25-25+7=0$
$(x+5)^{2}-18=0$
$(x+5)^{2} = 18$
Taking the square root on both sides:
$x+5 = \pm\sqrt{18}$
$x+5 = \pm 3\sqrt{2}$
$x = -5 \pm 3\sqrt{2}$
Therefore,the roots are $-5+3\sqrt{2}$ and $-5-3\sqrt{2}$.

(C) To solve $25 x^{2}-30 x+3=0$ by completing the square,we first isolate the constant term: $25 x^{2}-30 x = -3$.
Divide the entire equation by $25$ to make the coefficient of $x^2$ equal to $1$: $x^{2} - \frac{30}{25} x = -\frac{3}{25}$,which simplifies to $x^{2} - \frac{6}{5} x = -\frac{3}{25}$.
To complete the square,add $(\frac{1}{2} \times \text{coefficient of } x)^2 = (\frac{1}{2} \times -\frac{6}{5})^2 = (-\frac{3}{5})^2 = \frac{9}{25}$ to both sides:
$x^{2} - \frac{6}{5} x + \frac{9}{25} = -\frac{3}{25} + \frac{9}{25}$.
This gives $(x - \frac{3}{5})^2 = \frac{6}{25}$.
Taking the square root of both sides: $x - \frac{3}{5} = \pm \frac{\sqrt{6}}{5}$.
Therefore,$x = \frac{3}{5} \pm \frac{\sqrt{6}}{5}$,which results in $x = \frac{3 \pm \sqrt{6}}{5}$.
Thus,the roots are $\frac{3-\sqrt{6}}{5}$ and $\frac{3+\sqrt{6}}{5}$.

(D) To solve the quadratic equation $16 x^{2}-24 x-1=0$ by the method of completing the square,we first rewrite the equation as $16 x^{2}-24 x = 1$.
Divide the entire equation by the coefficient of $x^{2}$,which is $16$:
$x^{2} - \frac{24}{16} x = \frac{1}{16}$
$x^{2} - \frac{3}{2} x = \frac{1}{16}$
Now,add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{3}{2}$,so half of it is $-\frac{3}{4}$,and its square is $\frac{9}{16}$:
$x^{2} - \frac{3}{2} x + \frac{9}{16} = \frac{1}{16} + \frac{9}{16}$
This simplifies to:
$(x - \frac{3}{4})^{2} = \frac{10}{16}$
Taking the square root of both sides:
$x - \frac{3}{4} = \pm \sqrt{\frac{10}{16}}$
$x - \frac{3}{4} = \pm \frac{\sqrt{10}}{4}$
Solving for $x$:
$x = \frac{3}{4} \pm \frac{\sqrt{10}}{4}$
$x = \frac{3 \pm \sqrt{10}}{4}$
Thus,the roots are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$.

(A) Given equation: $3 y^{2}+7 y-20=0$
Divide the entire equation by $3$ to make the coefficient of $y^{2}$ equal to $1$:
$y^{2}+\frac{7}{3} y-\frac{20}{3}=0$
To complete the square,add and subtract the square of half the coefficient of $y$,which is $(\frac{1}{2} \times \frac{7}{3})^{2} = (\frac{7}{6})^{2} = \frac{49}{36}$:
$y^{2}+\frac{7}{3} y + \frac{49}{36} - \frac{49}{36} - \frac{20}{3} = 0$
$(y+\frac{7}{6})^{2} - (\frac{49}{36} + \frac{240}{36}) = 0$
$(y+\frac{7}{6})^{2} - \frac{289}{36} = 0$
$(y+\frac{7}{6})^{2} = (\frac{17}{6})^{2}$
Taking the square root on both sides:
$y+\frac{7}{6} = \pm \frac{17}{6}$
Case $1$: $y = \frac{17}{6} - \frac{7}{6} = \frac{10}{6} = \frac{5}{3}$
Case $2$: $y = -\frac{17}{6} - \frac{7}{6} = -\frac{24}{6} = -4$
Thus,the roots are $-4$ and $\frac{5}{3}$.

(B) Given equation: $5x^{2}-4x-10=0$
Divide the entire equation by $5$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2}-\frac{4}{5}x-2=0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{4}{5})^{2} = (\frac{2}{5})^{2} = \frac{4}{25}$:
$x^{2}-\frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - 2 = 0$
$(x-\frac{2}{5})^{2} - (\frac{4}{25} + 2) = 0$
$(x-\frac{2}{5})^{2} - (\frac{4+50}{25}) = 0$
$(x-\frac{2}{5})^{2} - \frac{54}{25} = 0$
$(x-\frac{2}{5})^{2} = \frac{54}{25}$
Taking the square root on both sides:
$x-\frac{2}{5} = \pm \sqrt{\frac{54}{25}} = \pm \frac{3\sqrt{6}}{5}$
$x = \frac{2}{5} \pm \frac{3\sqrt{6}}{5}$
$x = \frac{2 \pm 3\sqrt{6}}{5}$
Therefore,the roots are $\frac{2-3\sqrt{6}}{5}$ and $\frac{2+3\sqrt{6}}{5}$.

(C) Given quadratic equation is $m^{2}-18m+81=0$.
Comparing this with the standard form $a^{2}-2ab+b^{2}=(a-b)^{2}$,we can write:
$m^{2}-2(m)(9)+(9)^{2}=0$
This simplifies to:
$(m-9)^{2}=0$
Therefore:
$(m-9)(m-9)=0$
Setting each factor to zero:
$m-9=0$ or $m-9=0$
Solving for $m$:
$m=9$ or $m=9$
Thus,the roots of the given quadratic equation are $9$ and $9$.

(D) To solve $x^{2}-8x+12=0$ by completing the square:
$1$. Move the constant term to the side: $x^{2}-8x = -12$
$2$. Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-8$, half of it is $-4$, and its square is $(-4)^{2} = 16$.
$3$. $x^{2}-8x+16 = -12+16$
$4$. Rewrite the left side as a perfect square: $(x-4)^{2} = 4$
$5$. Take the square root of both sides: $x-4 = \pm 2$
$6$. Solve for $x$: $x = 4+2 = 6$ or $x = 4-2 = 2$.
Thus, the roots are $2$ and $6$.

(A) Given the quadratic equation: $x^{2}+2x-2=0$
Step $1$: Move the constant term to the right side: $x^{2}+2x=2$
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $2$,so half of it is $1$,and its square is $1^{2}=1$. Adding $1$ to both sides:
$x^{2}+2x+1=2+1$
Step $3$: Write the left side as a perfect square: $(x+1)^{2}=3$
Step $4$: Take the square root of both sides: $x+1 = \pm\sqrt{3}$
Step $5$: Solve for $x$: $x = -1 \pm \sqrt{3}$
Thus,the roots are $-1+\sqrt{3}$ and $-1-\sqrt{3}$.

(B) Given equation: $x^{2}-4x-8=0$
Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-4$,so half of it is $-2$,and its square is $(-2)^{2}=4$.
$x^{2}-4x+4=8+4$
Step $3$: Write the left side as a perfect square: $(x-2)^{2}=12$
Step $4$: Take the square root on both sides: $x-2 = \pm\sqrt{12}$
$x-2 = \pm 2\sqrt{3}$
Step $5$: Solve for $x$: $x = 2 \pm 2\sqrt{3}$
Thus,the roots are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

(A) Given equation: $4x^{2} + 20x + 23 = 0$.
Divide the entire equation by $4$: $x^{2} + 5x + \frac{23}{4} = 0$.
Rewrite as: $x^{2} + 5x = -\frac{23}{4}$.
Add the square of half the coefficient of $x$ (which is $(\frac{5}{2})^{2} = \frac{25}{4}$) to both sides: $x^{2} + 5x + \frac{25}{4} = -\frac{23}{4} + \frac{25}{4}$.
This simplifies to: $(x + \frac{5}{2})^{2} = \frac{2}{4} = \frac{1}{2}$.
Taking the square root on both sides: $x + \frac{5}{2} = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
Thus,$x = -\frac{5}{2} \pm \frac{\sqrt{2}}{2} = \frac{-5 \pm \sqrt{2}}{2}$.
Therefore,the roots are $\frac{-5+\sqrt{2}}{2}$ and $\frac{-5-\sqrt{2}}{2}$.

(A) Given equation: $4x^{2} + 4bx - (a^{2} - b^{2}) = 0$.
Divide by $4$: $x^{2} + bx - \frac{a^{2}-b^{2}}{4} = 0$.
Add and subtract $(\frac{b}{2})^{2}$ to complete the square: $x^{2} + bx + (\frac{b}{2})^{2} - (\frac{b}{2})^{2} - \frac{a^{2}-b^{2}}{4} = 0$.
$(x + \frac{b}{2})^{2} = \frac{b^{2}}{4} + \frac{a^{2}-b^{2}}{4} = \frac{a^{2}}{4}$.
Taking square root on both sides: $x + \frac{b}{2} = \pm \frac{a}{2}$.
Case $1$: $x = \frac{a-b}{2}$.
Case $2$: $x = \frac{-a-b}{2} = -(\frac{a+b}{2})$.
Thus,the roots are $-\left(\frac{a+b}{2}\right), \left(\frac{a-b}{2}\right)$.

(A) Given equation: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$.
To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{\sqrt{3}+1}{2})^{2}$.
$x^{2}-(\sqrt{3}+1) x + (\frac{\sqrt{3}+1}{2})^{2} - (\frac{\sqrt{3}+1}{2})^{2} + \sqrt{3} = 0$.
$(x - \frac{\sqrt{3}+1}{2})^{2} = (\frac{\sqrt{3}+1}{2})^{2} - \sqrt{3}$.
$(x - \frac{\sqrt{3}+1}{2})^{2} = \frac{3+1+2\sqrt{3}}{4} - \sqrt{3} = \frac{4+2\sqrt{3}-4\sqrt{3}}{4} = \frac{4-2\sqrt{3}}{4} = \frac{(\sqrt{3}-1)^{2}}{4}$.
Taking square root on both sides: $x - \frac{\sqrt{3}+1}{2} = \pm \frac{\sqrt{3}-1}{2}$.
Case $1$: $x = \frac{\sqrt{3}+1 + \sqrt{3}-1}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Case $2$: $x = \frac{\sqrt{3}+1 - (\sqrt{3}-1)}{2} = \frac{\sqrt{3}+1-\sqrt{3}+1}{2} = \frac{2}{2} = 1$.
Thus,the roots are $\sqrt{3}$ and $1$.

(B) Given equation: $x^{2}-4 \sqrt{2} x+6=0$.
To complete the square,we rewrite the equation as $x^{2}-2(x)(2 \sqrt{2}) = -6$.
Add $(2 \sqrt{2})^{2} = 8$ to both sides: $x^{2}-2(x)(2 \sqrt{2}) + (2 \sqrt{2})^{2} = -6 + 8$.
This simplifies to $(x-2 \sqrt{2})^{2} = 2$.
Taking the square root on both sides: $x-2 \sqrt{2} = \pm \sqrt{2}$.
Case $1$: $x = 2 \sqrt{2} + \sqrt{2} = 3 \sqrt{2}$.
Case $2$: $x = 2 \sqrt{2} - \sqrt{2} = \sqrt{2}$.
Thus,the roots are $\sqrt{2}$ and $3 \sqrt{2}$.

(C) Given equation: $3x^2 + 11x + 10 = 0$.
Divide the entire equation by $3$ to make the coefficient of $x^2$ equal to $1$: $x^2 + \frac{11}{3}x + \frac{10}{3} = 0$.
Shift the constant term to the right side: $x^2 + \frac{11}{3}x = -\frac{10}{3}$.
Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{11}{3}$,so half of it is $\frac{11}{6}$. The square is $(\frac{11}{6})^2 = \frac{121}{36}$.
$x^2 + \frac{11}{3}x + \frac{121}{36} = -\frac{10}{3} + \frac{121}{36}$.
$(x + \frac{11}{6})^2 = -\frac{120}{36} + \frac{121}{36} = \frac{1}{36}$.
Taking the square root on both sides: $x + \frac{11}{6} = \pm \frac{1}{6}$.
Case $1$: $x + \frac{11}{6} = \frac{1}{6} \implies x = \frac{1}{6} - \frac{11}{6} = -\frac{10}{6} = -\frac{5}{3}$.
Case $2$: $x + \frac{11}{6} = -\frac{1}{6} \implies x = -\frac{1}{6} - \frac{11}{6} = -\frac{12}{6} = -2$.
Thus,the roots are $-\frac{5}{3}$ and $-2$.

(N/A) Given equation: $2x^{2} + x + 4 = 0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x + 2 = 0$.
Rewrite the equation as $x^{2} + \frac{1}{2}x = -2$.
To complete the square,add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{1}{2}$,so half of it is $\frac{1}{4}$. The square is $(\frac{1}{4})^{2} = \frac{1}{16}$.
$x^{2} + \frac{1}{2}x + \frac{1}{16} = -2 + \frac{1}{16}$.
$(x + \frac{1}{4})^{2} = \frac{-32 + 1}{16} = -\frac{31}{16}$.
Since the square of a real number cannot be negative,there are no real roots for this equation.

(A) Given equation: $2x^{2} - 7x + 3 = 0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} - \frac{7}{2}x + \frac{3}{2} = 0$.
Shift the constant term to the right side:
$x^{2} - \frac{7}{2}x = -\frac{3}{2}$.
Add the square of half the coefficient of $x$ (which is $\frac{1}{2} \times \frac{7}{2} = \frac{7}{4}$) to both sides:
$x^{2} - \frac{7}{2}x + (\frac{7}{4})^{2} = -\frac{3}{2} + (\frac{7}{4})^{2}$.
$(x - \frac{7}{4})^{2} = -\frac{3}{2} + \frac{49}{16}$.
$(x - \frac{7}{4})^{2} = \frac{-24 + 49}{16} = \frac{25}{16}$.
Taking the square root on both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$.
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$.
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$.
Thus,the roots are $3$ and $\frac{1}{2}$.

(B) Given the quadratic equation: $x^{2}+8x+3=0$.
Step $1$: Shift the constant term to the right side: $x^{2}+8x = -3$.
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $8$,so half of it is $4$,and its square is $4^{2} = 16$.
$x^{2}+8x+16 = -3+16$.
Step $3$: Write the left side as a perfect square: $(x+4)^{2} = 13$.
Step $4$: Take the square root of both sides: $x+4 = \pm\sqrt{13}$.
Step $5$: Solve for $x$: $x = -4 \pm \sqrt{13}$.
Thus,the roots are $-4+\sqrt{13}$ and $-4-\sqrt{13}$.