The problem presents a worksheet containing questions related to quadratic equations. It involves identifying quadratic equations, finding solutions, transforming equations, finding parameters for specific solution types, determining the nature of equations, solving for x, and applying Vieta's formulas.

1. Problem Description

The problem presents a worksheet containing questions related to quadratic equations. It involves identifying quadratic equations, finding solutions, transforming equations, finding parameters for specific solution types, determining the nature of equations, solving for x, and applying Vieta's formulas.

2. Solution Steps

1. Which of the following are quadratic equations in x?

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a \ne 0$.

a) $x^2 + 3 = 0$ - Quadratic

b) $-x^2 + 2x = 7 \implies -x^2 + 2x - 7 = 0$ - Quadratic

c) $3x - 4y = 2$ - Linear (involves two variables, x and y)

d) $2 - 3x + 0x = 0 \implies 2 - 3x = 0$ - Linear

2. Find solutions for the following equations:

a) $(0 - 3)x^2 = 0 \implies -3x^2 = 0 \implies x^2 = 0 \implies x = 0$

b) $(3 - \sqrt{2})x^2 = 0 \implies x^2 = 0 \implies x = 0$ (since $3 - \sqrt{2} \ne 0$)

c) $mx^2 = 0$ for $m \ne 0 \implies x^2 = 0 \implies x = 0$

d) $(m - 1)x^2 = 0$ for $m = 1 \implies (1 - 1)x^2 = 0 \implies 0x^2 = 0$. This equation is true for all values of $x$. Therefore, x is any real number.

3. Transform equations in the general form: $ax^2+bx+c=0$

a) $\frac{2x^2+3x}{5} - 5x + \frac{x^2}{2} = 0 \implies \frac{2(2x^2+3x) - 10(5x) + 5(x^2)}{10} = 0 \implies 4x^2 + 6x - 50x + 5x^2 = 0 \implies 9x^2 - 44x = 0$

b) $(3x-5)^2 - (x+3)^2 = 16 \implies (9x^2 - 30x + 25) - (x^2 + 6x + 9) = 16 \implies 8x^2 - 36x + 16 = 16 \implies 8x^2 - 36x = 0$

c) $\frac{x}{x-1} + \frac{x}{x+1} = \frac{9}{4} \implies \frac{x(x+1) + x(x-1)}{(x-1)(x+1)} = \frac{9}{4} \implies \frac{x^2+x+x^2-x}{x^2-1} = \frac{9}{4} \implies \frac{2x^2}{x^2-1} = \frac{9}{4} \implies 8x^2 = 9x^2 - 9 \implies x^2 - 9 = 0$

4. Find the parameter $m$ so the equation $x^2 + 2(3-m)x + 2m - 3 = 0$ has a repeated real number solution.

For a repeated real root, the discriminant must be zero: $b^2 - 4ac = 0$.

Here, $a = 1, b = 2(3-m), c = 2m - 3$.

$(2(3-m))^2 - 4(1)(2m-3) = 0 \implies 4(9 - 6m + m^2) - 8m + 12 = 0 \implies 36 - 24m + 4m^2 - 8m + 12 = 0 \implies 4m^2 - 32m + 48 = 0 \implies m^2 - 8m + 12 = 0 \implies (m - 6)(m - 2) = 0$.

So $m = 6$ or $m = 2$.

5. Determine the nature of the equation $(k-1)x^2 + 2(k+2)x + k - 3 = 0$, depending on the parameter $k$.

The discriminant is $D = b^2 - 4ac = [2(k+2)]^2 - 4(k-1)(k-3) = 4(k^2 + 4k + 4) - 4(k^2 - 4k + 3) = 4k^2 + 16k + 16 - 4k^2 + 16k - 12 = 32k + 4$.

If $D > 0$, two distinct real roots. $32k + 4 > 0 \implies k > -1/8$.

If $D = 0$, one repeated real root. $32k + 4 = 0 \implies k = -1/8$.

If $D < 0$, two complex roots. $32k + 4 < 0 \implies k < -1/8$.

Also, if $k=1$, it becomes a linear equation $2(1+2)x + 1 - 3 = 0 \implies 6x - 2 = 0 \implies x = 1/3$

6. Solve for $x$ (to two decimal places where necessary)

a) $3x^2 - 23x + 14 = 0 \implies x = \frac{23 \pm \sqrt{23^2 - 4(3)(14)}}{2(3)} = \frac{23 \pm \sqrt{529 - 168}}{6} = \frac{23 \pm \sqrt{361}}{6} = \frac{23 \pm 19}{6} \implies x = \frac{42}{6} = 7$ or $x = \frac{4}{6} = \frac{2}{3} \approx 0.67$

7. Find the sides of the rectangle with surface $10 cm^2$ and perimeter $14 cm$ (use Vieta's formulas).

Let the sides be $l$ and $w$.

$lw = 10$ and $2(l+w) = 14 \implies l+w = 7$.

$l$ and $w$ are roots of the equation $x^2 - 7x + 10 = 0$.

$(x-5)(x-2) = 0 \implies x = 5$ or $x = 2$.

Thus, the sides are 5 cm and 2 cm.

8. Using the Vieta's formulas find which are the correct roots of the given quadratic equation:

Vieta's formulas: For $ax^2 + bx + c = 0$, the sum of roots is $-b/a$ and the product of roots is $c/a$.

a) 3 and 4, for $x^2 - 7x + 12 = 0$. Sum: $3 + 4 = 7$. Product: $3 \cdot 4 = 12$. Since $-b/a = -(-7)/1 = 7$ and $c/a = 12/1 = 12$, this is correct.

b) 3 and -4, for $x^2 + 7x - 12 = 0$. Sum: $3 + (-4) = -1$. Product: $3 \cdot (-4) = -12$. Since $-b/a = -7/1 = -7$ and $c/a = -12/1 = -12$, this is incorrect.

c) $1 \pm \sqrt{2}$, for $x^2 - 2x - 1 = 0$. Sum: $1 + \sqrt{2} + 1 - \sqrt{2} = 2$. Product: $(1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1$. Since $-b/a = -(-2)/1 = 2$ and $c/a = -1/1 = -1$, this is correct.

3. Final Answer

1. a) and b)

2. a) 0, b) 0, c) 0, d) any real number

3. a) $9x^2 - 44x = 0$, b) $8x^2 - 36x = 0$, c) $x^2 - 9 = 0$

4. $m = 6$ or $m = 2$

5. If $k > -1/8$, two distinct real roots. If $k = -1/8$, one repeated real root. If $k < -1/8$, two complex roots.

6. a) $x = 7$ or $x = \frac{2}{3}$

7. 5 cm and 2 cm

8. a) and c)