Solutions: Quadratic Equations - Test 4

Q1. (b) $a < 0$

A quadratic graph is a parabola. It opens downwards if the coefficient of $x^2$ is negative.

Q2. (c) 3

Substitute $x = -1/2$: $2(-1/2)^2 + k(-1/2) + 1 = 0 \Rightarrow 2(1/4) - k/2 + 1 = 0 \Rightarrow 1/2 - k/2 + 1 = 0 \Rightarrow k/2 = 3/2 \Rightarrow k = 3$.

Q3. (c) $2x^2 - 7x + 6 = 0$

Substitute $x=2$: $2(2)^2 - 7(2) + 6 = 8 - 14 + 6 = 0$.

Q4. (a) $-2 < k < 2$

For no real roots, $D < 0$. $(-k)^2 - 4(1)(1) < 0 \Rightarrow k^2 - 4 < 0 \Rightarrow k^2 < 4 \Rightarrow -2 < k < 2$.

Q5. (d) A is false but R is true.

$(x+1)^2 - x^2 = 0 \Rightarrow x^2 + 2x + 1 - x^2 = 0 \Rightarrow 2x + 1 = 0$. This is a linear equation with only 1 root. So A is false.

Q6. $px^2 - 3px + 9 = 0$. For equal roots, $D = 0$.
$(-3p)^2 - 4(p)(9) = 0 \Rightarrow 9p^2 - 36p = 0 \Rightarrow 9p(p-4) = 0$.
$p=0$ or $p=4$. Since $p=0$ makes it non-quadratic, $p=4$. p = 4 [2]

Q7. $\sqrt{2x+9} = 13 - x$. Square both sides: $2x+9 = (13-x)^2$.
$2x+9 = 169 + x^2 - 26x \Rightarrow x^2 - 28x + 160 = 0$.
$(x-20)(x-8) = 0 \Rightarrow x = 20, 8$.
Check: If $x=20$, $\sqrt{49}+20 = 27 \neq 13$. If $x=8$, $\sqrt{25}+8 = 13$. x = 8 [2]

Q8. Let integers be $x, x+2$. $x(x+2) = 483 \Rightarrow x^2 + 2x - 483 = 0$.
$D = 4 - 4(1)(-483) = 4 + 1932 = 1936$. $\sqrt{1936} = 44$.
$x = \frac{-2 \pm 44}{2}$. $x = 21$ or $x = -23$.
Integers: 21, 23 (or -23, -21). 21, 23 [2]

Q9. Let stream speed $x$. Upstream $18-x$, Downstream $18+x$.
$\frac{24}{18-x} - \frac{24}{18+x} = 1 \Rightarrow 24[\frac{18+x - (18-x)}{(18-x)(18+x)}] = 1$.
$24(2x) = 324 - x^2 \Rightarrow 48x = 324 - x^2 \Rightarrow x^2 + 48x - 324 = 0$.
$(x+54)(x-6) = 0 \Rightarrow x = 6$ (speed > 0). 6 km/h [3]

Q10. Let one pipe take $x$ min, other $x+5$. Total $100/9$ min.
$\frac{1}{x} + \frac{1}{x+5} = \frac{9}{100} \Rightarrow \frac{2x+5}{x^2+5x} = \frac{9}{100}$.
$200x + 500 = 9x^2 + 45x \Rightarrow 9x^2 - 155x - 500 = 0$.
$x = \frac{155 \pm \sqrt{24025 + 18000}}{18} = \frac{155 \pm 205}{18}$.
$x = 360/18 = 20$ or negative. So $x=20$.
Times: 20 min and 25 min. [3]

(i) The equation is quadratic with negative $t^2$ coefficient. Shape is a **Parabola opening downwards**. [1]

(ii) Hits ground when $h(t)=0$. $-t^2 + 2t + 3 = 0 \Rightarrow t^2 - 2t - 3 = 0$.
$(t-3)(t+1) = 0 \Rightarrow t = 3$ (time > 0). 3 seconds [2]

(iii) Max height at vertex $t = -b/2a = -2/-2 = 1$.
$h(1) = -1 + 2 + 3 = 4$ meters. [1]

(i) Dimensions of rug: $(12-2x)$ and $(8-2x)$.
Area: $(12-2x)(8-2x) = 60 \Rightarrow 4(6-x)(4-x) = 60 \Rightarrow (6-x)(4-x) = 15$.
$24 - 10x + x^2 = 15 \Rightarrow x^2 - 10x + 9 = 0$. [2]

(ii) $(x-9)(x-1) = 0 \Rightarrow x = 9, 1$.
If $x=9$, $8-2x$ is negative. So $x=1$. Width = 1 m. [1]

(iii) Length = $12 - 2(1) = 10$ m. Breadth = $8 - 2(1) = 6$ m. [1]