Quadratic Equations — Important Questions

Answer: (a) $-8$ — $D=b^2-4ac=(-4)^2-4(2)(3)=16-24=-8$. Since $D<0$, the equation has no real roots.

Answer: (a) $5,\ -2$ — $x^2-3x-10=(x-5)(x+2)=0$, so $x=5$ or $x=-2$.

Answer: (a) $\pm2\sqrt6$ — For equal roots $D=0$: $\ k^2-4(2)(3)=0\Rightarrow k^2=24\Rightarrow k=\pm2\sqrt6$.

Answer: (a) $7$ — Substituting $x=-5$: $\ 2(25)+p(-5)-15=0\Rightarrow 50-5p-15=0\Rightarrow 5p=35\Rightarrow p=7$.

Answer: (a) Both are true and R explains A. For $x^2+x+1=0$, $\ D=1^2-4(1)(1)=-3<0$, so it has no real roots — exactly the condition in R.

$D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31$. Since $D<0$, the equation has no real roots (the roots are non-real).

Equal roots $\Rightarrow D=0$: $\ k^2-4(2)(2)=0\Rightarrow k^2=16\Rightarrow k=\pm4$.

Let the integers be $x$ and $x+1$.

$x^2+(x+1)^2=365\Rightarrow 2x^2+2x+1=365\Rightarrow 2x^2+2x-364=0\Rightarrow x^2+x-182=0.$

$(x+14)(x-13)=0\Rightarrow x=13$ (rejecting $x=-14$ since the integers are positive).

The integers are $13$ and $14$.

Let the larger number be $x$ and the smaller be $y$. Given $x^2-y^2=180$ and $y^2=8x$.

Substituting: $x^2-8x=180\Rightarrow x^2-8x-180=0\Rightarrow (x-18)(x+10)=0\Rightarrow x=18$ (taking $x>0$ so that $y^2=8x>0$).

Then $y^2=8(18)=144\Rightarrow y=\pm12$.

The numbers are $18$ and $12$ (or $18$ and $-12$).

The coefficients sum to zero: $(b-c)+(c-a)+(a-b)=0$, so $x=1$ is always a root. For equal roots both roots must equal $1$, hence the product of the roots is $1$:
$\frac{a-b}{b-c}=1\Rightarrow a-b=b-c\Rightarrow a+c=2b.$
Hence proved.

(Equivalently, $D=(c-a)^2-4(b-c)(a-b)=0$ simplifies to $(a+c-2b)^2=0$.)

Let the speed be $x$ km/h, so the time taken is $\dfrac{360}{x}$ hours.

$\dfrac{360}{x}-\dfrac{360}{x+5}=1\Rightarrow 360\left(\dfrac{(x+5)-x}{x(x+5)}\right)=1\Rightarrow \dfrac{1800}{x^2+5x}=1.$

$x^2+5x-1800=0.$ Here $D=25+7200=7225=85^2$, so $x=\dfrac{-5\pm85}{2}$.

Taking the positive value, $x=\dfrac{80}{2}=40$.

The speed of the train is $40$ km/h.

Let the smaller tap take $x$ hours; then the larger takes $(x-10)$ hours. Together they fill the tank in $\dfrac{75}{8}$ hours, so
$\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}.$
$\dfrac{(x-10)+x}{x(x-10)}=\dfrac{8}{75}\Rightarrow 75(2x-10)=8(x^2-10x)\Rightarrow 8x^2-230x+750=0\Rightarrow 4x^2-115x+375=0.$

$D=115^2-4(4)(375)=13225-6000=7225=85^2$, so $x=\dfrac{115\pm85}{8}$, giving $x=25$ or $x=3.75$.

Since $x-10$ must be positive, reject $x=3.75$. Thus $x=25$.

Smaller tap: $25$ hours; larger tap: $15$ hours.

(i) Length $=(2x+1)$ m, so $x(2x+1)=528\Rightarrow 2x^2+x-528=0$.

(ii) $D=b^2-4ac=1^2-4(2)(-528)=1+4224=4225=65^2$.

(iii) $x=\dfrac{-1\pm65}{4}$. Taking the positive root, $x=\dfrac{64}{4}=16$. So breadth $=16$ m and length $=2(16)+1=33$ m. (Check: $16\times33=528$ ✓.)