ProTeacher.in

Q1. Find the nature of the roots of the following quadratic equations. If real roots exist, find them.

(i) $2x^2-3x+5=0$
Discriminant $\mathrm{\Delta }=b^2-4ac=(-3{)}^2-4(2)(5)=9-40=-31<0$.
Nature: $\mathrm{\Delta }<\mathrm{0 }$⇒ no real roots (two complex conjugate roots).

(ii) $3x^2-4\sqrt{3}x+4=0$
$\mathrm{\Delta }=(-4\sqrt{3}{)}^2-4(3)(4)=48-48=0.$
Nature: $\mathrm{\Delta }=0$ ⇒ two equal real roots (a repeated root).
Root: $x={\displaystyle \frac{-b}{2a}}={\displaystyle \frac{4\sqrt{3}}{6}}={\displaystyle \frac{2\sqrt{3}}{3}}\mathrm{.}$

(iii) $2x^2-6x+3=0$
$\mathrm{\Delta }=(-6{)}^2-4(2)(3)=36-24=12>0$

Nature: $\mathrm{\Delta }>0$ ⇒ two distinct real roots.
Roots: $x={\displaystyle \frac{6\pm \sqrt{12}}{4}}={\displaystyle \frac{6\pm 2\sqrt{3}}{4}}={\displaystyle \frac{3\pm \sqrt{3}}{2}}\mathrm{.}$

---

Q2. Find the values of $k$ so that the quadratic has two equal roots.

(i) $2x^2+kx+3=0$.
For equal roots $\mathrm{\Delta }=0\Rightarrow k^2-4(2)(3)=0\Rightarrow k^2-24=0.$
So $k=\pm \sqrt{24}=\pm 2\sqrt{6}\mathrm{}$

(ii) $kx(x-2)+6=0$.
Expand: $k{x}^2-2kx+6=0$. Here $a=k,b=-2k,c=6$. For equal roots $\mathrm{\Delta }=0$

$$(-2k{)}^2-4(k)(6)=0\Rightarrow 4k^2-24k=4k(k-6)=0.$$

This gives $k=0$ or $k=6$.
But $k=0$ makes the equation $6=0$ (not a quadratic), so discard $k=0$.
Valid value: $k=\mathrm{6 }$(gives a quadratic with a repeated root).

---

Q3. Mango grove: length = twice breadth, area $=800{\text{m}}^2$. Find length and breadth.

Let breadth $=b$. Length $=2b$. Area: $2b^2=800\Rightarrow b^2=400\Rightarrow b=\mathrm{20 }$(positive).
Length $=2b=40$
Answer: Breadth $20\text{ m}$, length $40\text{ m}$.

---

Q4. Ages problem: Sum of ages $=20$. Four years ago product of ages was $48$. Is this possible? If so, find present ages.

Let present ages be $x$ and $20-x$. Four years ago their ages were $x-4$ and $16-x$. Given:

$$(x-4)(16-x)=48.$$$\mathrm{\Delta }<0$ ⇒ no real solution.
Conclusion: The situation is not possible (no pair of real ages satisfies the conditions).

---

Q5. Park: perimeter $=80$ m and area $=400{\text{m}}^2$. Is this possible? If so, find length and breadth.

Let length $=l$, breadth $=b$. From perimeter $l+b=40$. Area gives $l(40-l)=400$.
So $-l^2+40l-400=0\Rightarrow l^2-40l+400=0.$
Discriminant $\mathrm{\Delta }={40}^2-4\cdot 1\cdot 400=1600-1600=0.$
$\mathrm{\Delta }=0$ ⇒ equal roots: $l={\displaystyle \frac{40}{2}}=20$. Then $b=40-l=20$
Answer: Yes — the park is $20\text{ m}\times 20\text{ m}$ (a square).

---