Quadratic Equations - Exercise 4.3

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Overview

Master the discriminant $D = b^2 - 4ac$ and determine the nature of roots for $ax^2 + bx + c = 0$. Solve problems involving real distinct roots ($D > 0$), real equal roots ($D = 0$), and no real roots ($D < 0$), alongside geometric feasibility scenarios.

NCERT Exercise Practice questions with detailed step-by-step solutions

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Q1: Nature of Roots Q2: Equal Roots ($k$) Q3: Mango Grove Q4: Age Feasibility Q5: Park Perimeter
Question 1 Nature of Roots & Discriminant
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2 - 3x + 5 = 0$
(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$
(iii) $2x^2 - 6x + 3 = 0$
(i) $2x^2 - 3x + 5 = 0$
Comparing with $ax^2 + bx + c = 0$: $a = 2, b = -3, c = 5$.
Discriminant $D = b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$.
Since $D < 0$, no real roots exist for this equation.
(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$
Comparing with $ax^2 + bx + c = 0$: $a = 3, b = -4\sqrt{3}, c = 4$.
Discriminant $D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$.
Since $D = 0$, two equal real roots exist.
Using the quadratic formula: $x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-(-4\sqrt{3}) \pm 0}{2(3)} = \dfrac{4\sqrt{3}}{6} = \dfrac{2\sqrt{3}}{3} = \dfrac{2}{\sqrt{3}}$.
$\therefore$ The roots are $\dfrac{2\sqrt{3}}{3}$ and $\dfrac{2\sqrt{3}}{3}$ (or $\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}$).
(iii) $2x^2 - 6x + 3 = 0$
Comparing with $ax^2 + bx + c = 0$: $a = 2, b = -6, c = 3$.
Discriminant $D = b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$.
Since $D > 0$, two distinct real roots exist.
Using the quadratic formula:
$x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-(-6) \pm \sqrt{12}}{2(2)} = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}$.
$\therefore$ The roots are $\dfrac{3 + \sqrt{3}}{2}$ and $\dfrac{3 - \sqrt{3}}{2}$.
Question 2 Equal Roots Condition ($D = 0$)
Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots:
(i) $2x^2 + kx + 3 = 0$
(ii) $kx(x - 2) + 6 = 0$
(i) $2x^2 + kx + 3 = 0$
Here $a = 2, b = k, c = 3$.
For equal roots, $D = 0 \implies b^2 - 4ac = 0$:
$k^2 - 4(2)(3) = 0 \implies k^2 - 24 = 0$
$k^2 = 24 \implies k = \pm\sqrt{24} = \pm 2\sqrt{6}$.
$\therefore k = \pm 2\sqrt{6}$.
(ii) $kx(x - 2) + 6 = 0$
Simplifying into standard form: $kx^2 - 2kx + 6 = 0$.
Here $a = k, b = -2k, c = 6$.
For equal roots, $D = 0 \implies b^2 - 4ac = 0$:
$(-2k)^2 - 4(k)(6) = 0$
$4k^2 - 24k = 0 \implies 4k(k - 6) = 0$
$\implies k = 0$ or $k = 6$.
If $k = 0$, the $x^2$ coefficient vanishes ($a = 0$), so the equation ceases to be a quadratic equation.
Therefore, $k \neq 0$.
$\therefore k = 6$.
Question 3 Geometric Feasibility
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ m}^2$? If so, find its length and breadth.
Step 1: Set up algebraic dimensions
Let the breadth of the rectangular grove be $x\text{ metres}$.
Then its length $= 2x\text{ metres}$.
Step 2: Form the quadratic equation and check discriminant
$\text{Area} = \text{Length} \times \text{Breadth} = (2x)(x) = 2x^2$
Given Area $= 800\text{ m}^2$:
$2x^2 = 800 \implies x^2 = 400 \implies x^2 - 400 = 0$
Comparing with $ax^2 + bx + c = 0$: $a = 1, b = 0, c = -400$.
$D = b^2 - 4ac = 0^2 - 4(1)(-400) = 1600 > 0$.
Since $D > 0$, real roots exist. Thus, the situation is possible.
Step 3: Solve for dimensions
$x^2 = 400 \implies x = \pm 20$
Since breadth cannot be negative, $x = 20\text{ m}$.
$\text{Length} = 2x = 2(20) = 40\text{ m}$.
Yes, it is possible. Length = 40 m and Breadth = 20 m.
Question 4 Age Problem & Real Root Analysis
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is $20$ years. Four years ago, the product of their ages in years was $48$.
Step 1: Assign variables and express past ages
Let the present age of the first friend be $x\text{ years}$.
Then the present age of the second friend $= (20 - x)\text{ years}$.
Four years ago:
First friend's age $= (x - 4)\text{ years}$
Second friend's age $= (20 - x - 4) = (16 - x)\text{ years}$
Step 2: Formulate quadratic equation
Given product of ages four years ago was $48$:
$(x - 4)(16 - x) = 48$
$16x - x^2 - 64 + 4x = 48$
$-x^2 + 20x - 64 = 48 \implies -x^2 + 20x - 112 = 0 \implies x^2 - 20x + 112 = 0$
Step 3: Check discriminant
Here $a = 1, b = -20, c = 112$.
$D = b^2 - 4ac = (-20)^2 - 4(1)(112) = 400 - 448 = -48$.
Since $D < 0$, no real roots exist for this equation.
No, the situation is not possible because the discriminant is negative ($D = -48 < 0$), meaning no real ages satisfy this condition.
Question 5 Perimeter & Area Feasibility
Is it possible to design a rectangular park of perimeter $80\text{ m}$ and area $400\text{ m}^2$? If so, find its length and breadth.
Step 1: Relate length and breadth using perimeter
$\text{Perimeter} = 2(l + b) = 80\text{ m} \implies l + b = 40\text{ m}$.
Let length $= l\text{ m}$, then breadth $b = (40 - l)\text{ m}$.
Step 2: Formulate quadratic equation from area
$\text{Area} = l \times b = l(40 - l) = 400$
$40l - l^2 = 400 \implies l^2 - 40l + 400 = 0$
Here $a = 1, b = -40, c = 400$.
$D = b^2 - 4ac = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$.
Since $D = 0$, real and equal roots exist, so the park design is possible.
Step 3: Solve for dimensions
$(l - 20)^2 = 0 \implies l = 20\text{ m}$.
$\text{Breadth } b = 40 - 20 = 20\text{ m}$.
Yes, it is possible. The park is a square with Length = 20 m and Breadth = 20 m.
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