GSEB Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.4

Question 1. Find the discriminant 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) \( 2x^2 - 6x + 3 = 0 \)
Answer:
(i) We have the equation: \( 2x^2 - 3x + 5 = 0 \).
Here, the coefficients are \( a = 2 \), \( b = -3 \), and \( c = 5 \).
Now, let's calculate the discriminant \( D \), which is given by the formula \( D = b^2 - 4ac \).
\( D = (-3)^2 - 4(2)(5) \)
\( D = 9 - 40 \)
\( D = -31 \)
Since \( D < 0 \), the discriminant is negative. This means the given quadratic equation has no real roots.
(ii) We have the equation: \( 3x^2 - 4\sqrt{3}x + 4 = 0 \).
Here, the coefficients are \( a = 3 \), \( b = -4\sqrt{3} \), and \( c = 4 \).
Now, let's calculate the discriminant \( D \), which is given by the formula \( D = b^2 - 4ac \).
\( D = (-4\sqrt{3})^2 - 4(3)(4) \)
\( D = (16 \times 3) - 48 \)
\( D = 48 - 48 \)
\( D = 0 \)
Since \( D = 0 \), the discriminant is zero. Therefore, the given quadratic equation has two equal real roots.
To find the roots, we use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-4\sqrt{3}) \pm \sqrt{0}}{2(3)} \)
\( x = \frac{4\sqrt{3}}{6} \)
\( x = \frac{2\sqrt{3}}{3} \)
Thus, the roots are \( \frac{2\sqrt{3}}{3} \) and \( \frac{2\sqrt{3}}{3} \).
(iii) We have the equation: \( 2x^2 - 6x + 3 = 0 \).
Here, the coefficients are \( a = 2 \), \( b = -6 \), and \( c = 3 \).
Now, let's calculate the discriminant \( D \), which is given by the formula \( D = b^2 - 4ac \).
\( D = (-6)^2 - 4(2)(3) \)
\( D = 36 - 24 \)
\( D = 12 \)
Since \( D > 0 \), the discriminant is positive. This means the given quadratic equation has two distinct real roots.
To find the roots, we use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \).
\( x = \frac{-(-6) \pm \sqrt{12}}{2(2)} \)
\( x = \frac{6 \pm 2\sqrt{3}}{4} \)
Now, we can simplify this expression.
\( x = \frac{2(3 \pm \sqrt{3})}{4} \)
\( x = \frac{3 \pm \sqrt{3}}{2} \)
Thus, the two roots are \( \frac{3 + \sqrt{3}}{2} \) and \( \frac{3 - \sqrt{3}}{2} \).
In simple words: First, calculate the discriminant to see if real roots exist. If it's negative, there are no real roots. If it's zero, there are two identical real roots. If it's positive, there are two distinct real roots. Then, use the quadratic formula to find the roots.

Exam Tip: Remember to simplify square roots like \( \sqrt{12} \) to \( 2\sqrt{3} \) to ensure your final answer is in its simplest form. Also, always check the sign of the discriminant first, as it tells you immediately about the nature of the roots.

Question 2. Find the value of k for each of the following quadratic equations so that they have equal roots.
(i) \( 2x^2 + kx + 3 = 0 \)
(ii) \( kx(x - 2) + 6 = 0 \)
Answer:
(i) We have the quadratic equation: \( 2x^2 + kx + 3 = 0 \).
Here, the coefficients are \( a = 2 \), \( b = k \), and \( c = 3 \).
For equal roots, the discriminant \( D \) must be equal to zero, i.e., \( D = b^2 - 4ac = 0 \).
So, \( k^2 - 4(2)(3) = 0 \)
\( k^2 - 24 = 0 \)
\( k^2 = 24 \)
\( k = \pm \sqrt{24} \)
\( k = \pm \sqrt{4 \times 6} \)
\( k = \pm 2\sqrt{6} \)
Therefore, the value of \( k \) for the given equation to have equal roots is \( \pm 2\sqrt{6} \).
(ii) We have the equation: \( kx(x - 2) + 6 = 0 \).
First, expand and rewrite it in the standard quadratic form \( ax^2 + bx + c = 0 \).
\( kx^2 - 2kx + 6 = 0 \)
Here, the coefficients are \( a = k \), \( b = -2k \), and \( c = 6 \).
For equal roots, the discriminant \( D \) must be equal to zero, i.e., \( D = b^2 - 4ac = 0 \).
So, \( (-2k)^2 - 4(k)(6) = 0 \)
\( 4k^2 - 24k = 0 \)
Now, we can factor out \( 4k \).
\( 4k(k - 6) = 0 \)
This gives us two possible values for \( k \):
\( 4k = 0 \implies k = 0 \)
Or \( k - 6 = 0 \implies k = 6 \)
However, if \( k = 0 \), the original equation \( kx^2 - 2kx + 6 = 0 \) becomes \( 0x^2 - 0x + 6 = 0 \), which simplifies to \( 6 = 0 \). This is not a quadratic equation and has no solution, so \( k \) cannot be 0.
Therefore, the value of \( k \) for the given equation to have equal roots is \( 6 \).
In simple words: To find the value of 'k' for equal roots, set the discriminant (\( b^2 - 4ac \)) to zero. Then, solve the resulting equation for 'k'. Remember that if \( k \) makes the original equation not quadratic, that value of \( k \) is not valid.

Exam Tip: Always remember to check for extraneous solutions, especially when \( k \) appears as a coefficient of \( x^2 \). If \( k=0 \) turns the equation into a non-quadratic one, it's not a valid solution.

Question 3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²? If so, find its length and breadth.
Answer: Let the breadth of the rectangular mango grove be \( x \) meters.
According to the problem, the length is twice its breadth, so the length will be \( 2x \) meters.
The area of a rectangle is given by Length \( \times \) Breadth.
Given that the area is 800 m².
So, \( (2x)(x) = 800 \)
\( 2x^2 = 800 \)
Divide both sides by 2:
\( x^2 = 400 \)
To check if this situation is possible, we can form a quadratic equation from \( x^2 = 400 \) as \( x^2 - 400 = 0 \).
Here, the coefficients are \( a = 1 \), \( b = 0 \) (since there's no \( x \) term), and \( c = -400 \).
Let's calculate the discriminant \( D \): \( D = b^2 - 4ac \).
\( D = (0)^2 - 4(1)(-400) \)
\( D = 0 + 1600 \)
\( D = 1600 \)
Since \( D = 1600 > 0 \), the discriminant is positive, meaning real roots exist. Thus, designing such a rectangular mango grove is possible.
Now, let's find the values of \( x \).
\( x^2 = 400 \)
\( x = \pm \sqrt{400} \)
\( x = \pm 20 \)
So, \( x = 20 \) or \( x = -20 \).
Since breadth cannot be negative, we discard \( x = -20 \).
Therefore, the breadth of the rectangular mango grove is \( x = 20 \) meters.
The length is \( 2x = 2 \times 20 = 40 \) meters.
In simple words: First, set up equations for length and breadth based on the problem. Then, use the area to find an equation for the breadth. Check if real solutions exist by looking at the discriminant. If yes, calculate the breadth and then the length using the positive value.

Exam Tip: When dealing with dimensions like length or breadth, always remember that physical measurements cannot be negative. Discard any negative solutions obtained from the quadratic equation.

Question 4. 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.
Answer: Let the present age of the first friend be \( x \) years.
Since the sum of their ages is 20 years, the present age of the second friend will be \( (20 - x) \) years.
Four years ago:
Age of the first friend was \( (x - 4) \) years.
Age of the second friend was \( (20 - x - 4) \) years, which simplifies to \( (16 - x) \) years.
According to the problem, the product of their ages four years ago was 48.
So, \( (x - 4)(16 - x) = 48 \)
Expand the left side:
\( 16x - x^2 - 64 + 4x = 48 \)
Combine like terms:
\( -x^2 + 20x - 64 = 48 \)
Move 48 to the left side to set the equation to zero:
\( -x^2 + 20x - 64 - 48 = 0 \)
\( -x^2 + 20x - 112 = 0 \)
Multiply by -1 to make the \( x^2 \) coefficient positive:
\( x^2 - 20x + 112 = 0 \)
Now, let's check if this situation is possible by calculating the discriminant \( D \).
Here, \( a = 1 \), \( b = -20 \), and \( c = 112 \).
\( D = b^2 - 4ac \)
\( D = (-20)^2 - 4(1)(112) \)
\( D = 400 - 448 \)
\( D = -48 \)
Since \( D = -48 < 0 \), the discriminant is negative. This means the quadratic equation has no real roots.
Therefore, the given situation is not possible.
In simple words: Use variables to represent the friends' current ages and their ages four years ago. Set up an equation based on the product of their past ages. Solve this quadratic equation. If the discriminant is negative, the situation is not possible because real ages must exist.

Exam Tip: Always translate word problems into algebraic equations carefully. After forming the quadratic equation, calculate the discriminant first to determine if real solutions exist before attempting to find the roots, especially for "is it possible?" questions.