Exercise 4.3 answers: Quadratic Equations

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(i) 2x² − 3x + 5 = 0

1. Compare with ax² + bx + c = 0: a = 2, b = −3, c = 5.
2. Discriminant = b² − 4ac = (−3)² − 4(2)(5) = 9 − 40 = −31.
3. Since the discriminant is negative, there are no real roots.

AnswerNo real roots exist.

(ii) 3x² − 4√3 x + 4 = 0

1. Here a = 3, b = −4√3, c = 4.
2. Discriminant = (−4√3)² − 4(3)(4) = 48 − 48 = 0.
3. A zero discriminant means the roots are real and equal.
4. The equal root is x = −b/2a = 4√3/6 = 2√3/3.

AnswerReal and equal roots; x = 2√3/3 (twice).

(iii) 2x² − 6x + 3 = 0

1. Here a = 2, b = −6, c = 3.
2. Discriminant = (−6)² − 4(2)(3) = 36 − 24 = 12.
3. A positive discriminant means the roots are real and distinct.
4. By the quadratic formula: x = [6 ± √12]/4 = [6 ± 2√3]/4 = (3 ± √3)/2.

AnswerReal and distinct roots; x = (3 + √3)/2 or (3 − √3)/2.

Watch this explained “Naming it: the discriminant”, 4:27 into Why the sign of b² − 4ac settles how many real roots exist

Open NCERT p. 47Matches NCERT’s answer

(i) 2x² + kx + 3 = 0

1. Here a = 2, b = k, c = 3.
2. For equal roots the discriminant must be 0: k² − 4(2)(3) = 0, i.e. k² = 24.
3. So k = ±√24 = ±2√6.

Answerk = 2√6 or k = −2√6.

(ii) kx (x – 2) + 6 = 0

1. Expand: kx² − 2kx + 6 = 0. This is quadratic only if k ≠ 0.
2. Here a = k, b = −2k, c = 6. Discriminant = (−2k)² − 4(k)(6) = 4k² − 24k.
3. For equal roots: 4k² − 24k = 0, i.e. 4k(k − 6) = 0, so k = 0 or k = 6.
4. Reject k = 0, since then the x² term disappears and the equation stops being quadratic.

Answerk = 6.

Watch this explained “Choosing the coefficient”, 9:41 into Why the sign of b² − 4ac settles how many real roots exist

Open NCERT p. 47Checked by computer

1. Let the breadth be x metres; the length is twice the breadth, so it is 2x metres.
2. Area = 2x × x = 2x², which must equal 800: 2x² − 800 = 0, i.e. x² − 400 = 0.
3. The discriminant of x² − 400 = 0 is 0² − 4(1)(−400) = 1600, which is positive, so a real breadth exists — the design is possible.
4. Solve: x² = 400, so x = 20 or x = −20. Reject the negative value.

AnswerYes, it is possible. Breadth = 20 m, length = 40 m.

Watch this explained “Three situations, three signs”, 11:03 into Why the sign of b² − 4ac settles how many real roots exist

Open NCERT p. 47Checked by computer

1. Let one friend's present age be x years; since the ages add to 20, the other's present age is (20 − x) years.
2. Four years ago their ages were (x − 4) and (16 − x). Their product then was 48: (x − 4)(16 − x) = 48.
3. Expand: −x² + 20x − 64 = 48, i.e. x² − 20x + 112 = 0.
4. Discriminant = (−20)² − 4(1)(112) = 400 − 448 = −48, which is negative.
5. A negative discriminant means no real value of x satisfies the equation.

AnswerNo, this situation is not possible — there are no real ages that fit both conditions.

Watch this explained “Three situations, three signs”, 11:03 into Why the sign of b² − 4ac settles how many real roots exist

Open NCERT p. 47Matches NCERT’s answer

1. Let the length be l metres and breadth b metres. Perimeter = 2(l + b) = 80, so l + b = 40, i.e. l = 40 − b.
2. Area = l × b = (40 − b)b = 400, so b² − 40b + 400 = 0.
3. Discriminant = (−40)² − 4(1)(400) = 1600 − 1600 = 0, so a real (repeated) breadth exists — the design is possible, and in only one way.
4. Solve: b² − 40b + 400 = (b − 20)² = 0, so b = 20. Then l = 40 − 20 = 20.

AnswerYes, it is possible. Length = 20 m and breadth = 20 m (the park is a square).

Watch this explained “Three situations, three signs”, 11:03 into Why the sign of b² − 4ac settles how many real roots exist