Class X Mathematics worksheet on quadratic equations with ten problems for practice.

Let’s solve each problem one by one, step by step.

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Problem 1:
Find the roots of the quadratic equation:
x² + 5x – (a+1)(a+6) = 0, where a is a constant.

We can use factoring or the quadratic formula. Let’s try factoring.

We need two numbers that multiply to –(a+1)(a+6) and add up to 5.

Notice:
(a+6) – (a+1) = 5 → perfect!

So we can write:

x² + (a+6)x – (a+1)x – (a+1)(a+6) = 0
→ x(x + a + 6) – (a+1)(x + a + 6) = 0
→ [x – (a+1)](x + a + 6) = 0

So roots are:
x = a+1 and x = –(a+6)

✔ Final Answer for Q1: x = a + 1, x = –a – 6

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Problem 2:
If mx² + 2x + m = 0 has 2 equal roots, find value(s) of m.

For equal roots, discriminant D = 0.

D = b² – 4ac = (2)² – 4(m)(m) = 4 – 4m²

Set D = 0:
4 – 4m² = 0 → 4m² = 4 → m² = 1 → m = ±1

But wait — if m = 0, it’s not quadratic anymore. But here m = ±1, both non-zero → valid.

✔ Final Answer for Q2: m = 1 or m = –1

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Problem 3:
If 1 is a root of ay² + ay + 3 = 0 and y² + y + b = 0, then find ab.

Plug y=1 into first equation:
a(1)² + a(1) + 3 = 0 → a + a + 3 = 0 → 2a = –3 → a = –3/2

Plug y=1 into second equation:
(1)² + 1 + b = 0 → 1 + 1 + b = 0 → b = –2

Then ab = (–3/2)(–2) = 3

✔ Final Answer for Q3: ab = 3

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Problem 4:
Check nature of roots of x² – 4x + 3√2 = 0

Discriminant D = b² – 4ac = (–4)² – 4(1)(3√2) = 16 – 12√2

√2 ≈ 1.414 → 12×1.414 ≈ 16.968 → so D ≈ 16 – 16.968 = –0.968 < 0

Since D < 0 → roots are imaginary (not real)

✔ Final Answer for Q4: Roots are imaginary (no real roots)

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Problem 5:
Find roots of ax² + a = a²x + x

Bring all terms to left:

ax² + a – a²x – x = 0
→ ax² – (a² + 1)x + a = 0

Try factoring:

Look for two numbers that multiply to a·a = a² and add to –(a² + 1)? Not obvious.

Use quadratic formula:

x = [ (a² + 1) ± √{(a² + 1)² – 4·a·a} ] / (2a)

Compute discriminant:

(a² + 1)² – 4a² = a⁴ + 2a² + 1 – 4a² = a⁴ – 2a² + 1 = (a² – 1)²

So sqrt(D) = |a² – 1| → but since squared, we can write as (a² – 1) with sign handled by ±

Thus:

x = [ (a² + 1) ± (a² – 1) ] / (2a)

Case 1: + → [a² + 1 + a² – 1]/(2a) = (2a²)/(2a) = a

Case 2: – → [a² + 1 – a² + 1]/(2a) = 2/(2a) = 1/a

✔ Final Answer for Q5: x = a, x = 1/a

*(Note: Assume a ≠ 0)*

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Problem 6:
Determine whether –2√3 is a solution of x² – 3√3 x + 6 = 0

Plug x = –2√3:

Left side = (–2√3)² – 3√3(–2√3) + 6
= 4×3 + 6×3 + 6
= 12 + 18 + 6 = 36 ≠ 0

Wait — let me recalculate:

(–2√3)² = 4 × 3 = 12
–3√3 × (–2√3) = +6 × (√3 × √3) = 6 × 3 = 18
+6

Total: 12 + 18 + 6 = 36 ≠ 0 → NOT a solution

But maybe I misread? Let me check again.

Equation: x² – 3√3 x + 6 = 0

At x = –2√3:

Term 1: x² = 12
Term 2: –3√3 * x = –3√3 * (–2√3) = +6 * 3 = 18
Term 3: +6

Sum: 12 + 18 + 6 = 36 → definitely not zero.

Wait — perhaps the question meant +2√3? Or maybe I made mistake in sign?

No — the question says “–2√3”, and calculation shows 36 ≠ 0.

Alternatively, maybe they want us to test if it satisfies — answer is no.

But let me double-check arithmetic:

√3 × √3 = 3 → correct
–3√3 × –2√3 = (+6) × 3 = 18 → correct
x² = 4×3=12 → correct
+6 → total 36 → yes.

So –2√3 is NOT a solution.

✔ Final Answer for Q6: No, –2√3 is not a solution

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Problem 7:
Polygon of n sides has n(n–3)/2 diagonals. How many sides if diagonals = 54?

Set: n(n–3)/2 = 54
Multiply both sides by 2: n(n–3) = 108
→ n² – 3n – 108 = 0

Factor: Find two numbers multiplying to –108, adding to –3 → –12 and 9

So: (n – 12)(n + 9) = 0 → n = 12 or n = –9

Reject negative → n = 12

✔ Final Answer for Q7: 12 sides

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Problem 8:
Rewrite as quadratic in x:
4/x – 3 = 5/(2x+3), x ≠ 0, –3/2

Multiply both sides by x(2x+3) to eliminate denominators:

Left: [4/x – 3] * x(2x+3) = 4(2x+3) – 3x(2x+3)
Right: [5/(2x+3)] * x(2x+3) = 5x

So:

4(2x+3) – 3x(2x+3) = 5x
Expand:

8x + 12 – [6x² + 9x] = 5x
→ 8x + 12 – 6x² – 9x = 5x
→ –6x² – x + 12 = 5x
Bring all to left:

–6x² – x + 12 – 5x = 0
→ –6x² – 6x + 12 = 0
Multiply by –1: 6x² + 6x – 12 = 0
Divide by 6: x² + x – 2 = 0

✔ Final Answer for Q8: x² + x – 2 = 0

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Problem 9:
Check if (x+3)² = (x+2)² + 1 is quadratic.

Expand both sides:

Left: x² + 6x + 9
Right: x² + 4x + 4 + 1 = x² + 4x + 5

Set equal:
x² + 6x + 9 = x² + 4x + 5
Subtract x² from both sides:
6x + 9 = 4x + 5
→ 2x + 4 = 0 → linear equation!

So after simplifying, highest degree is 1 → NOT quadratic.

✔ Final Answer for Q9: No, it reduces to a linear equation

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Problem 10:
Determine whether √7 y² – 6y – 13√7 = 0 has roots.

This is a quadratic in y: Ay² + By + C = 0, where A = √7, B = –6, C = –13√7

Discriminant D = B² – 4AC = (–6)² – 4(√7)(–13√7)
= 36 + 4×13×(√7)²
= 36 + 52×7
= 36 + 364 = 400 > 0

Since D > 0 → two distinct real roots exist.

✔ Final Answer for Q10: Yes, it has two distinct real roots

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Now compiling all final answers:

Final Answer:
1. x = a + 1, x = –a – 6
2. m = 1 or m = –1
3. ab = 3
4. Roots are imaginary
5. x = a, x = 1/a
6. No
7. 12
8. x² + x – 2 = 0
9. No
10. Yes