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87 Quadratic Equation. 100 Important Quadratic Equations Practice Questions for SSC Class 10
Part 1: Top 10 Solved Quadratic Equations
Part 2: Practice Questions (Q11 - Q100)
Solve the following Quadratic Equations (Factorization/Formula):
- \(x^2 - 4x - 5 = 0\)
- \(x^2 + 8x + 15 = 0\)
- \(x^2 - 7x + 12 = 0\)
- \(2x^2 - 5x + 2 = 0\)
- \(3x^2 - x - 10 = 0\)
- \(x^2 - 11x + 24 = 0\)
- \(x^2 + 2x - 48 = 0\)
- \(5m^2 = 22m + 15\)
- \(2x^2 - 2x + \frac{1}{2} = 0\)
- \(6x - \frac{2}{x} = 1\)
- \(x^2 - 25 = 0\)
- \(3y^2 = 15y\)
- \(x^2 + 4x + 1 = 0\)
- \(m^2 - 5m - 3 = 0\)
- \(x^2 + 5x + 5 = 0\)
- \(y^2 + \frac{1}{3}y = 2\)
- \(5x^2 + 13x + 8 = 0\)
- \(x^2 + 10x + 24 = 0\)
- \(x^2 - x - 72 = 0\)
- \(x^2 - 16x + 63 = 0\)
- \(2x^2 + 9x + 10 = 0\)
- \(x^2 - 2x - 3 = 0\)
- \(x^2 + 6x + 9 = 0\)
- \(x^2 - 10x + 25 = 0\)
- \(x^2 + 14x + 49 = 0\)
- \(4x^2 - 4x + 1 = 0\)
- \(x^2 - 1 = 0\)
- \(2x^2 - 7x + 6 = 0\)
- \(3x^2 + 8x + 5 = 0\)
- \(x^2 - 12x + 32 = 0\)
Find the Discriminant and State Nature of Roots:
- \(x^2 + x + 1 = 0\)
- \(2x^2 - 5x - 3 = 0\)
- \(x^2 - 6x + 9 = 0\)
- \(3x^2 + 2x - 1 = 0\)
- \(x^2 + 4x + 5 = 0\)
- \(2x^2 - 7x + 3 = 0\)
- \(x^2 - 2x + 1 = 0\)
- \(x^2 + 5x + 6 = 0\)
- \(4x^2 - 12x + 9 = 0\)
- \(x^2 - 8x + 15 = 0\)
- \(2x^2 + 5x + 5 = 0\)
- \(x^2 - x - 1 = 0\)
- \(x^2 + 10x + 25 = 0\)
- \(3x^2 + 7x + 2 = 0\)
- \(x^2 + 2x + 3 = 0\)
- \(5x^2 - 4x + 1 = 0\)
- \(x^2 - 4x + 3 = 0\)
- \(2x^2 - 6x + 3 = 0\)
- \(x^2 + 12x + 36 = 0\)
- \(x^2 - 5x + 7 = 0\)
Form Quadratic Equations from Roots:
- Roots: 2, 5
- Roots: -3, -4
- Roots: 0, 4
- Roots: 1/2, 1/2
- Roots: \(\sqrt{2}, -\sqrt{2}\)
- Roots: 7, -7
- Roots: 6, 1
- Roots: -5, 2
- Roots: 8, 3
- Roots: -1, -1
- Roots: 10, -2
- Roots: 0, 0
- Roots: 5, 5
- Roots: -6, 6
- Roots: 4, -3
- Roots: 1, 9
- Roots: -2, -8
- Roots: 3, -3
- Roots: 1/3, 3
- Roots: -10, -10
Advanced and Word-Based Conditions:
- Find \(k\) if roots of \(x^2 + kx + 12 = 0\) are real and equal.
- Find \(k\) if one root of \(x^2 - kx + 18 = 0\) is 6.
- Sum of roots is 10 and product is 21. Find equation.
- One root is \(2 + \sqrt{3}\), find the other root.
- Solve \(x^4 - 5x^2 + 4 = 0\) (Reducible to quadratic).
- Solve \((x-3)(x+4) = 0\).
- Solve \(x^2 = 49\).
- Solve \(5x^2 = 20\).
- Find \(k\) if \(\Delta = 0\) for \(kx(x-2) + 6 = 0\).
- Solve \(x + 1/x = 2.5\).
- Product of two consecutive natural numbers is 20.
- Find \(x\) if \(x^2 - 9x + 20 = 0\).
- Solve \(y^2 + 10y + 21 = 0\).
- Solve \(x^2 - 11x + 30 = 0\).
- Solve \(x^2 - 2x - 8 = 0\).
- Roots are 1 and -1. Form equation.
- Roots are 4 and 0. Form equation.
- Find \(\Delta\) for \(x^2 + 5x + 5 = 0\).
- Solve \(x^2 - 3x = 0\).
- Solve \(2x^2 = 8\).
Answer Key (Q11 - Q100)
10 Important Quadratic Equations Questions and Solutions for Class 10
10 Quadratic Equations Questions with Solution
Solutions
Solve the quadratic equation $x^2 - 5x + 6 = 0$ by factorization method.
Given equation: $x^2 - 5x + 6 = 0$
To factorize, we look for two numbers whose sum is $-5$ and product is $6$. These numbers are $-2$ and $-3$.
$x^2 - 2x - 3x + 6 = 0$
$x(x - 2) - 3(x - 2) = 0$
$(x - 2)(x - 3) = 0$
Therefore, $x - 2 = 0$ or $x - 3 = 0$
$x = 2$ or $x = 3$
Solve the following equation using the Quadratic Formula: $2x^2 + 7x + 5 = 0$.
Comparing $2x^2 + 7x + 5 = 0$ with $ax^2 + bx + c = 0$, we get:
$a = 2, b = 7, c = 5$
Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(5)}}{2(2)}$
$x = \frac{-7 \pm \sqrt{49 - 40}}{4}$
$x = \frac{-7 \pm \sqrt{9}}{4} = \frac{-7 \pm 3}{4}$
Case 1: $x = \frac{-7 + 3}{4} = \frac{-4}{4} = -1$
Case 2: $x = \frac{-7 - 3}{4} = \frac{-10}{4} = -2.5$
Roots: $x = -1, -2.5$
Determine the nature of the roots of the quadratic equation: $3x^2 - 4x + 1 = 0$.
Here, $a = 3, b = -4, c = 1$
Discriminant ($D$) $= b^2 - 4ac$
$D = (-4)^2 - 4(3)(1)$
$D = 16 - 12 = 4$
Since $D > 0$ and $D$ is a perfect square, the roots are real, rational, and unequal.
Solve $x^2 + 6x + 9 = 0$.
Given: $x^2 + 6x + 9 = 0$
This is in the form of $(a + b)^2 = a^2 + 2ab + b^2$.
$(x)^2 + 2(x)(3) + (3)^2 = 0$
$(x + 3)^2 = 0$
$x + 3 = 0$
$x = -3, -3$ (Equal roots)
Find the value of $k$ if one root of the quadratic equation $kx^2 - 14x + 8 = 0$ is $2$.
Since $x = 2$ is a root, it must satisfy the equation.
$k(2)^2 - 14(2) + 8 = 0$
$4k - 28 + 8 = 0$
$4k - 20 = 0$
$4k = 20$
$k = 5$
The sum of two numbers is 15 and the sum of their reciprocals is $3/10$. Find the numbers.
Let the numbers be $x$ and $15 - x$.
According to the condition: $\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}$
$\frac{15 - x + x}{x(15 - x)} = \frac{3}{10}$
$\frac{15}{15x - x^2} = \frac{3}{10}$
$150 = 3(15x - x^2)$
$50 = 15x - x^2$ (dividing by 3)
$x^2 - 15x + 50 = 0$
$(x - 10)(x - 5) = 0$
The numbers are 10 and 5.
Solve: $x^2 - 2x - 15 = 0$
$x^2 - 5x + 3x - 15 = 0$
$x(x - 5) + 3(x - 5) = 0$
$(x - 5)(x + 3) = 0$
$x = 5$ or $x = -3$
Form a quadratic equation whose roots are $4$ and $-3$.
Sum of roots ($\alpha + \beta$) $= 4 + (-3) = 1$
Product of roots ($\alpha\beta$) $= 4 \times (-3) = -12$
Equation: $x^2 - (\text{Sum})x + (\text{Product}) = 0$
$x^2 - (1)x + (-12) = 0$
$x^2 - x - 12 = 0$
Solve $4x^2 - 20x + 25 = 0$ using factorization.
$4x^2 - 10x - 10x + 25 = 0$
$2x(2x - 5) - 5(2x - 5) = 0$
$(2x - 5)(2x - 5) = 0$
$2x - 5 = 0 \Rightarrow 2x = 5$
$x = 5/2$
Solve: $x + \frac{1}{x} = 2.5$
Multiply the whole equation by $x$:
$x^2 + 1 = 2.5x$
$x^2 - 2.5x + 1 = 0$
Multiply by 2 to remove decimals: $2x^2 - 5x + 2 = 0$
$2x^2 - 4x - x + 2 = 0$
$2x(x - 2) - 1(x - 2) = 0$
$(x - 2)(2x - 1) = 0$
$x = 2$ or $x = 1/2$
10 Quadratic Equation Questions Important for Board Exam
Part 1: 10 Important Solved Questions
SSC Class 10 Algebra Quadratic Equations Important Solved Questions and Practice Set
Given: \(x^2 - 15x + 54 = 0\)
We need to find two numbers whose sum is \(-15\) and product is \(54\).
The numbers are \(-9\) and \(-6\).
\(x^2 - 9x - 6x + 54 = 0\)
\(x(x - 9) - 6(x - 9) = 0\)
\((x - 9)(x - 6) = 0\)
\(x - 9 = 0\) or \(x - 6 = 0\)
\(x = 9\) or \(x = 6\)
Roots: 9, 6
Comparing with \(ax^2 + bx + c = 0\), we get:
\(a = 1, b = 6, c = 5\)
Discriminant \(\Delta = b^2 - 4ac = (6)^2 - 4(1)(5) = 36 - 20 = 16\)
Using formula \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\):
\(x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}\)
\(x = \frac{-6 + 4}{2} = \frac{-2}{2} = -1\) or \(x = \frac{-6 - 4}{2} = \frac{-10}{2} = -5\)
Roots: -1, -5
\(a = 2, b = -5, c = 7\)
\(\Delta = b^2 - 4ac = (-5)^2 - 4(2)(7) = 25 - 56 = -31\)
Since \(\Delta < 0\), the roots are not real.
Equation: \(kx^2 - 2kx + 6 = 0\)
For real and equal roots, \(\Delta = 0\).
\(a = k, b = -2k, c = 6\)
\((-2k)^2 - 4(k)(6) = 0\)
\(4k^2 - 24k = 0\)
\(4k(k - 6) = 0\)
Since \(k \neq 0\) (as it is a quadratic equation), \(k - 6 = 0 \implies \mathbf{k = 6}\).
Let \(\alpha = 3\) and \(\beta = -10\).
Sum of roots \(\alpha + \beta = 3 + (-10) = -7\)
Product of roots \(\alpha\beta = 3 \times (-10) = -30\)
Equation: \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(x^2 - (-7)x + (-30) = 0\)
Equation: \(x^2 + 7x - 30 = 0\)
\(\alpha + \beta = -b/a = -5\), \(\alpha\beta = c/a = -1\)
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(= (-5)^3 - 3(-1)(-5)\)
\(= -125 - 15 = -140\)
Value: -140
Let present age be \(x\).
\((x - 2)(x + 3) = 84\)
\(x^2 + 3x - 2x - 6 = 84\)
\(x^2 + x - 90 = 0\)
\((x + 10)(x - 9) = 0\)
\(x = -10\) (Discarded) or \(x = 9\).
Present age: 9 years
\(x^2 + x = 20\)
Add \((1/2 \times \text{coeff of } x)^2 = (1/2)^2 = 1/4\) to both sides:
\(x^2 + x + 1/4 = 20 + 1/4\)
\((x + 1/2)^2 = 81/4\)
Taking square root: \(x + 1/2 = \pm 9/2\)
\(x = 9/2 - 1/2 = 4\) or \(x = -9/2 - 1/2 = -5\)
Roots: 4, -5
Let roots be \(\alpha, \beta\). Given \(|\alpha - \beta| = 1\).
\(\alpha + \beta = p, \alpha\beta = q\).
We know \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
\(1^2 = p^2 - 4q\)
\(1 = p^2 - 4q \implies \mathbf{p^2 = 4q + 1}\). (Hence Proved)
Let original speed be \(x\) km/hr. Time \(T_1 = 360/x\).
New speed \(x + 5\). Time \(T_2 = 360/(x + 5)\).
\(T_1 - T_2 = 1\)
\(360/x - 360/(x + 5) = 1\)
\(360(x + 5 - x) = x(x + 5)\)
\(1800 = x^2 + 5x \implies x^2 + 5x - 1800 = 0\)
\((x + 45)(x - 40) = 0\)
Speed cannot be negative, so \(x = 40\).
Speed: 40 km/hr
Part 2: 50 Practice Questions
ЁЯФС Answer Key
2: (-2, -5)
3: (2, 3)
4: (3, -3)
5: (3, 0.5)
6: (5, 5)
7: (1, 2/3)
8: (2, -4)
9: (1, -1)
10: (4, 2)
11: (-1, -4)
12: (1.5, -2)
13: (5, -2)
14: (5, 6)
15: (-3, -5)
16: (5, 8)
17: (2, -2)
18: (4, -0.4)
19: (-2, -2)
20: (4, 8)
21: (3, -0.5)
22: (5, -3)
23: (-4, -5)
24: (3, 4)
25: (3, 1/3)
26: (7, 7)
27: (4, -4)
28: (-1, -2.5)
29: (4, -3)
30: (7, 8)
31: (-3, -7)
32: (2, 6)
33: (0.5, 0.5)
34: (5, -5)
35: (2, -3)
36: (7, -2)
37: (4, 7)
38: (-3, -3)
39: (1, 0.5)
40: (10, -10)
41: (8, 9)
42: (-5, -7)
43: (1, 1)
44: (-1/3, -1)
45: (4, 5)
46: (6, -6)
47: (3, -5)
48: (6, 7)
49: (8, -8)
50: (1, -0.5)
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