Quadratic Equations MCQ Test — Class 10 CBSE

Solution:

Option A: (x + 1)² = 2x + 3. Expanding the left side gives x² + 2x + 1 = 2x + 3. Simplifying, we get x² - 2 = 0, which is a quadratic equation (a=1, b=0, c=-2).

Option B: x(x + 2) = x² + 5. Expanding gives x² + 2x = x² + 5. Simplifying, we get 2x - 5 = 0, which is a linear equation.

Option C: x³ - 4x² + 5 = 0. This is a cubic equation, not quadratic, as the highest power of x is 3.

Option D: (x - 2)(x + 2) = x² - 4. Expanding gives x² - 4 = x² - 4. Simplifying, we get 0 = 0, which is an identity and not an equation in x.

Solution:

By definition, a 'root' or 'solution' of an equation is a value that, when substituted for the variable (x in this case), makes the equation true.

Therefore, if 'p' is a root of ax² + bx + c = 0, then substituting x = p into the equation must result in ap² + bp + c = 0.

The other options are either specific conditions (like sum of coefficients, formula for equal roots, or condition for distinct real roots) and not generally true for any root 'p'.

Solution:

For the equation x² - 7x + 12 = 0, we need to find two numbers whose product is 12 (coefficient of x² × constant term) and whose sum is -7 (coefficient of x).

The numbers -3 and -4 satisfy these conditions: (-3) × (-4) = 12 and (-3) + (-4) = -7.

Ravi's split, x² - 3x - 4x + 12 = 0, is therefore correct. The next logical step in factorization would be to group terms: x(x-3) - 4(x-3) = 0.

Solution:

Let the first positive integer be x. Since the integers are consecutive, the next positive integer will be x + 1.

The problem states that their product is 306. So, we can write the equation: x(x + 1) = 306.

Expand the left side: x² + x = 306.

To form a standard quadratic equation (ax² + bx + c = 0), bring all terms to one side: x² + x - 306 = 0.

Solution:

For an expression of the form x² + bx, to complete the square, we need to add (b/2)².

In the given expression x² - 10x, the coefficient of x is b = -10.

So, we need to add (-10/2)² = (-5)² = 25.

Thus, x² - 10x + 25 can be written as (x - 5)².

Solution:

The quadratic formula is a universal method for finding the roots of any quadratic equation of the form ax² + bx + c = 0. It works for all cases, including those that are difficult or impossible to factorize.

Option A is incorrect because while it's often preferred when factorization is difficult, it's not restricted to only those cases.

Option B is incorrect; real roots exist when b² - 4ac ≥ 0. If b² - 4ac < 0, there are no real roots.

Option D is incorrect; b² - 4ac (the discriminant) can be positive, zero, or negative, determining the nature of the roots.

Solution:

The nature of the roots of a quadratic equation ax² + bx + c = 0 is determined by its discriminant D = b² - 4ac.

If D > 0, the equation has two distinct real roots.

If D = 0, the equation has two equal real roots (also called coincident roots).

If D < 0, the equation has no real roots (or imaginary roots).

Based on these rules, only the statement 'If D = 0, the equation has two equal real roots' is true.

Solution:

A quadratic equation ax² + bx + c = 0 has equal roots if its discriminant (D = b² - 4ac) is equal to 0.

In the given equation 9x² + 3kx + 4 = 0, we have a = 9, b = 3k, and c = 4.

Set the discriminant to zero: (3k)² - 4(9)(4) = 0.

Simplify and solve for k: 9k² - 144 = 0 => 9k² = 144 => k² = 144 / 9 => k² = 16 => k = ±√16 => k = ±4.

Solution:

Factorization is generally suitable for equations with simple integer or rational roots. When roots are irrational or complex, or when the coefficients make factorization non-obvious, the quadratic formula is the most suitable and reliable method.

Option A: Simple integer roots are ideal for factorization.

Option B: Large coefficients can make both methods cumbersome, but factorization might still be hard to spot. The quadratic formula is always applicable.

Option C: If the discriminant is a perfect square, the roots will be rational, meaning factorization is usually straightforward.

Solution:

We need to find the equation where the discriminant D = b² - 4ac is less than 0 (D < 0).

Option A: x² - 4x + 4 = 0. Here a=1, b=-4, c=4. D = (-4)² - 4(1)(4) = 16 - 16 = 0. (Equal real roots)

Option B: 2x² + 3x + 1 = 0. Here a=2, b=3, c=1. D = (3)² - 4(2)(1) = 9 - 8 = 1. (Distinct real roots)

Option C: x² + x + 1 = 0. Here a=1, b=1, c=1. D = (1)² - 4(1)(1) = 1 - 4 = -3. Since D < 0, this equation has no real roots.

Option D: x² - 5x + 6 = 0. Here a=1, b=-5, c=6. D = (-5)² - 4(1)(6) = 25 - 24 = 1. (Distinct real roots)