NCERT Solutions
Class 10 - Mathematics - Chapter 4: QUADRATIC EQUATIONS - Exercise 4.1
Question 1

Question. 1

Check whether the following are quadratic equations:

(i) \((x + 1)^2 = 2(x - 3)\)

(ii) \(x^2 - 2x = (-2)(3 - x)\)

(iii) \((x - 2)(x + 1) = (x - 1)(x + 3)\)

(iv) \((x - 3)(2x + 1) = x(x + 5)\)

(v) \((2x - 1)(x - 3) = (x + 5)(x - 1)\)

(vi) \(x^2 + 3x + 1 = (x - 2)^2\)

(vii) \((x + 2)^3 = 2x(x^2 - 1)\)

(viii) \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

Answer:

(i) Yes

(ii) Yes

(iii) No

(iv) Yes

(v) Yes

(vi) No

(vii) No

(viii) Yes

Video Explanation:

Detailed Answer with Explanation:

Quick reminder: A quadratic equation in \(x\) is any equation that can be written in the form \(ax^2 + bx + c = 0\) with \(a \neq 0\). That means, after expanding and bringing all terms to one side, the highest power of \(x\) must be 2.


(i) Check \((x + 1)^2 = 2(x - 3)\)

Step 1: Expand the left-hand side: \((x + 1)^2 = x^2 + 2x + 1\).

Step 2: Expand the right-hand side: \(2(x - 3) = 2x - 6\).

Step 3: Bring all terms to the left: \(x^2 + 2x + 1 - (2x - 6) = 0\), i.e. \(x^2 + 2x + 1 - 2x + 6 = 0\).

Step 4: Simplify: \(x^2 + 7 = 0\). Highest power is 2, so this is a quadratic.

(ii) Check \(x^2 - 2x = (-2)(3 - x)\)

Step 1: Expand the right-hand side: \((-2)(3 - x) = -6 + 2x\).

Step 2: Bring everything to the left: \(x^2 - 2x - (-6 + 2x) = 0\), i.e. \(x^2 - 2x + 6 - 2x = 0\).

Step 3: Combine like terms: \(x^2 - 4x + 6 = 0\). Highest power is 2, so this is a quadratic.

(iii) Check \((x - 2)(x + 1) = (x - 1)(x + 3)\)

Step 1: Expand LHS: \((x - 2)(x + 1) = x^2 - x - 2\).

Step 2: Expand RHS: \((x - 1)(x + 3) = x^2 + 2x - 3\).

Step 3: Move RHS to LHS: \(x^2 - x - 2 - (x^2 + 2x - 3) = 0\).

Step 4: Simplify: \(x^2 - x - 2 - x^2 - 2x + 3 = -3x + 1 = 0\). This is \(-3x + 1 = 0\), degree 1, so it is not quadratic.

(iv) Check \((x - 3)(2x + 1) = x(x + 5)\)

Step 1: Expand LHS: \((x - 3)(2x + 1) = 2x^2 - 5x - 3\).

Step 2: Expand RHS: \(x(x + 5) = x^2 + 5x\).

Step 3: Bring all terms to LHS: \(2x^2 - 5x - 3 - (x^2 + 5x) = 0\).

Step 4: Simplify: \(2x^2 - 5x - 3 - x^2 - 5x = x^2 - 10x - 3 = 0\). Highest power is 2, so this is a quadratic.

(v) Check \((2x - 1)(x - 3) = (x + 5)(x - 1)\)

Step 1: Expand LHS: \((2x - 1)(x - 3) = 2x^2 - 7x + 3\).

Step 2: Expand RHS: \((x + 5)(x - 1) = x^2 + 4x - 5\).

Step 3: Bring RHS to LHS: \(2x^2 - 7x + 3 - (x^2 + 4x - 5) = 0\).

Step 4: Simplify: \(2x^2 - 7x + 3 - x^2 - 4x + 5 = x^2 - 11x + 8 = 0\). Highest power is 2, so this is a quadratic.

(vi) Check \(x^2 + 3x + 1 = (x - 2)^2\)

Step 1: Expand RHS: \((x - 2)^2 = x^2 - 4x + 4\).

Step 2: Move RHS to LHS: \(x^2 + 3x + 1 - (x^2 - 4x + 4) = 0\).

Step 3: Simplify: \(x^2 + 3x + 1 - x^2 + 4x - 4 = 7x - 3 = 0\). Degree is 1, so this is not quadratic.


(vii) Check \((x + 2)^3 = 2x(x^2 - 1)\)

Step 1: Expand LHS: \((x + 2)^3 = x^3 + 6x^2 + 12x + 8\).

Step 2: Expand RHS: \(2x(x^2 - 1) = 2x^3 - 2x\).

Step 3: Bring RHS to LHS: \(x^3 + 6x^2 + 12x + 8 - (2x^3 - 2x) = 0\).

Step 4: Simplify: \(x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = -x^3 + 6x^2 + 14x + 8 = 0\). Degree is 3, so this is not quadratic.


(viii) Check \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

Step 1: Expand RHS: \((x - 2)^3 = x^3 - 6x^2 + 12x - 8\).

Step 2: Move RHS to LHS: \(x^3 - 4x^2 - x + 1 - (x^3 - 6x^2 + 12x - 8) = 0\).

Step 3: Simplify carefully: \(x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 2x^2 - 13x + 9 = 0\).

Step 4: Highest power of \(x\) is 2, so this is a quadratic.

Bottom line: For each equation: (1) expand both sides, (2) move all terms to one side, (3) simplify, and (4) look at the highest power of \(x\). If it is 2 and the \(x^2\) coefficient is not zero, the equation is quadratic.

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