NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 4: Quadatric Equation

Exercise 4.1

To check if an equation is quadratic, remember: a quadratic equation must have the highest power of \(x\) as 2 (like \(ax^2+bx+c=0\), where \(a \neq 0\)).

(A) Expand the right side:

\((4-x)^2 + 3 = (x^2 - 8x + 16) + 3 = x^2 - 8x + 19\).

The left side is \(x^2 + 2x + 1\). Subtract the right side from the left:

\( (x^2 + 2x + 1) - (x^2 - 8x + 19) = 10x - 18 = 0 \).

This is a linear equation (highest power of \(x\) is 1).

(B) Expand the right side:

\((5 - x)\left(2x - \dfrac{2}{5}\right) = (5)(2x) + (5)\left(-\dfrac{2}{5}\right) + (-x)(2x) + (-x)\left(-\dfrac{2}{5}\right)\).

= \(10x - 2 - 2x^2 + \dfrac{2}{5}x\).

= \(-2x^2 + \dfrac{52}{5}x - 2\).

The left side is \(-2x^2\). If we move terms, the \(-2x^2\) cancels and we are left with a linear equation.

(C) Substitute \(k = -1\):

Equation becomes \((k+1)x^2 + \dfrac{3}{2}x = 7 \Rightarrow (0)x^2 + \dfrac{3}{2}x = 7\).

So it reduces to \( \dfrac{3}{2}x = 7 \), which is a linear equation (no \(x^2\) term left).

(D) Expand the right side:

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

The left side is \(x^3 - x^2\). Subtract RHS from LHS:

\((x^3 - x^2) - (x^3 - 3x^2 + 3x - 1) = 2x^2 - 3x + 1 = 0\).

This is a quadratic equation because the highest power of \(x\) is 2.

Therefore, the correct answer is option (D).

Step 1: Recall what a quadratic equation is.

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where:

• \(a, b, c\) are numbers,
• \(a \neq 0\),
• The highest power of \(x\) is 2.

Step 2: Check each option.

(A) \(2(x-1)^2 = 4x^2 - 2x + 1\)

Expand left side: \(2(x^2 - 2x + 1) = 2x^2 - 4x + 2\).

Equation: \(2x^2 - 4x + 2 = 4x^2 - 2x + 1\).

Bring all terms to one side: \(-2x^2 - 2x + 1 = 0\) or \(2x^2 + 2x - 1 = 0\).

This is quadratic (highest power is 2).

(B) \(2x - x^2 = x^2 + 5\)

Rearrange: \(-x^2 + 2x = x^2 + 5\).

Bring terms together: \(-2x^2 + 2x - 5 = 0\) or \(2x^2 - 2x + 5 = 0\).

This is also quadratic (highest power is 2).

(C) \((\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x\)

Expand: \(2x^2 + 2\sqrt{6}x + 3 + x^2 = 3x^2 - 5x\).

So LHS = \(3x^2 + 2\sqrt{6}x + 3\).

Equation: \(3x^2 + 2\sqrt{6}x + 3 = 3x^2 - 5x\).

Cancel \(3x^2\) on both sides → \(2\sqrt{6}x + 3 = -5x\).

Combine terms: \((2\sqrt{6} + 5)x + 3 = 0\).

This is linear (highest power is 1).

(D) \((x^2 + 2x)^2 = x^4 + 3 + 4x^3\)

Expand LHS: \(x^4 + 4x^3 + 4x^2\).

Equation: \(x^4 + 4x^3 + 4x^2 = x^4 + 4x^3 + 3\).

Cancel \(x^4\) and \(4x^3\) on both sides → \(4x^2 = 3\).

This is quadratic (highest power is 2).

Final Answer: Option (C) is not quadratic because after simplification, it becomes a linear equation.

We are asked to check which equation has \(x = 2\) as a root. A number is a root of an equation if, after putting that number in place of \(x\), the left side becomes \(0\).

Option A: \(x^2 - 4x + 5\) Put \(x = 2\): \(2^2 - 4(2) + 5 = 4 - 8 + 5 = 1\). Since \(1 \neq 0\), this is not a root.

Option B: \(x^2 + 3x - 12\) Put \(x = 2\): \(2^2 + 3(2) - 12 = 4 + 6 - 12 = -2\). Since \(-2 \neq 0\), this is not a root.

Option C: \(2x^2 - 7x + 6\) Put \(x = 2\): \(2(2^2) - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0\). Since the value is exactly \(0\), \(x = 2\) is a root of this equation.

Option D: \(3x^2 - 6x - 2\) Put \(x = 2\): \(3(2^2) - 6(2) - 2 = 3(4) - 12 - 2 = 12 - 12 - 2 = -2\). Since \(-2 \neq 0\), this is not a root.

So, only Option C works. Therefore, the correct answer is C.

We are told that \(\tfrac{1}{2}\) is a root of the quadratic equation:

\(x^2 + kx - \tfrac{5}{4} = 0\)

Step 1: If a number is a root of the equation, it means that if we put that number in place of \(x\), the whole equation will become zero.

Step 2: Put \(x = \tfrac{1}{2}\) in the equation:

\(\left(\tfrac{1}{2}\right)^2 + k\left(\tfrac{1}{2}\right) - \tfrac{5}{4} = 0\)

Step 3: Simplify each term:

• \(\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\)
• \(k \times \tfrac{1}{2} = \tfrac{k}{2}\)
• Constant term = \(-\tfrac{5}{4}\)

So the equation becomes:

\(\tfrac{1}{4} + \tfrac{k}{2} - \tfrac{5}{4} = 0\)

Step 4: Combine the fractions \(\tfrac{1}{4} - \tfrac{5}{4}\):

\(\tfrac{1 - 5}{4} = -\tfrac{4}{4} = -1\)

Now the equation looks like:

\(\tfrac{k}{2} - 1 = 0\)

Step 5: Add 1 to both sides:

\(\tfrac{k}{2} = 1\)

Step 6: Multiply both sides by 2 to find \(k\):

\(k = 2\)

Final Answer: \(k = 2\)

Step 1: The general form of a quadratic equation is \(ax^2 + bx + c = 0\).

Step 2: The formula for the sum of the roots is:

\[ \text{Sum of roots} = -\dfrac{b}{a} \]

Here, \(a\) is the coefficient of \(x^2\), and \(b\) is the coefficient of \(x\).

Step 3: Check each option one by one.

(A) \(2x^2 - 3x + 6 = 0\): Here, \(a = 2\), \(b = -3\). So, \(-b/a = -(-3)/2 = 3/2\). Not equal to 3.

(B) \(-x^2 + 3x - 3 = 0\): Here, \(a = -1\), \(b = 3\). So, \(-b/a = -(3)/(-1) = 3\). Yes, this matches the required sum.

(C) \(\sqrt{2}\,x^2 - \dfrac{3}{\sqrt{2}}x + 1 = 0\): Here, \(a = \sqrt{2}\), \(b = -\dfrac{3}{\sqrt{2}}\). So, \(-b/a = -\left(-\dfrac{3}{\sqrt{2}}\right) / \sqrt{2} = \dfrac{3}{2}\). Not equal to 3.

(D) \(3x^2 - 3x + 3 = 0\): Here, \(a = 3\), \(b = -3\). So, \(-b/a = -(-3)/3 = 1\). Not equal to 3.

Step 4: Therefore, the correct option is B.

We are given the quadratic equation: \(2x^2 - kx + k = 0\).

For a quadratic equation \(ax^2 + bx + c = 0\), the condition for equal roots is:

\(D = b^2 - 4ac = 0\), where \(D\) is the discriminant.

Step 1: Identify \(a\), \(b\), and \(c\) from the equation:

• \(a = 2\) (coefficient of \(x^2\))
• \(b = -k\) (coefficient of \(x\))
• \(c = k\) (constant term)

Step 2: Write discriminant formula:

\(D = b^2 - 4ac\)

Substitute values:

\(D = (-k)^2 - 4(2)(k)\)

\(D = k^2 - 8k\)

Step 3: Set discriminant equal to zero (for equal roots):

\(k^2 - 8k = 0\)

Step 4: Factorize:

\(k(k - 8) = 0\)

Step 5: Solve each factor:

• If \(k = 0\)
• If \(k - 8 = 0\Rightarrow k = 8\)

Final Answer: The values of \(k\) are 0 and 8.

Step 1: Look at the quadratic part: \(9x^2 + \dfrac{3}{4}x\).

Factor out \(9\) from these two terms:

\(9x^2 + \dfrac{3}{4}x = 9\Big(x^2 + \dfrac{1}{12}x\Big)\).

Step 2: To complete the square for \(x^2 + \dfrac{1}{12}x\), we use the rule: add \(\Big(\dfrac{coefficient\ of\ x}{2}\Big)^2\).

Here the coefficient of \(x\) is \(\dfrac{1}{12}\).

Step 3: Half of \(\dfrac{1}{12}\) is \(\dfrac{1}{24}\).

Now square it: \(\Big(\dfrac{1}{24}\Big)^2 = \dfrac{1}{576}\).

Step 4: We added \(\dfrac{1}{576}\) inside the bracket, but since the whole bracket is multiplied by 9, the real addition to the expression is:

\(9 \times \dfrac{1}{576} = \dfrac{9}{576} = \dfrac{1}{64}\).

Step 5: Therefore, the constant we add and subtract is \(\dfrac{1}{64}\).

Answer: Option B (\(\dfrac{1}{64}\))

We are solving the quadratic equation \(2x^2 - \sqrt{5}x + 1 = 0\).

To check how many real roots it has, we use the discriminant formula:

\(D = b^2 - 4ac\), where \(a=2\), \(b=-\sqrt{5}\), and \(c=1\).

Now, calculate step by step:

\(b^2 = (-\sqrt{5})^2 = 5\).

\(4ac = 4 \times 2 \times 1 = 8\).

So, \(D = 5 - 8 = -3\).

Because \(D < 0\), the quadratic has no real roots. Instead, it will have complex roots.

To check if a quadratic equation has two distinct real roots, we use the discriminant.

The discriminant is given by: \(D = b^2 - 4ac\).

• If \(D > 0\), the equation has two distinct real roots.
• If \(D = 0\), the equation has two equal real roots.
• If \(D < 0\), the equation has no real roots.

Now check each option:

(A) \(2x^2 - 3\sqrt{2}x + \tfrac{9}{4} = 0\)

Here, \(a = 2, b = -3\sqrt{2}, c = \tfrac{9}{4}\).

So, \(D = (-3\sqrt{2})^2 - 4(2)(\tfrac{9}{4}) = 18 - 18 = 0\).

Since \(D = 0\), this equation has equal roots, not distinct.

(B) \(x^2 + x - 5 = 0\)

Here, \(a = 1, b = 1, c = -5\).

So, \(D = (1)^2 - 4(1)(-5) = 1 + 20 = 21\).

Since \(D > 0\), this equation has two distinct real roots.

(C) \(x^2 + 3x + 2\sqrt{2} = 0\)

Here, \(a = 1, b = 3, c = 2\sqrt{2}\).

So, \(D = (3)^2 - 4(1)(2\sqrt{2}) = 9 - 8\sqrt{2}\).

Since \(8\sqrt{2} \approx 11.3\), we get \(D \approx -2.3 < 0\).

This means no real roots.

(D) \(5x^2 - 3x + 1 = 0\)

Here, \(a = 5, b = -3, c = 1\).

So, \(D = (-3)^2 - 4(5)(1) = 9 - 20 = -11\).

Since \(D < 0\), there are no real roots.

Conclusion: Only option (B) has \(D > 0\). So, the correct answer is Option B.

To check whether a quadratic equation has real roots or not, we use the discriminant formula:

\(D = b^2 - 4ac\).

- If \(D > 0\), there are two different real roots.

- If \(D = 0\), there are two equal real roots.

- If \(D < 0\), there are no real roots.

Option (A): \(x^2 - 4x + 3\sqrt{2} = 0\)

Here, \(a = 1, b = -4, c = 3\sqrt{2}\).

So, \(D = (-4)^2 - 4(1)(3\sqrt{2}) = 16 - 12\sqrt{2}\).

Since \(12\sqrt{2} \approx 16.97\), we get \(16 - 16.97 < 0\).

That means \(D < 0\). So no real roots.

Option (B): \(x^2 + 4x - 3\sqrt{2} = 0\)

Here, \(a = 1, b = 4, c = -3\sqrt{2}\).

So, \(D = (4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2}\).

This is a positive number, so there are real roots.

Option (C): \(x^2 - 4x - 3\sqrt{2} = 0\)

Here, \(a = 1, b = -4, c = -3\sqrt{2}\).

So, \(D = (-4)^2 - 4(1)(-3\sqrt{2}) = 16 + 12\sqrt{2}\).

This is also positive, so there are real roots.

Option (D): \(3x^2 + 4\sqrt{3}x + 4 = 0\)

Here, \(a = 3, b = 4\sqrt{3}, c = 4\).

So, \(D = (4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0\).

This means the roots are real and equal.

✅ Therefore, the equation in Option (A) has no real roots.

Step 1: Start with the given equation:

\((x^2 + 1)^2 - x^2 = 0\)

Step 2: To make it easier, put \(y = x^2\). Then the equation becomes:

\((y + 1)^2 - y = 0\)

Step 3: Expand \((y + 1)^2\):

\(y^2 + 2y + 1\)

Step 4: Substitute back:

\(y^2 + 2y + 1 - y = 0\)

Step 5: Simplify:

\(y^2 + y + 1 = 0\)

Step 6: This is a quadratic equation. The discriminant \(D = b^2 - 4ac\). Here, \(a = 1, b = 1, c = 1\).

So, \(D = 1^2 - 4(1)(1) = 1 - 4 = -3\).

Step 7: Since the discriminant is negative (\(D < 0\)), this quadratic has no real solutions for \(y\).

Step 8: But \(y = x^2\). If \(y\) has no real value, then \(x^2\) also has no real value. That means \(x\) has no real solution.

Final Answer: The equation has no real roots.