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

Exercise 4.2

Key idea: For a quadratic equation \(ax^2 + bx + c = 0\), we calculate the discriminant:

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

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there are two equal real roots (not distinct).
• If \(D < 0\), there are no real roots.

(i) \(x^2 - 3x + 4 = 0\)

a = 1, b = -3, c = 4

D = (-3)^2 - 4(1)(4) = 9 - 16 = -7

Since D < 0, there are no real roots.

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

a = 2, b = 1, c = -1

D = (1)^2 - 4(2)(-1) = 1 + 8 = 9

D > 0, so there are two distinct real roots.

(iii) \(2x^2 - 6x + 9/2 = 0\)

a = 2, b = -6, c = 9/2

D = (-6)^2 - 4(2)(9/2) = 36 - 36 = 0

D = 0, so there are real roots but they are equal, not distinct.

(iv) \(3x^2 - 4x + 1 = 0\)

a = 3, b = -4, c = 1

D = (-4)^2 - 4(3)(1) = 16 - 12 = 4

D > 0, so there are two distinct real roots.

(v) \((x+4)^2 - 8x = 0\)

Expand: (x+4)^2 = x^2 + 8x + 16

Equation: x^2 + 8x + 16 - 8x = x^2 + 16 = 0

a = 1, b = 0, c = 16

D = 0^2 - 4(1)(16) = -64

D < 0, so there are no real roots.

(vi) \((x - √2)^2 - 2(x+1) = 0\)

Expand: (x - √2)^2 = x^2 - 2√2x + 2

Equation: x^2 - 2√2x + 2 - 2x - 2 = x^2 - (2√2 + 2)x

a = 1, b = -(2√2 + 2), c = 0

D = b^2 - 4ac = (-(2√2+2))^2 = (2√2+2)^2

This is positive, so there are two distinct real roots.

(vii) \(√2x^2 - (3/√2)x + (1/√2) = 0\)

Multiply whole equation by √2 to remove fractions:

2x^2 - 3x + 1 = 0

a = 2, b = -3, c = 1

D = (-3)^2 - 4(2)(1) = 9 - 8 = 1

D > 0, so two distinct real roots.

(viii) \(x(1-x) - 2 = 0\)

Expand: x - x^2 - 2 = 0

Rearrange: -x^2 + x - 2 = 0 → x^2 - x + 2 = 0

a = 1, b = -1, c = 2

D = (-1)^2 - 4(1)(2) = 1 - 8 = -7

D < 0, so no real roots.

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

Expand: (x-1)(x+2) = x^2 + x - 2

Add +2: x^2 + x - 2 + 2 = x^2 + x = 0

a = 1, b = 1, c = 0

D = (1)^2 - 4(1)(0) = 1

D > 0, so two distinct real roots.

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

Expand: (x+1)(x-2) = x^2 - x - 2

Add +x: x^2 - x - 2 + x = x^2 - 2

a = 1, b = 0, c = -2

D = (0)^2 - 4(1)(-2) = 0 + 8 = 8

D > 0, so two distinct real roots.

(i) A quadratic equation can have 0, 1, or 2 real roots. Example: \(x^2+1=0\) has 0 real roots, \((x-2)^2=0\) has 1 real root, and \(x^2-1=0\) has 2 real roots. So, it is not true that every quadratic has exactly one root.

(ii) Some quadratics do not have any real root. Example: \(x^2+1=0\). Here, no real number squared gives \(-1\). So, not every quadratic has at least one real root. This is false.

(iii) By the Fundamental Theorem of Algebra, a quadratic always has 2 roots (they may be real or complex). Example: \(x^2+1=0\) has two roots: \(i\) and \(-i\), which are complex. So the statement “every quadratic has at least two roots” is true if we count complex roots as well.

(iv) Since the highest power of \(x\) is 2, a quadratic can never have more than 2 roots. So, “at most 2 roots” is true.

(v) If the coefficient of \(x^2\) (say \(a\)) and the constant term (say \(c\)) have opposite signs, then their product \(a \times c < 0\). This means the parabola will cut the x-axis at two points, giving two real roots. Example: \(x^2 - 4 = 0\). Here \(a=1\), \(c=-4\), opposite signs → real roots \(x=\pm 2\). So this is true.

(vi) If \(a\) and \(c\) have the same sign, and coefficient of \(x\) (that is \(b\)) is 0, then the quadratic looks like \(ax^2 + c = 0\). Example: \(x^2 + 4 = 0\). Here both \(a\) and \(c\) are positive. This has no real solution because \(x^2 = -4\) is impossible for real numbers. So this statement is true.

Let us check step by step:

1. A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a, b, c\) are numbers and \(a \neq 0\).
2. "Integral coefficients" means that \(a, b, c\) are whole numbers (positive, negative, or zero).
3. The statement says: "If the coefficients are integers, then the roots must also be integers."
4. To check this, we try an example: \(x^2 - 2 = 0\).
5. Here, \(a = 1, b = 0, c = -2\). All are integers, so coefficients are integral.
6. Now solve it: \(x^2 - 2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2}\).
7. But \(\sqrt{2}\) is not an integer (it is an irrational number).
8. This shows that even if the coefficients are integers, the roots may not be integers.

Therefore, the statement is false.

Let us check step by step:

1. A quadratic equation is of the form \(ax^2 + bx + c = 0\), where \(a, b, c\) are numbers called coefficients.
2. Rational coefficients means \(a, b, c\) are either integers or fractions (numbers that can be written as \(\frac{p}{q}\)).
3. Irrational roots means the solutions (values of \(x\)) cannot be written as a simple fraction, like \(\sqrt{2}, \sqrt{3}, \pi\), etc.
4. Consider the equation: \(x^2 - 2 = 0\).
5. Here, the coefficients are:
   • \(a = 1\)
   • \(b = 0\)
   • \(c = -2\)
   All of these are rational numbers.
6. Now solve: \(x^2 - 2 = 0 \implies x^2 = 2\).
7. Taking square root: \(x = \pm \sqrt{2}\).
8. But \(\sqrt{2}\) is an irrational number.

So this quadratic equation has rational coefficients but irrational roots. Hence, the answer is Yes.

Let's understand step by step:

1. A quadratic equation is in the form: \(ax^2 + bx + c = 0\), where \(a, b, c\) are called coefficients.
2. Coefficients can be any real numbers: they may be rational (like \(2, -3, \frac{1}{2}\)) or irrational (like \(\sqrt{2}, \pi\)).
3. We want to check: is it possible that the coefficients are irrational but the solutions (roots) are rational numbers?

Example:

• Take the equation: \(\sqrt{2}x^2 - \sqrt{2} = 0\).
• Here, the coefficient of \(x^2\) is \(\sqrt{2}\) (irrational) and the constant term is \(-\sqrt{2}\) (irrational).

Now solve it:

1. \(\sqrt{2}x^2 - \sqrt{2} = 0\)
2. Take out \(\sqrt{2}\) as common: \(\sqrt{2}(x^2 - 1) = 0\)
3. So, \(x^2 - 1 = 0\)
4. Which gives: \(x^2 = 1\)
5. Therefore, \(x = \pm 1\)

Thus, the roots are \(1\) and \(-1\), both rational numbers.

Conclusion: Yes, it is possible. The equation \(\sqrt{2}x^2 - \sqrt{2} = 0\) has irrational coefficients but rational roots.

To check if 0.2 is a root, we must put \(x = 0.2\) in the equation \(x^2 - 0.4 = 0\).

Step 1: Write the equation with \(x = 0.2\):

\((0.2)^2 - 0.4\)

Step 2: Find \((0.2)^2\):

\(0.2 \times 0.2 = 0.04\)

Step 3: Subtract 0.4 from 0.04:

\(0.04 - 0.4 = -0.36\)

Step 4: The result is \(-0.36\). For a number to be a root, the result must be exactly 0.

Since \(-0.36 \neq 0\), 0.2 is not a root of the equation.

Step 1: Start with the quadratic equation:

\(x^2 + bx + c = 0\)

Here, \(b = 0\). So the equation becomes:

\(x^2 + c = 0\)

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Step 2: Rearrange the equation:

\(x^2 = -c\)

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Step 3: We are told \(c < 0\) (c is negative). That means \(-c\) is positive.

So, \(-c = |c|\) (the absolute value of c).

Thus, \(x^2 = |c|\).

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Step 4: Take the square root on both sides:

\(x = +\sqrt{|c|}\) or \(x = -\sqrt{|c|}\)

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Step 5: These two roots have the same size (same numerical value) but one is positive and the other is negative.

That means they are equal in magnitude and opposite in sign.