CBSE Class 10 Mathematics Quadratic Equations Assignment Set 03

Keypoint

Question 1. The general form of quadratic equation is \(ax^2 + bx + c = 0\), where \(a \neq 0\). Example: \(3x^2 + 7x + 4 = 0\)
Answer: A quadratic equation in variable \(x\) is an equation of degree 2, meaning the highest power of \(x\) is 2. The standard form is \(ax^2 + bx + c = 0\), where \(a, b, c\) are real numbers and \(a \neq 0\). In the example \(3x^2 + 7x + 4 = 0\), we have \(a = 3\), \(b = 7\), and \(c = 4\).
In simple words: A quadratic equation is an equation where the highest exponent of the variable is 2, and the coefficient of the squared term cannot be zero.

Exam Tip: Always verify that the coefficient of \(x^2\) is not zero (i.e., \(a \neq 0\)), otherwise the equation becomes a linear equation.

Question 2. Discriminant (\(D\)) = \(b^2 - 4ac\)
Answer: The discriminant \(D\) is a part of the quadratic formula that determines the nature of the roots of a quadratic equation. It is calculated as \(D = b^2 - 4ac\), using the coefficients from the general form \(ax^2 + bx + c = 0\).
In simple words: The discriminant helps us find out whether the roots of a quadratic equation are real or imaginary without fully solving it.

Exam Tip: Memorize the discriminant formula \(D = b^2 - 4ac\) thoroughly as it is essential for solving quadratic equations and determining root characteristics.

Question 3. Nature of Roots
(i) If \(D > 0\), roots are real and unequal.
(ii) If \(D = 0\), roots are real and equal.
(iii) If \(D < 0\), no real roots
Answer: The value of the discriminant \(D = b^2 - 4ac\) dictates the nature of the roots:
1. If \(D > 0\), the equation has two distinct real roots.
2. If \(D = 0\), the equation has two equal real roots (one repeated real root).
3. If \(D < 0\), the equation has no real roots (the roots are complex numbers).
In simple words: Positive discriminant means two different real answers, zero discriminant means one repeated real answer, and negative discriminant means no real answers.

Exam Tip: When a question asks to find the condition for "real and equal roots", always set \(D = 0\) (\(b^2 - 4ac = 0\)).

Question 4. Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Where \(a =\) coefficient of \(x^2\)
\(b =\) coefficient of \(x\)
\(c =\) constant term
Example: Solve \(3x^2 - 7x + 4 = 0\)
Solution given:
\(a = 3, b = -7, c = 4\)
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 3 \times 4}}{2 \times 3}\)
\(x = \frac{7 \pm \sqrt{49 - 48}}{6}\)
\(x = \frac{7 \pm \sqrt{1}}{6}\)
\(x = \frac{7 \pm 1}{6}\)
Roots are \(x = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}\) and \(x = \frac{7 - 1}{6} = \frac{6}{6} = 1\)
Answer: The quadratic formula provides a universal method to find the roots of any quadratic equation \(ax^2 + bx + c = 0\).
Step 1: Identify coefficients: \(a = 3\), \(b = -7\), \(c = 4\).
Step 2: Substitute values into the quadratic formula:
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(4)}}{2(3)}\)
Step 3: Simplify the expression under the square root (discriminant):
\(x = \frac{7 \pm \sqrt{49 - 48}}{6}\)
\(x = \frac{7 \pm \sqrt{1}}{6}\)
\(x = \frac{7 \pm 1}{6}\)
Step 4: Separate into two cases for plus and minus:
Case (+): \(x = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}\)
Case (-): \(x = \frac{7 - 1}{6} = \frac{6}{6} = 1\)
In simple words: Plug the values of \(a\), \(b\), and \(c\) into the quadratic formula and simplify step-by-step to get the two possible values of \(x\).

Exam Tip: Be very careful with negative signs when substituting \(b\) and \(c\) into the quadratic formula, especially when squaring negative numbers like \((-7)^2 = 49\).