Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum 2023 / PECTAA 2026 Syllabus
📖 What's Inside: This review exercise covers MCQs, solving quadratic equations by factorization, completing square, and quadratic formula, nature of roots, forming equations, and formula rearrangement. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Unit 2: Quadratic Equations & Inequalities
Class 10 Math Unit 2 Review Exercise 2 – Quadratic Equations: Complete Guide
Review Exercise 2 of Unit 2 consolidates all key concepts of quadratic equations. This comprehensive review covers MCQs, solving quadratic equations by three methods (factorization, completing square, and quadratic formula), examining the nature of roots, forming equations from roots, and rearranging formulas.
What You Will Learn
By working through Review Exercise 2, students master: solving quadratic equations using factorization, completing the square, and the quadratic formula, determining the nature of roots using the discriminant (\(b^2 - 4ac\)), finding the sum and product of roots, forming quadratic equations from given roots, and rearranging formulas to make a variable the subject.
Topics Covered in This Review Exercise
- MCQs on Quadratic Equations – types, discriminant, solution sets, sum/product of roots, roots of equations, nature of roots, and subject of formula.
- Solving Quadratic Equations – by factorization, completing the square, and quadratic formula (2 sub-questions).
- Forming Quadratic Equations – from given roots using sum and product of roots.
- Nature of Roots – examining discriminant for two equations.
- Rearranging Formulas – making velocity the subject of the height formula.
- Equal Roots – finding the value of k for equal roots.
Why Review Exercise 2 Is Important
Review Exercise 2 is crucial for board exam preparation as it covers all key concepts of quadratic equations. The MCQs test conceptual understanding, while the solving questions develop procedural fluency. This exercise serves as excellent revision material before exams.
Punjab Board Preparation
Students preparing for board exams under any of the 10 BISE Punjab boards should prioritise this review exercise. It covers frequently tested topics including MCQs, solving methods, and nature of roots.
📖 Multiple Choice Questions (Unit 2 Review)
(i) The type of the equation \(2x^2 - x + 1 = 0\) is:
✅ Correct Answer: (A) quadratic
The equation is \(2x^2 - x + 1 = 0\). The highest power of the variable \(x\) is 2. Therefore, it is a quadratic equation.
The equation is \(2x^2 - x + 1 = 0\). The highest power of the variable \(x\) is 2. Therefore, it is a quadratic equation.
(ii) What is the discriminant of \(x^2 + 5x - 5 = 0\)?
✅ Correct Answer: (D) 45
For \(x^2+5x-5=0\), we have \(a=1,\ b=5,\ c=-5\). \[\text{Disc.} = b^2-4ac = 5^2-4(1)(-5) = 25+20 = 45\]
For \(x^2+5x-5=0\), we have \(a=1,\ b=5,\ c=-5\). \[\text{Disc.} = b^2-4ac = 5^2-4(1)(-5) = 25+20 = 45\]
(iii) The solution set of \(3x^2 - 9 = 0\) is:
✅ Correct Answer: (C) \(\{\pm\sqrt{3}\}\)
\[\begin{aligned} 3x^2-9 &= 0 \\ 3x^2 &= 9 \\ x^2 &= 3 \\ x &= \pm\sqrt{3} \end{aligned}\] The solution set is \(\{\pm\sqrt{3}\}\).
\[\begin{aligned} 3x^2-9 &= 0 \\ 3x^2 &= 9 \\ x^2 &= 3 \\ x &= \pm\sqrt{3} \end{aligned}\] The solution set is \(\{\pm\sqrt{3}\}\).
(iv) Sum of the roots of \(3x^2 + 5x - 12 = 0\) is:
✅ Correct Answer: (B) \(-\dfrac{5}{3}\)
For \(3x^2+5x-12=0\), we have \(a=3,\ b=5,\ c=-12\). \[Sum\ of\ roots = -\frac{b}{a} = -\frac{5}{3}\]
For \(3x^2+5x-12=0\), we have \(a=3,\ b=5,\ c=-12\). \[Sum\ of\ roots = -\frac{b}{a} = -\frac{5}{3}\]
(v) Product of the roots of \(3x^2 + 5x - 12 = 0\) is:
✅ Correct Answer: (A) \(-4\)
For \(3x^2+5x-12=0\), we have \(a=3,\ b=5,\ c=-12\). \[Product\ of\ roots = \frac{c}{a} = \frac{-12}{3} = -4\]
For \(3x^2+5x-12=0\), we have \(a=3,\ b=5,\ c=-12\). \[Product\ of\ roots = \frac{c}{a} = \frac{-12}{3} = -4\]
(vi) What are the roots of \((x-3)(x+3) = 0\)?
✅ Correct Answer: (A) \(3,\; -3\)
Solve \((x-3)(x+3)=0\). Set each factor to zero: \(x-3=0 \Rightarrow x=3\) and \(x+3=0 \Rightarrow x=-3\).
Solve \((x-3)(x+3)=0\). Set each factor to zero: \(x-3=0 \Rightarrow x=3\) and \(x+3=0 \Rightarrow x=-3\).
(vii) 3 and 2 are the roots of:
✅ Correct Answer: (C) \(x^2-5x+6=0\)
\[\begin{aligned} S &= 3+2 = 5 \\ P &= 3 \times 2 = 6 \end{aligned}\] By using \(x^2 - Sx + P = 0\) \[\Rightarrow x^2 - 5x + 6 = 0\]
\[\begin{aligned} S &= 3+2 = 5 \\ P &= 3 \times 2 = 6 \end{aligned}\] By using \(x^2 - Sx + P = 0\) \[\Rightarrow x^2 - 5x + 6 = 0\]
(viii) If \(b^2-4ac > 0\) and is a perfect square, then the roots of \(ax^2+bx+c=0\) are:
✅ Correct Answer: (B) unequal
If the discriminant \(b^2-4ac > 0\) and is a perfect square, then the roots are real, rational, and unequal (distinct).
If the discriminant \(b^2-4ac > 0\) and is a perfect square, then the roots are real, rational, and unequal (distinct).
(ix) If \(b^2-4ac = 0\), then the roots of \(ax^2+bx+c=0\) are:
✅ Correct Answer: (D) equal
If the discriminant \(b^2-4ac = 0\), then the roots are real, rational, and equal.
If the discriminant \(b^2-4ac = 0\), then the roots are real, rational, and equal.
(x) Subject \(c\) of \(x - 2c = b\) is:
✅ Correct Answer: (C) \(\dfrac{x-b}{2}\)
\[\begin{aligned} x - 2c &= b \\ x - b &= 2c \\ \frac{x-b}{2} &= c \\ c &= \frac{x-b}{2} \end{aligned}\]
\[\begin{aligned} x - 2c &= b \\ x - b &= 2c \\ \frac{x-b}{2} &= c \\ c &= \frac{x-b}{2} \end{aligned}\]
📖 Review Exercise 2 – Solved Problems
2(i)Solve \(8x^2 = x + 7\) by Factorization, Completing the Square, and Quadratic Formula.
Method 1: Factorization
\[\begin{aligned}
8x^2 &= x + 7 \\
8x^2 - x - 7 &= 0 \\
8x^2 - 8x + 7x - 7 &= 0 \\
8x(x-1) + 7(x-1) &= 0 \\
(x-1)(8x+7) &= 0
\end{aligned}\]
either
\[\begin{aligned} x-1&=0 \\ x&=1 \end{aligned}\]or
\[\begin{aligned} 8x+7&=0 \\ 8x&=-7 \\ x&=-\tfrac{7}{8} \end{aligned}\]\(S.S = \left\{1,\ -\dfrac{7}{8}\right\}\)
Method 2: Completing the Square
\[\begin{aligned}
8x^2 - x - 7 &= 0
\end{aligned}\]
Dividing both sides by 8
\[\begin{aligned} x^2 - \frac{x}{8} - \frac{7}{8} &= 0 \\ x^2 - \frac{1}{8}x &= \frac{7}{8} \end{aligned}\]Multiplying coefficient of \(x\) by \(\dfrac{1}{2}\): \(\dfrac{1}{2} \cdot \dfrac{1}{8} = \dfrac{1}{16}\). Adding \(\left(\dfrac{1}{16}\right)^2\) to both sides
\[\begin{aligned} x^2 - \frac{1}{8}x + \left(\frac{1}{16}\right)^2 &= \frac{7}{8} + \frac{1}{256} \\ \left(x - \frac{1}{16}\right)^2 &= \frac{224}{256} + \frac{1}{256} \\ \left(x - \frac{1}{16}\right)^2 &= \frac{225}{256} \end{aligned}\]Taking square root on both sides
\[\begin{aligned} x - \frac{1}{16} &= \pm\frac{15}{16} \\ x &= \frac{1}{16} \pm \frac{15}{16} \end{aligned}\]either
\[\begin{aligned} x &= \frac{1+15}{16} = \frac{16}{16} = 1 \end{aligned}\]or
\[\begin{aligned} x &= \frac{1-15}{16} = \frac{-14}{16} = -\frac{7}{8} \end{aligned}\]\(S.S = \left\{1,\ -\dfrac{7}{8}\right\}\)
Method 3: Quadratic Formula
\[8x^2 - x - 7 = 0\]
Here \(a=8,\ b=-1,\ c=-7\)
\[\begin{aligned} x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\[6pt] x &= \frac{-(-1) \pm \sqrt{(-1)^2-4(8)(-7)}}{2(8)} \\[6pt] x &= \frac{1 \pm \sqrt{1+224}}{16} \\[6pt] x &= \frac{1 \pm \sqrt{225}}{16} \\[6pt] x &= \frac{1 \pm 15}{16} \end{aligned}\]either
\[\begin{aligned} x &= \frac{1+15}{16} = \frac{16}{16} = 1 \end{aligned}\]or
\[\begin{aligned} x &= \frac{1-15}{16} = \frac{-14}{16} = -\frac{7}{8} \end{aligned}\]\(S.S = \left\{1,\ -\dfrac{7}{8}\right\}\)
2(ii)Solve \(2x^2 - x - 10 = 0\) by Factorization, Completing the Square, and Quadratic Formula.
Method 1: Factorization
\[\begin{aligned}
2x^2 - x - 10 &= 0 \\
2x^2 - 5x + 4x - 10 &= 0 \\
x(2x-5) + 2(2x-5) &= 0 \\
(2x-5)(x+2) &= 0
\end{aligned}\]
either
\[\begin{aligned} 2x-5&=0 \\ 2x&=5 \\ x&=\tfrac{5}{2} \end{aligned}\]or
\[\begin{aligned} x+2&=0 \\ x&=-2 \end{aligned}\]\(S.S = \left\{\dfrac{5}{2},\ -2\right\}\)
Method 2: Completing the Square
\[2x^2 - x - 10 = 0\]
Dividing both sides by 2
\[\begin{aligned} x^2 - \frac{1}{2}x - 5 &= 0 \\ x^2 - \frac{1}{2}x &= 5 \end{aligned}\]Multiplying coefficient of \(x\) by \(\dfrac{1}{2}\): \(\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\). Adding \(\left(\dfrac{1}{4}\right)^2\) to both sides
\[\begin{aligned} x^2 - \frac{1}{2}x + \left(\frac{1}{4}\right)^2 &= 5 + \frac{1}{16} \\ \left(x - \frac{1}{4}\right)^2 &= \frac{80}{16} + \frac{1}{16} \\ \left(x - \frac{1}{4}\right)^2 &= \frac{81}{16} \end{aligned}\]Taking square root on both sides
\[\begin{aligned} x - \frac{1}{4} &= \pm\frac{9}{4} \\ x &= \frac{1}{4} \pm \frac{9}{4} \end{aligned}\]either
\[\begin{aligned} x &= \frac{1+9}{4} = \frac{10}{4} = \frac{5}{2} \end{aligned}\]or
\[\begin{aligned} x &= \frac{1-9}{4} = \frac{-8}{4} = -2 \end{aligned}\]\(S.S = \left\{\dfrac{5}{2},\ -2\right\}\)
Method 3: Quadratic Formula
\[2x^2 - x - 10 = 0\]
Here \(a=2,\ b=-1,\ c=-10\)
\[\begin{aligned} x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\[6pt] x &= \frac{-(-1) \pm \sqrt{(-1)^2-4(2)(-10)}}{2(2)} \\[6pt] x &= \frac{1 \pm \sqrt{1+80}}{4} \\[6pt] x &= \frac{1 \pm \sqrt{81}}{4} \\[6pt] x &= \frac{1 \pm 9}{4} \end{aligned}\]either
\[\begin{aligned} x &= \frac{1+9}{4} = \frac{10}{4} = \frac{5}{2} \end{aligned}\]or
\[\begin{aligned} x &= \frac{1-9}{4} = \frac{-8}{4} = -2 \end{aligned}\]\(S.S = \left\{\dfrac{5}{2},\ -2\right\}\)
3Form a quadratic equation whose roots are \(6\) and \(\dfrac{3}{2}\).
\[\begin{aligned}
Sum\ of\ roots &= 6 + \frac{3}{2} = \frac{12}{2} + \frac{3}{2} = \frac{15}{2} \\[4pt]
S &= \frac{15}{2} \\[8pt]
Product\ of\ roots &= 6 \times \frac{3}{2} = \frac{18}{2} \\[4pt]
P &= 9
\end{aligned}\]
By using \(x^2 - Sx + P = 0\)
\[\begin{aligned} x^2 - \frac{15}{2}x + 9 &= 0 \end{aligned}\]Multiplying both sides by 2
\[\begin{aligned} 2x^2 - 15x + 18 &= 0 \end{aligned}\]\(\boldsymbol{2x^2 - 15x + 18 = 0}\)
4Examine the nature of the roots of the following equations.
(i) \(15x^2 + 11x + 2 = 0\)
Here \(a=15,\ b=11,\ c=2\)
\[\begin{aligned} Disc. &= b^2 - 4ac \\ &= (11)^2 - 4(15)(2) \\ &= 121 - 120 \\ &= 1 \\ &= 1^2 > 0 \end{aligned}\]As Disc. \(> 0\) and a perfect square, the roots are rational and unequal.
(ii) \(x^2 - x - 1 = 0\)
Here \(a=1,\ b=-1,\ c=-1\)
\[\begin{aligned} Disc. &= b^2 - 4ac \\ &= (-1)^2 - 4(1)(-1) \\ &= 1 + 4 \\ &= 5 > 0 \end{aligned}\]As Disc. \(> 0\) and not a perfect square, the roots are irrational and unequal.
5If a ball is thrown upward with velocity \(v\), the maximum height is \(h = \dfrac{v^2}{2g}\). Rearrange the formula to make \(v\) the subject.
\[\begin{aligned}
h &= \frac{v^2}{2g} \\
2gh &= v^2 \\
v^2 &= 2gh \\
\sqrt{v^2} &= \sqrt{2gh} \\
\boldsymbol{v} &= \boldsymbol{\sqrt{2gh}}
\end{aligned}\]
\(v = \sqrt{2gh}\)
6If the equation \(x^2 + 2(1+k)x + k^2 = 0\) has equal roots, find the value of \(k\).
Here \(a=1,\ b=2(1+k),\ c=k^2\)
\[\begin{aligned} Disc. &= b^2 - 4ac \\ &= [2(1+k)]^2 - 4(1)(k^2) \\ &= 4(k+1)^2 - 4k^2 \\ &= 4(k^2 + 2k + 1) - 4k^2 \\ &= 4k^2 + 8k + 4 - 4k^2 \\ &= 8k + 4 \end{aligned}\]Since the roots are equal, so \(Disc. = 0\)
\[\begin{aligned} 8k + 4 &= 0 \\ 8k &= -4 \\ k &= -\frac{4}{8} \\ \boldsymbol{k} &= \boldsymbol{-\frac{1}{2}} \end{aligned}\]\(k = -\dfrac{1}{2}\)
📈 Key Concepts – Quadratic Equations Review
- Quadratic Equation: ax² + bx + c = 0, a ≠ 0.
- Discriminant: D = b² - 4ac determines nature of roots.
- Sum of roots: -b/a, Product: c/a.
- Solving methods: Factorization, Completing Square, Quadratic Formula.
❓ Frequently Asked Questions
What is covered in Unit 2 Review Exercise 2 of Class 10 Math?
Unit 2 Review Exercise 2 covers MCQs on quadratic equations, solving quadratic equations by three methods (factorization, completing square, and quadratic formula), examining the nature of roots using discriminant, forming quadratic equations from given roots, and rearranging formulas.
How many questions are there in Unit 2 Review Exercise 2?
Review Exercise 2 has 6 main questions covering MCQs (10 sub-questions), solving quadratic equations by three methods (2 sub-questions), forming a quadratic equation, examining nature of roots (2 sub-questions), rearranging a formula, and finding the value of k for equal roots.
What is the discriminant formula in quadratic equations?
The discriminant formula is \(D = b^2 - 4ac\). It determines the nature of roots: if \(D > 0\) and perfect square → rational and unequal; if \(D > 0\) and not perfect square → irrational and unequal; if \(D = 0\) → real and equal; if \(D < 0\) → imaginary.
What are the three methods to solve quadratic equations?
The three methods are: (1) Factorization Method - factoring the quadratic expression into linear factors, (2) Completing the Square Method - converting the equation into a perfect square trinomial, (3) Quadratic Formula Method - using \(x = (-b \pm \sqrt{b^2-4ac})/(2a)\).
Is this solution according to the PECTAA 2026 syllabus?
Yes, these solutions are prepared according to the PECTAA 2026 / National Curriculum 2023 syllabus for Class 10 Mathematics.
Is this Review Exercise 2 solution valid for all Punjab Boards?
Yes, the content follows the unified Punjab textbook and is applicable to students of all 10 BISE Punjab boards.
Are solved PDF notes available for Review Exercise 2?
Yes, a complete solved PDF for Review Exercise 2 is embedded on this page and available to download for free.
Can I download the Unit 2 Review Exercise 2 solution as a PDF?
Yes, use the Download PDF button on this page to save the complete solved Review Exercise 2 notes to your device.
Is Review Exercise 2 important for Class 10 board exams?
Yes, Review Exercise 2 is very important for board exams as it covers all key concepts of quadratic equations including MCQs, solving methods, and nature of roots. It serves as comprehensive preparation material.
Who prepared these Class 10 Math Unit 2 notes?
These notes were prepared by Muhammad Tayyab, Subject Specialist Mathematics at Govt Christian High School Daska, for Hira Science Academy.