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
1. Four possible answers are given for the following questions. Choose the correct answer.
| Q. No | Question | A | B | C | D |
|---|---|---|---|---|---|
| (i) | The type of the equation \(2x^2 - x + 1 = 0\) is: | quadratic | linear | third degree | pure quadratic |
| (ii) | What is the discriminant of \(x^2 + 5x - 5 = 0\)? | \(-20\) | \(20\) | \(25\) | \(45\) |
| (iii) | The solution set of \(3x^2 - 9 = 0\) is: | \(\{3\}\) | \(\pm 3\) | \(\{\pm\sqrt{3}\}\) | \(\{3\}\) |
| (iv) | Sum of the roots of \(3x^2 + 5x - 12 = 0\) is: | \(\dfrac{5}{3}\) | \(-\dfrac{5}{3}\) | \(\dfrac{12}{3}\) | \(\dfrac{3}{5}\) |
| (v) | Product of the roots of \(3x^2 + 5x - 12 = 0\) is: | \(-4\) | \(3\) | \(4\) | \(5\) |
| (vi) | What are the roots of \((x-3)(x+3) = 0\)? | \(3,\; -3\) | \(3,\; 3\) | \(-3,\; -3\) | \(9,\; 0\) |
| (vii) | 3 and 2 are the roots of: | \(x^2+5x+6=0\) | \(x^2+6x+5=0\) | \(x^2-5x+6=0\) | \(x^2+6x-5=0\) |
| (viii) | If \(b^2-4ac > 0\) and is a perfect square, then the roots of \(ax^2+bx+c=0\) are: | equal | unequal | imaginary | irrational |
| (ix) | If \(b^2-4ac = 0\), then the roots of \(ax^2+bx+c=0\) are: | unequal | irrational | imaginary | equal |
| (x) | Subject \(c\) of \(x - 2c = b\) is: | \(x+b\) | \(b-x\) | \(\dfrac{x-b}{2}\) | \(\dfrac{b-x}{2}\) |
Answer Key with Explanations
| Q. No | Correct Answer | Explanation |
|---|---|---|
| (i) | (a) | The equation is \(2x^2 - x + 1 = 0\). The highest power of the variable \(x\) is 2. Therefore, it is a quadratic equation. |
| (ii) | (d) | 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) | (c) | \[\begin{aligned} 3x^2-9 &= 0 \\ 3x^2 &= 9 \\ x^2 &= 3 \\ x &= \pm\sqrt{3} \end{aligned}\] The solution set is \(\{\pm\sqrt{3}\}\). |
| (iv) | (b) | For \(3x^2+5x-12=0\), we have \(a=3,\ b=5,\ c=-12\). \[Sum\ of\ roots = -\frac{b}{a} = -\frac{5}{3}\] |
| (v) | (a) | 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) | (a) | 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) | (c) | \[\begin{aligned} S &= 3+2 = 5 \\ P &= 3 \times 2 = 6 \end{aligned}\] By using \(x^2 - Sx + P = 0\) \[\Rightarrow x^2 - 5x + 6 = 0\] |
| (viii) | (b) | If the discriminant \(b^2-4ac > 0\) and is a perfect square, then the roots are real, rational, and unequal (distinct). |
| (ix) | (d) | If the discriminant \(b^2-4ac = 0\), then the roots are real, rational, and equal. |
| (x) | (c) | \[\begin{aligned} x - 2c &= b \\ x - b &= 2c \\ \frac{x-b}{2} &= c \\ c &= \frac{x-b}{2} \end{aligned}\] |
2. Solve the following quadratic equations by Factorization, Completing the Square, and Quadratic Formula.
i\(8x^2 = x + 7\)
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
\[\begin{aligned} x &= \frac{1-15}{16} = \frac{-14}{16} = -\frac{7}{8} \end{aligned}\]\(S.S = \left\{1,\ -\dfrac{7}{8}\right\}\)
ii\(2x^2 - x - 10 = 0\)
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.