Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
๐ Based on National Curriculum 2023 / PECTAA 2026 Syllabus
๐ What's Inside: This exercise covers forming quadratic equations from given roots, finding equations with transformed roots, and determining conditions for roots. Perfect for Punjab Boards exam preparation.
๐ Related Resources โ Unit 2: Quadratic Equations & Inequalities
1
Form a quadratic equation whose roots are given below:
(i) \(-4,9\)
\[
\begin{aligned}
\text{Sum } S &= -4 + 9 = 5,\\
\text{Product } P &= (-4)(9) = -36.
\end{aligned}
\]
\[
x^{2} - Sx + P = 0 \quad \Rightarrow \quad x^{2} - 5x - 36 = 0.
\]
(ii) \(5, -7\)
\[
\begin{aligned}
S &= 5 + (-7) = -2,\\
P &= (5)(-7) = -35.
\end{aligned}
\]
\[
x^{2} - Sx + P = 0 \;\Rightarrow\; x^{2} - (-2)x + (-35) = 0 \;\Rightarrow\; x^{2} + 2x - 35 = 0.
\]
(iii) \(\frac{-7}{5}, \frac{-6}{5}\)
\[
\begin{aligned}
S &= \frac{-7}{5} + \frac{-6}{5} = \frac{-13}{5},\\
P &= \left(\frac{-7}{5}\right)\left(\frac{-6}{5}\right) = \frac{42}{25}.
\end{aligned}
\]
\[
x^{2} - \left(\frac{-13}{5}\right)x + \frac{42}{25} = 0
\;\Longrightarrow\;
x^{2} + \frac{13}{5}x + \frac{42}{25} = 0.
\]
Multiplying by 25:
\[
25x^{2} + 65x + 42 = 0.
\]
(iv) \(\frac{-3}{2}, \frac{7}{2}\)
\[
\begin{aligned}
S &= \frac{-3}{2} + \frac{7}{2} = \frac{4}{2} = 2,\\
P &= \frac{-21}{4}.
\end{aligned}
\]
\[
x^{2} - 2x + \left(\frac{-21}{4}\right) = 0
\;\Longrightarrow\;
x^{2} - 2x - \frac{21}{4} = 0.
\]
Multiplying by 4:
\[
4x^{2} - 8x - 21 = 0.
\]
(v) \(3+\sqrt{5}, \; 3-\sqrt{5}\)
\[
\begin{aligned}
S &= (3+\sqrt{5}) + (3-\sqrt{5}) = 6,\\
P &= (3+\sqrt{5})(3-\sqrt{5}) = 3^{2} - (\sqrt{5})^{2} = 9-5 = 4.
\end{aligned}
\]
\[
x^{2} - 6x + 4 = 0.
\]
(vi) \(-2+\sqrt{3}, \; -2-\sqrt{3}\)
\[
\begin{aligned}
S &= (-2+\sqrt{3}) + (-2-\sqrt{3}) = -4,\\
P &= (-2+\sqrt{3})(-2-\sqrt{3}) = (-2)^{2} - (\sqrt{3})^{2} = 4-3 = 1.
\end{aligned}
\]
\[
x^{2} - (-4)x + 1 = 0 \;\Longrightarrow\; x^{2} + 4x + 1 = 0.
\]
2
Find the quadratic equation with roots exceeding by 2 than those of roots of \(x^{2} + 9x + 20 = 0\)
\[
x^{2} + 9x + 20 = 0 \quad (a=1,\; b=9,\; c=20).
\]
Let \(\alpha,\beta\) be the roots of given equation. New roots: \(\alpha+2,\; \beta+2\).
\[
\begin{aligned}
S_{\text{new}} &= (\alpha+2)+(\beta+2) = \alpha+\beta+4 = -9+4 = -5,\\
P_{\text{new}} &= (\alpha+2)(\beta+2) = \alpha\beta + 2(\alpha+\beta) + 4 = 20 + 2(-9) + 4 = 6.
\end{aligned}
\]
\[
x^{2} - Sx + P = 0 \;\Longrightarrow\; x^{2} - (-5)x + 6 = 0 \;\Longrightarrow\; x^{2} + 5x + 6 = 0.
\]
3
Find the equation whose roots are double the roots of \(x^{2} - px + q = 0\)
\[
x^{2} + px + q = 0 \quad \text{(note: given as } x^{2} - px + q = 0 \text{ in statement, but solved as } x^{2}+px+q=0).
\]
Let \(\alpha,\beta\) be the roots of given equation. New roots: \(2\alpha, 2\beta\).
\[
\begin{aligned}
S_{\text{new}} &= 2\alpha+2\beta = 2(\alpha+\beta) = 2(p),\\
P_{\text{new}} &= (2\alpha)(2\beta) = 4\alpha\beta = 4(q).
\end{aligned}
\]
\[
x^{2} - 2p x + 4q = 0.
\]
4
If \(\alpha,\beta\) are the roots of \(x^{2} + 2x + 4 = 0\), then find the equation whose roots are:
\[
\alpha + \beta = -2,\qquad \alpha\beta = 4.
\]
(i) \(\frac{1}{\alpha},\; \frac{1}{\beta}\)
\[
\begin{aligned}
S &= \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{-2}{4} = -\frac{1}{2},\\
P &= \frac{1}{\alpha\beta} = \frac{1}{4}.
\end{aligned}
\]
\[
x^{2} - \left(-\frac{1}{2}\right)x + \frac{1}{4} = 0
\;\Longrightarrow\;
x^{2} + \frac{1}{2}x + \frac{1}{4} = 0.
\]
Multiply by 4: \(\;4x^{2} + 2x + 1 = 0.\)
(ii) \(\frac{\alpha}{\beta},\; \frac{\beta}{\alpha}\)
\[
\begin{aligned}
S &= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}
= \frac{(\alpha+\beta)^{2} - 2\alpha\beta}{\alpha\beta}
= \frac{(-2)^{2} - 2(4)}{4}
= \frac{4-8}{4} = -1,\\
P &= \frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha} = 1.
\end{aligned}
\]
\[
x^{2} - (-1)x + 1 = 0 \;\Longrightarrow\; x^{2} + x + 1 = 0.
\]
(iii) \(2\alpha - \frac{1}{\beta},\; 2\beta - \frac{1}{\alpha}\)
\[
\begin{aligned}
S &= 2\alpha + 2\beta - \left(\frac{1}{\beta}+\frac{1}{\alpha}\right)
= 2(\alpha+\beta) - \frac{\alpha+\beta}{\alpha\beta}\\
&= 2(-2) - \frac{-2}{4} = -4 + \frac{1}{2} = -\frac{7}{2},\\[6pt]
P &= \left(2\alpha-\frac{1}{\beta}\right)\left(2\beta-\frac{1}{\alpha}\right)
= 4\alpha\beta - 2 - 2 + \frac{1}{\alpha\beta}
= 4(4) - 4 + \frac{1}{4} = 16 - 4 + \frac{1}{4} = \frac{49}{4}.
\end{aligned}
\]
\[
x^{2} - \left(-\frac{7}{2}\right)x + \frac{49}{4} = 0
\;\Longrightarrow\;
x^{2} + \frac{7}{2}x + \frac{49}{4} = 0.
\]
Multiply by 4: \(\;4x^{2} + 14x + 49 = 0.\)
(iv) \(\alpha^{2},\; \beta^{2}\)
\[
\begin{aligned}
S &= \alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = (-2)^{2} - 2(4) = 4-8 = -4,\\
P &= \alpha^{2}\beta^{2} = (\alpha\beta)^{2} = 4^{2} = 16.
\end{aligned}
\]
\[
x^{2} - (-4)x + 16 = 0 \;\Longrightarrow\; x^{2} + 4x + 16 = 0.
\]
(v) \(2\alpha-1,\; 2\beta-1\)
\[
\begin{aligned}
S &= (2\alpha-1)+(2\beta-1) = 2(\alpha+\beta) - 2 = 2(-2) - 2 = -6,\\
P &= (2\alpha-1)(2\beta-1) = 4\alpha\beta - 2(\alpha+\beta) + 1 = 4(4) - 2(-2) + 1 = 16+4+1 = 21.
\end{aligned}
\]
\[
x^{2} - (-6)x + 21 = 0 \;\Longrightarrow\; x^{2} + 6x + 21 = 0.
\]
5
Find the condition that roots of \(ax^{2} + bx + c = 0\) should be reciprocals of each other.
\[
ax^{2} + bx + c = 0,\quad a=a,\; b=b,\; c=c.
\]
Let one root be \(\alpha\), then the other is \(\frac{1}{\alpha}\).
\[
\alpha \cdot \frac{1}{\alpha} = \frac{c}{a} \;\Longrightarrow\; 1 = \frac{c}{a} \;\Longrightarrow\; a = c.
\]
Condition: \(\boxed{a = c}\).
6
Find the value of \(k\), given that one root of \(x^{2} - (2k+4)x + (7k+1) = 0\) is \(3\).
\[
x^{2} - (2k+4)x + (7k+1) = 0.
\]
Since \(3\) is a root, put \(x=3\):
\[
\begin{aligned}
3^{2} - (2k+4)(3) + (7k+1) &= 0,\\
9 - 6k - 12 + 7k + 1 &= 0,\\
k - 2 &= 0 \;\Longrightarrow\; k = 2.
\end{aligned}
\]
\(\boxed{k = 2}\)
7
Find the value of \(m\) in \(2x^{2} + 3x + m = 0\) when sum of its roots is equal to double the product of its roots.
\[
2x^{2} + 3x + m = 0,\quad a=2,\; b=3,\; c=m.
\]
Let \(\alpha,\beta\) be the roots.
\[
\alpha + \beta = -\frac{3}{2},\qquad \alpha\beta = \frac{m}{2}.
\]
Given: sum = double product
\[
\alpha + \beta = 2(\alpha\beta) \;\Longrightarrow\; -\frac{3}{2} = 2\left(\frac{m}{2}\right) \;\Longrightarrow\; -\frac{3}{2} = m.
\]
\(\boxed{m = -\frac{3}{2}}\)
8
If \(\alpha,\beta\) are roots of \(x^{2} + ax + b = 0\) and \(\alpha^{2},\beta^{2}\) are roots of \(x^{2} + Ax + B = 0\), then prove that \(A = 2b - a^{2},\; B = b^{2}\).
\[
x^{2} + ax + b = 0 \quad \Rightarrow \quad \alpha+\beta = -a,\quad \alpha\beta = b.
\]
For \(x^{2} + Ax + B = 0\), roots \(\alpha^{2},\beta^{2}\):
\[
\begin{aligned}
\alpha^{2}+\beta^{2} &= (\alpha+\beta)^{2} - 2\alpha\beta = (-a)^{2} - 2b = a^{2} - 2b,\\
\text{but } \alpha^{2}+\beta^{2} &= -A \quad \Rightarrow \quad -A = a^{2} - 2b \;\Longrightarrow\; A = 2b - a^{2}.\\[6pt]
\alpha^{2}\beta^{2} &= (\alpha\beta)^{2} = b^{2} \;\Longrightarrow\; B = b^{2}.
\end{aligned}
\]
\(\boxed{A = 2b - a^{2}},\quad \boxed{B = b^{2}}\) (Proved).
9
If \(\alpha,\beta\) are roots of \(x^{2} + px + q = 0\), then find the condition that
\[
x^{2} + px + q = 0 \quad \Rightarrow \quad \alpha+\beta = -p,\; \alpha\beta = q.
\]
(i) \(\alpha = \beta\)
\[
\begin{aligned}
\alpha - \beta &= 0 \;\Rightarrow\; (\alpha-\beta)^{2} = 0 \;\Rightarrow\; (\alpha+\beta)^{2} - 4\alpha\beta = 0,\\
(-p)^{2} - 4q &= 0 \;\Longrightarrow\; p^{2} = 4q.
\end{aligned}
\]
\(\boxed{p^{2} = 4q}\)
(ii) \(\alpha = \frac{1}{\beta}\)
\[
\alpha\beta = 1 \;\Longrightarrow\; q = 1.
\]
\(\boxed{q = 1}\)
Est. 1882
๐ Key Concepts
- Quadratic from roots: \(x^2 - Sx + P = 0\)
- Sum & Product: \(S = \alpha+\beta = -\frac{b}{a}\), \(P = \alpha\beta = \frac{c}{a}\)
- Transformation: New roots can be formed by shifting, scaling, reciprocals, etc.
- Conditions: Equal roots (\(b^2=4ac\)), reciprocal roots (\(a=c\))