On this page, you can find the complete solutions of the third exercise of the Quadratic Equations chapter from Basic Mathematics Grade XIII book published and distributed by Sukunda Pustak Bhawan.
In the above-mentioned book, permutation and combination is the first chapter and has four exercises only. Out of which, this is the solution to the third exercise.
Disclaimer:
Exercise 2
1 a) Show that each pair of the following equations has a common root:
a) $\rm x^2 - 8x + 15 = 0 \ and \ 2x^2 - x - 15 = 0$
Solution:
Equation (i)
\begin{equation} \rm x^2 - 8x + 15 = 0 \\ or, (x - 5)(x - 3) = 0 \\ \alpha = 5 \ and \ \beta = 3 \end{equation}
Put $\rm x = \alpha = 5$ in equation (ii),
\begin{equation} \rm 2(5)^2 - (5) - 15 = 50 - 5 - 15 \neq 0 \end{equation}
Put $\rm x = \beta = 3$ in equation (ii),
\begin{equation} \rm 2(3)^2 - (3) - 15 = 18 - 3 - 15 = 0 \end{equation}
Hence, above equations have a common root, i.e., 3.
b) $\rm 3x^2 - 8x + 4 = 0 \ and \ 4x^2 - 7x - 2 = 0$
Solution:
Equation (i)
\begin{equation} \rm 3x^2 - 8x + 4 = 0 \\ or, (3x - 2)(x - 2) = 0 \\ \alpha = \frac{2}{3} \ and \ \beta = 2 \end{equation}
Put $\rm x = \alpha = \frac{2}{3}$ in equation (ii),
\begin{equation} \rm 4(\frac{2}{3})^2 - 7(\frac{2}{3}) - 2 = \frac{16 - 42 - 18}{9} \neq 0 \end{equation}
Put $\rm x = \beta = 2$ in equation (ii),
\begin{equation} \rm 4(2)^2 - 7(2) - 2 = 16 - 14 - 2 = 0 \end{equation}
Hence, the above equations have a common root, i.e., 2.
2. Find the value of p so that each pair of the equations may have one root common.
a) $\rm 4x^2 + px - 12 = 0 \ and \ 4x^2 + 3px -4 = 0$
Solution:
For one root common
\begin{equation}
\rm (bc' - b'c)(ab' - a'b) = (ca' - c'a)^2 \\
or,
(p(-4) - (3p)(-12))(4(3p) - 4(p)) = ((-12)4 - (-4)4)^2 \\
or, (-4p +
36p)(12p - 4p) = (-48 + 16)^2 \\
or, (32p)(8p) = 32^2 \\
or, 8p^2 =
32 \\
or, p^2 = 4 \\
\therefore p = \pm 2
\end{equation}
b) $\rm 2x^2 + px - 1 = 0 \ and \ 3x^2 - 2x - 5 = 0$
Solution:
For one root common
\begin{equation}
\rm (bc' - b'c)(ab' - a'b) = (ca' - c'a)^2 \\
or,
(p(-5) - (-2)(-1))(2(-2) - 3(p)) = (2(-5) - 3(-1))^2 \\
or, (-5p - 2)(-4
- 3p) = (-10 + 3)^2 \\
or, (5p + 2)(4 + 3p) = (-7)^2 \\
or, 20p + 8
+ 15p^2 + 6p = 49 \\
or, 15p^2 + 26p - 41 = 0 \\
Solving \ the \
quadratic \ equation \ in \ p, \ we \ get,\\
\therefore p = 1 \ or \ p =
- \frac{41}{15}
\end{equation}
3. If the quadratic equations $\rm x^2 + px + q = 0 \ and \ x^2 + p'x + q' = 0$ have a common root, show that it must either be $\frac{pq' - p'q}{q - q'}$ or $\frac{q - q'}{p' - p}$
Solution:
Solving the given equations by using the rule of cross multiplication,
\begin{equation} \rm \frac{x^2}{pq' - p'q} = \frac{x}{q(1) - q'(1)} =
\frac{1}{(1)p' - (1)p}\\
Either \\
\frac{x^2}{pq' - p'q} =
\frac{x}{q - q'} \\
\Rightarrow x = \frac{pq' - p'q}{q - q'} \\
Or,
\\
\frac{x}{q - q'} = \frac{1}{p' - p} \\
\Rightarrow x = \frac{q =
q'}{p' - p}
\end{equation}
4. If the equations $x^2 + px + q = 0 \ and \ x^2 + qx + p = 0$ have a common root, prove that either p = q or p + q + 1 = 0.
Solution:
When two quadratic equations have a common root,
\begin{equation}
\rm (bc' - b'c)(ab' - a'b) = (ca' - c'a)^2 \\
or,
(p(p) - (q)(q))(1(q) - 1(p)) = (q(1) - p(1))^2 \\
or, (p^2 - q^2)(q - p)
= (q - p)^2 \\
or, (p^2 - q^2) = (q - p) \\
or, (p + q)(p - q) + (p
- q) = 0 \\
or, (p - q)(p + q +1) = 0 \\
Either \\
p - q = 0
\Rightarrow p = q\\
Or \\
p + q + 1 = 0
\end{equation}
5. If the quadratic equations $\rm ax^2 + bx + c = 0 \ and \ bx^2 + cx + a = 0$ have a common root, then either $\rm a + b + c = 0 \ or \ a = b = c$.
Solution:
If the above quadratic equations have a common root then,
Solving the given equations by using the rule of cross multiplication,
\begin{equation} \rm \frac{x^2}{ba - c^2} = \frac{x}{bc - a^2} = \frac{1}{ac - b^2}\\
Either \\
\frac{x^2}{ba - c^2} = \frac{x}{bc - a^2} \\
\Rightarrow x = \frac{ba - c^2}{bc - a^2} \\
Or, \\
\frac{x}{bc - a^2} = \frac{1}{ac - b^2} \\
\Rightarrow x = \frac{bc - a^2}{ac - b^2}\\
Both \ values \ of \ x \ should \ be \ the \ same \ to \ have \ a \ common \ root
\Rightarrow \frac{ba - c^2}{bc - a^2} = \frac{bc - a^2}{ac - b^2}\\
or, (ba - c^2)(ac - b^2) = (bc - a^2)^2 \\
or, a^2bc - ab^3 - ac^3 + b^2c^2 = b^2c^2 - 2a^2bc + a^4\\
or, a^4 + ab^3 + ac^3 - 3a^2bc = 0\\
or, a(a^3 + b^3 + c^3 - 3abc) = 0 \\
or, a (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0 \\
Since \ a \ neq 0\\
Either\\
a + b + c = 0\\
Or\\
(a^2 + b^2 + c^2 - ab - bc - ca = 0\\
or, 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 \\
or, a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca - a^2 = 0\\
or, (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \\
It \ is \ possible \ when \\
a - b = 0 \Rightarrow a = b\\
b - c = 0 \Rightarrow b = c\\
c - a = 0 \Rightarrow a = a\\
\Rightarrow a = b = c
\end{equation}
6. Prove that if the equations $\rm x^2 + bx + ca = 0$ and $\rm x^2 + cx + ab = 0$ have a common root, their other roots will satisfy $\rm x^2 + ax + bc = 0$.
Solution:
Solving for the common root of two equations using the rule of cross-multiplication,
\begin{equation} \rm
\frac{x^2}{b(ab) - ca(c)} = \frac{x}{ca(1) - ab(1)} = \frac{1}{(1)c - 1(b)} \\
\frac{x^2}{a(b^2 - c^2)} = \frac{x}{a(c - b)} = \frac{1}{(c - b)} \\
Taking \ first \ two \ equations,
x = \frac{a(b^2 - c^2)}{a (c -b)} \\
x = \frac{- (c^2 - b^2)}{(c -b)}\\
x = \frac{- (c - b)(c + b)}{(c-b)}\\
x = -(c+b)\\
Taking \ last \ two \ equations,
x = \frac{a(c-b)}{(c-b)}\\
\\
\Rightarrow -(c+b) = a \\
a + b + c = 0
\end{equation}
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