Algebra - Solved Examples

1.    If (x + 2)(x - a) = px2$x^2$ + qx + 8, what are the values of the constants a, p and q?
Solution: (x + 2)(x - a) = px2$p{x}^2$ + qx + 8
⇒x2$\Rightarrow x^2$ + 2x - ax - 2a = px2$p{x}^2$ + qx + 8
⇒x2$\Rightarrow x^2$ + (2 - a) x - 2a = px2$p{x}^2$ + qx + 8
Equating the coefficients of x2$x^2$ , x and constant terms on both sides.
p = 1; q = 2 - a
-2a = 8
Solving, we get a = -4, p = 1, q = 6

2.    (x +1 )(2x - 2)(3x + 3) =  ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$ What are the values of a, b, c and d?
Solution: (x + 1)(2x - 2)(3x + 3) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
6(x + 1)(x - 1)(x + 1) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
6(x2$x^2$ - 1)(x + 1) = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
⇒6(x3+x2−x−1)$\Rightarrow 6\left(x^3+x^2-x-1\right)$ = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
⇒6x3+6x2−6x−6$\Rightarrow 6x^3+6x^2-6x-6$ = ax3+bx2+cx+d$a{x}^3+b{x}^2+cx+d$
    Equating the coefficients of like learns on both the sides, a = 6, b = 6, c = -6 and d = -6.

3.    Find for what value(s) of k would there be an unique solution for the given set of questions.
    2x - 3y = 1 and kx + 5y = 7
Solution: If two equations ax + by = M and cx + dy = N have a unique solution, then ac≠bd${\displaystyle \frac{a}{c}\ne {\displaystyle \frac{b}{d}}}$
 So in the above problem, k ≠−103$\ne {\displaystyle \frac{-10}{3}}$

4.    Find the value(s) of k for which there is no solution for the given set of equations.
    2x - ky = -3 and 3x + 2y = 1
Solution: 3x + 2y = 1
If there are no set of solutions, then 2−k=32${\displaystyle \frac{2}{-k}={\displaystyle \frac{3}{2}}}$ or k = −43$-{\displaystyle \frac{4}{3}}$
    For k = −43$-{\displaystyle \frac{4}{3}}$, the two lines in the coordinate planes are parallel to each other.

5.    Find the value of k for which there are infinite solutions for the given set of equations.
    5x + 2y = k and 10x + 4y = 3
Solution:  For the two sets of equation to have infinite solutions, we have 510=24=k3${\displaystyle \frac{5}{10}={\displaystyle \frac{2}{4}={\displaystyle \frac{k}{3}}}}$
    . Hence, k = 32${\displaystyle \frac{3}{2}}$

6.    What is the solution of the following simultaneous equations?
    x + y + z = 6, x + 2y + 3z = 14 and x + 3y + z = 10
Solution:  x + y + z = 6        .... (i)
x + 2y + 3z = 14    .... (ii)
x + 3y + z = 10
From (i), we get z = 6 - x - y
Substitute it in (ii) and (iii)
We have from (ii)
x + 2y + 18 - 3x - 3y = 14 or 2x + y = 4
Similarily, from (iii)
x + 3y + 6 - x - y = 10 or 2y = 4 or y = 2
On solving, we get y = 2, x = 1, z = 3.


7.    The number of roots common between the two equations x3+3x2+4x+5=0$x^3+3x^2+4x+5=0$ and x3+2x2+7x+3=0$x^3+2x^2+7x+3=0$ is
Solution:  Here f(x) = x3+3x2+4x+5=0$x^3+3x^2+4x+5=0$ and g(x) = x3+2x2+7x+3=0$x^3+2x^2+7x+3=0$
To find the common roots we have to solve the equation f(x) - g(x) = 0
i.e. (x3+3x2+4x+5)−(x3+2x2+7x+3)=0$\left(x^3+3x^2+4x+5\right)-\left(x^3+2x^2+7x+3\right)=0$
x2$x^2$ - 3x + 2 = 0
x = 2, x = 1
These are the points of intersection f(x) and g(x). Whether these points are also the common root will have to be checked by putting these vales in f(x) = 0 and g(x) = 0.
For x = 2;
f(2) = g(2) = 43 ≠$\ne$ 0. Hence 2 is not a common root but only a point of intersection.
For x = 1;
f(1) = g(1) = 13 ≠$\ne$ 0. Again 1 is not a common root but a point of intersection. Hence, we find that the two equations do not have any common root between them.

9.    For what values of k is the system of equations independent? The equations are as follows:
    kx + 5y = 3; 3x + 4y = 9
Solution: For the equations to be independent, conditions is a1a2≠b1b2${\displaystyle \frac{a_1}{a_2}\ne {\displaystyle \frac{b_1}{b_2}}}$
    k3≠54${\displaystyle \frac{k}{3}\ne {\displaystyle \frac{5}{4}}}$ or k ≠154$\ne {\displaystyle \frac{15}{4}}$

10.    Gopi gives Rs. 90 plus one turban as salary to his servant for one year. The servant leaves after 9 months and receives Rs. 65 and the turban. Find the price of the turban.
Solution: Let the price of turban be x.
Thus, for one year the salary = (90 + x)
For 9 months he should earn 34${\displaystyle \frac{3}{4}}$ (90 + x).
Now he gets one turban and Rs. 65.
Thus, 34${\displaystyle \frac{3}{4}}$ (90 + x) = 65 + x  or 270 + 3x = 260 + 4x
    or x = 10

11.    Ranjit went to the market with Rs. 100. If he buys 3 pens and 6 pencils he uses up all his money. On the other hand, if he buys 3 pencils and 6 pens he would fall short by 20%. If he wants to buy equal number of pens and pencils, how many pencils can he buy?
Solution: Let price of a pen = x and pencil = y
3x + 6y = 100    ....(i)
and 6x + 3y = 125    ....(ii)
    Adding (i) and (ii)
9(x + y) = 225
x + y = 25
multiply by 4
4x + 4y = 100
    He can by 4 pens and 4 pencils in Rs. 100.

12.    x3−ax2+3x−b=0$x^3-a{x}^2+3x-b=0$ has one factor as (x - 2). When the equation is divided by (x + 1), it leaves a remainder -12. What are the values of ‘a’ and ‘b’?
Solution: x3−ax2+3x−b=0$x^3-a{x}^2+3x-b=0$ has one factor as (x - 2). So for x = 2 the equation will satisfy or we can say if we substitute the value of x in the equation as 2 it will result into 0.
    8 - 4a + 6 - b = 0 or 4a + b = 14    ....(i)
Now if we say that by dividing the equation by (x + 1) we get the remainder as -12 then if we put x = -1 in the equation then it will result in -12.
    -1 - a - 3 - b = -12 or a + b = 8    ....(ii)
    By solving (I) and (II) a = 2 and b = 6.

13.    For the given equation x9+5x8−x3+7x+2=0,$x^9+5x^8-x^3+7x+2=0,$ how many maximum real roots are possible?
Solution:  f(x) = x9+5x8−x3+7x+2=0$x^9+5x^8-x^3+7x+2=0$
In f(x), there are 2 changes of sign. So, there are two positive roots.
f(-x) = - x9+5x8−x3+7x+2=0$x^9+5x^8-x^3+7x+2=0$
In f(-x), there are 3 changes of sign. So, there are three negative roots.
So, in all there are 5 real roots possible (2 positive and 3 negative) and as degree of the given equation is     9, there are total 9 roots. So, remaining 4 roots will be imaginary.
    Hence answer option is (b).

14.    If  x3−6x2+11x−6=0$x^3-6x^2+11x-6=0$ and x3+3x2−6x−8=0$x^3+3x^2-6x-8=0$  have one common root between them. What is the value of that root?
Solution: There are two equations,
x3−6x2+11x−6=0$x^3-6x^2+11x-6=0$    ....(i)
x3+3x2−6x−8=0$x^3+3x^2-6x-8=0$    ....(ii)
So both (i) and (ii) have one root in common and we know that at the root the value of both the equations is 0. Suppose that common root is x then,
x3−6x2+11x−6=x3+3x2−6x−8⇒9x2−17x−2=0$x^3-6x^2+11x-6=x^3+3x^2-6x-8\Rightarrow 9x^2-17x-2=0$

    (x - 2)(9x + 1) = 0 or x = 2 or -1/9. As by applying the Descartes rule in (i) we can see it cannot have any negative root so x = 2. Hence answer is option (e).
    Why are we getting two values x = 2 and x = -1/9?
    x = 2 and x = -1/9 are the pint of intersections of these two equations out of which x = 2 is the point of intersection at x axis therefore this point (x = 2) is the common root as well.