Q5. If $x^{4}+2x^{3}+ax^{2}+bx+9$ is a perfect square, where $a$ and $b$ are positive real numbers, then the value of $a$ and $b$ are:
- $a=5, b=6$
- $a=6, b=7$
- $a=7, b=7$
- $a=7, b=6$
Hint: Assume the expression is $[(x-\alpha)(x-\beta)]^2$.
Solution:
Given:
$x^{4}+2x^{3}+ax^{2}+bx+9$
Let the expression be $[(x-\alpha)(x-\beta)]^2$
So, the roots are $\alpha, \alpha, \beta, \beta$
Sum of roots = $-\frac{\text{coefficient of } x^3}{\text{coefficient of } x^4}$
$⇒ \alpha+\alpha+\beta+\beta = -2$
$⇒ 2(\alpha+\beta) = -2$
$⇒ \alpha+\beta = -1$
Product of roots = $\frac{\text{constant term}}{\text{coefficient of } x^4}$
$⇒ \alpha \cdot \alpha \cdot \beta \cdot \beta = 9$
$⇒ (\alpha\beta)^2 = 9$
$⇒ \alpha\beta = \pm 3$
Now, sum of roots taken two at a time = coefficient of $x^2$
$a = \alpha^2 + \beta^2 + 4\alpha\beta$
$⇒ a = (\alpha+\beta)^2 + 2\alpha\beta$
$⇒ a = (-1)^2 + 2(\pm 3)$
$⇒ a = 1 \pm 6$
Since $a$ is positive,
$⇒ a = 7$
Now, sum of roots taken three at a time = $-\frac{b}{1}$
$⇒ -b = 2\alpha^2\beta + 2\alpha\beta^2$
$⇒ -b = 2\alpha\beta(\alpha+\beta)$
$⇒ -b = 2(3)(-1)$
$⇒ -b = -6$
$⇒ b = 6$
Hence, the correct answer is $a=7, b=6$.
Q6. If $a^3+3 a^2+3 a=63$, then the value of $a^2+2 a$ is:
- 22
- 19
- 15
- 8
Hint: Use $(a+1)^3=a^3+3a^2+3a+1$.
Solution:
Given:
$a^3+3a^2+3a=63$
Add 1 on both sides:
$a^3+3a^2+3a+1 = 63+1$
$⇒ (a+1)^3 = 64$
Take cube root on both sides:
$a+1 = 4$
$⇒ a = 4 - 1$
$⇒ a = 3$
Now find required value:
$a^2+2a = 3^2 + 2(3)$
$⇒ a^2+2a = 9 + 6$
$⇒ a^2+2a = 15$
Hence, the correct answer is 15.
Q7. If $(x-2)$ is a factor of $x^2+3Qx - 2Q$, then the value of $Q$ is:
- 2
- -2
- 1
- -1
Hint: If $(x-a)$ is a factor, then $P(a)=0$.
Solution:
Given polynomial:
$P(x)=x^2+3Qx-2Q$
Since $(x-2)$ is a factor,
$⇒ P(2)=0$
Substitute $x=2$:
$⇒ 2^2 + 3Q(2) - 2Q = 0$
$⇒ 4 + 6Q - 2Q = 0$
$⇒ 4 + 4Q = 0$
$⇒ 4Q = -4$
$⇒ Q = -1$
Hence, the correct answer is $-1$.
Q8. If $x^{3}+2x^{2}-5x+k$ is divisible by $x+1$, then the value of $k$ is:
- -6
- -1
- 0
- 6
Hint: Use $f(-1)=0$.
Solution:
Given:
$f(x)=x^3+2x^2-5x+k$
Since divisible by $x+1$,
$⇒ f(-1)=0$
Substitute $x=-1$:
$⇒ (-1)^3 + 2(-1)^2 - 5(-1) + k = 0$
$⇒ -1 + 2(1) + 5 + k = 0$
$⇒ -1 + 2 + 5 + k = 0$
$⇒ 6 + k = 0$
$⇒ k = -6$
Hence, the correct answer is $-6$.
Q9. Which of the following equations has 7 as a root?
- $3x^2-6x+2=0$
- $x^2-9x+14=0$
- $x^2-7x+10=0$
- $x^2+3x-12=0$
Hint: Substitute $x=7$ or factorize.
Solution:
Check option 1:
$3x^2-6x+2=0$
Substitute $x=7$:
$⇒ 3(49) - 6(7) + 2 = 147 - 42 + 2 = 107 \neq 0$
Not a root.
Check option 2:
$x^2-9x+14=0$
Factorize:
$⇒ x^2 - 7x - 2x + 14 = 0$
$⇒ x(x-7) -2(x-7) = 0$
$⇒ (x-7)(x-2)=0$
$⇒ x=7$ or $x=2$
Thus, 7 is a root.
Hence, the correct answer is $x^2-9x+14=0$.






