Quadratic Equations jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 11 :

The set of all real numbers $x$ for which $x^{2} -$ $| x + 2 |$ $+ x > 0,$ is

$(- \infty, - 2) \cup (2, \infty)$
$(- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$
$(- \infty, - 1) \cup (1, \infty)$
$(\sqrt{2}, \infty)$

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(2)

$x^{2} -$ $| x + 2 |$ $+ x > 0$

When $x + 2 > 0$ ,

$Then$ $x^{2} - (x + 2) + x > 0,$ $x^{2} - x - 2 + x > 0 \Rightarrow x^{2} - 2 > 0$

$(x - \sqrt{2}) (x + \sqrt{2}) > 0$

$x \in (- 2, - \sqrt{2}) \cup (+ \sqrt{2}, \infty)$ ...(i)

When $x + 2 < 0$ , $| x + 2 |$ $= - (x + 2)$ ,

$\Rightarrow x^{2} + x + 2 + x > 0$ $\forall x \in R$ ...(ii)

From (i) and (ii), we get $x \in (- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$

Q 12 :

Let $α$ and $β$ be the roots of the equation $x^{2} - 6 x - 2 = 0 .$ If $a_{n} = α^{n} - β^{n}$ , for $n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is equal to

3
- 3
6
- 6

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(1)

$α is a root of x^{2} - 6 x - 2 = 0$

$Then α^{2} - 6 α - 2 = 0$

Multiplying by $α^{n}$ , it becomes,

$α^{n + 2} - 6 α^{n + 1} - 2 α^{n} = 0$ ...(i)

Similarly, $β^{n + 2} - 6 β^{n + 1} - 2 β^{n} = 0$ ...(ii)

Subtracting (ii) from (i), we get

$(α^{n + 2} - β^{n + 2}) - 6 (α^{n + 1} - β^{n + 1}) - 2 (α^{n} - β^{n}) = 0$

$i.e., a_{n + 2} - 6 a_{n + 1} - 2 a_{n} = 0$

Thus, $\frac{a_{n + 2} - 2 a_{n}}{2 a_{n + 1}} = 3$

Set $n = 8$ to obtain the desired value $\frac{a_{10} - 2 a_{8}}{2 a_{9}} = 3$

Q 13 :

If one root of a quadratic equation is $\frac{1}{1 + \sqrt{3}},$ then the quadratic equation is

$2 x^{2} + x - 1 = 0$
$2 x^{2} - 2 x - 1 = 0$
$2 x^{2} + 2 x + 1 = 0$
$2 x^{2} + 2 x - 1 = 0$

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(4)

$One root = \frac{1}{1 + \sqrt{3}} = \frac{1 - \sqrt{3}}{- 2} = - \frac{1}{2} + \frac{\sqrt{3}}{2}$

$∴$ $Other root = - \frac{1}{2} - \frac{\sqrt{3}}{2}$

$Sum = - 1, product = \frac{1}{4} - \frac{3}{4} = - \frac{1}{2}$

$∴$ $Required equation is x^{2} + x - \frac{1}{2} = 0$

$or$ $2 x^{2} + 2 x - 1 = 0$

Q 14 :

Let $α$ and $β$ be the roots of the equation, $5 x^{2} + 6 x - 2 = 0 .$ If $S_{n} = α^{n} + β^{n},$ $n = 1, 2, 3, \dots,$ then

$6 S_{6} + 5 S_{5} = 2 S_{4}$
$6 S_{6} + 5 S_{5} + 2 S_{4} = 0$
$5 S_{6} + 6 S_{5} = 2 S_{4}$
$5 S_{6} + 6 S_{5} + 2 S_{4} = 0$

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(3)

$Given α, β are the roots of the equation$ $5 x^{2} + 6 x - 2 = 0 .$

$\Rightarrow α + β = - \frac{6}{5} and α β = - \frac{2}{5}$

$Now, S_{n} = α^{n} + β^{n} \Rightarrow S_{n + 1} = α^{n + 1} + β^{n + 1}$

$S_{n + 1} = (α^{n} + β^{n}) (α + β) - α β (α^{n - 1} + β^{n - 1})$

$= S_{n} (- \frac{6}{5}) - (- \frac{2}{5}) S_{n - 1}$ $[∵ α + β = - \frac{6}{5}, α β = - \frac{2}{5} and α^{n} + β^{n} = S_{n}]$

$\Rightarrow S_{n + 1} = - \frac{6}{5} S_{n} + \frac{2}{5} S_{n - 1}$ $\Rightarrow S_{n + 1} = \frac{2}{5} (S_{n - 1} - 3 S_{n})$

$\Rightarrow 5 S_{n + 1} = 2 (S_{n - 1} - 3 S_{n}),$ $which satisfies option (3).$

Q 15 :

Let $a \in R$ and let $α, β$ be the roots of the equation $x^{2} + 60^{\frac{1}{4}} x + a = 0 .$ If $α^{4} + β^{4} = - 30,$ then the product of all possible values of $a$ is _______.

(45)

$x^{2} + 60^{\frac{1}{4}} x + a = 0$ ...(i)

As we have, $α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}$

$= ({{(α + β)}^{2} - 2 α β)}^{2} - 2 α^{2} β^{2}$ ...(ii)

Now, $α + β = - 60^{\frac{1}{4}} and α β = a$ $[From (i)]$

${(α + β)}^{2} = {(60)}^{\frac{1}{2}} = \sqrt{60} and α^{2} \cdot β^{2} = a^{2}$ ...(iii)

Also, $α^{4} + β^{4} = - 30$ ...(iv)

So, substituting values from (iii) and (iv) in (ii), we get

$a^{2} - 2 \sqrt{60} a + 45 = 0$ $\Rightarrow a^{2} - 4 \sqrt{15} a + 45 = 0$

$\Rightarrow (a - 3 \sqrt{15}) (a - \sqrt{15}) = 0$

$Product of values of a = 3 \sqrt{15} \times \sqrt{15} = 3 \times 15 = 45$

Q 16 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is:

$x^{2} - 195 x + 9466 = 0$
$x^{2} - 180 x + 9506 = 0$
$x^{2} - 195 x + 9506 = 0$
$x^{2} - 190 x + 9466 = 0$

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(3)

$We have,$ $x^{2} + 2 \sqrt{2} x - 1 = 0$

$∴$ $α + β = - 2 \sqrt{2} and α β = - 1$

Now, $α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = {(- 2 \sqrt{2})}^{2} - 2 (- 1)$ $= 8 + 2 = 10$

Also, $α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 {(α β)}^{2} = {(10)}^{2} - 2 (1)$ $= 100 - 2 = 98$

and $α^{6} + β^{6} = {(α^{2} + β^{2})}^{3} - 3 α^{2} β^{2} (α^{2} + β^{2}) = {(10)}^{3} - 3 (1) (10)$ $= 1000 - 30 = 970$

So, $\frac{1}{10} (α^{6} + β^{6}) = 97$

Hence, equation whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6})$ is

$x^{2} - (98 + 97) x + 98 \times 97 = 0$ i.e., $x^{2} - 195 x + 9506 = 0$

Q 17 :

The solution set of the inequation $\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ is

$(- 7, - 4)$
$(- 7, - 4) \cup (4, 1)$
$(- 7, 1)$
$(- 7, - 4) \cup (- 4, 1)$

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(4)

$Given inequation is$

$\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ ...(i)

Since $| x + 4 | > 0$ $\forall x \in R except for x = - 4$

$∵$ $for x = - 4, (i) does not exist.$

$∴$ $x^{2} + 6 x - 7 < 0 for given inequation$

$or x^{2} + 7 x - x - 7 < 0$

$or (x + 7) (x - 1) < 0 or x \in (- 7, 1)$

$∴$ $Solution set of (i) is (- 7, - 4) \cup (- 4, 1)$

Q 18 :

Sum of the roots of the equation $| x - 3 |^{2}$ $+$ $| x - 3 |$ $- 2 = 0$ is equal to

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(3)

$We have,$ $| x - 3 |^{2}$ $+$ $| x - 3 |$ $- 2 = 0$

$\Rightarrow$ $| x - 3 |$ $= 1 \Rightarrow x - 3 = \pm 1 \Rightarrow x = 2, 4$

Q 19 :

If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0,$ then $α^{101} + β^{107}$ is equal to

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- 1
0

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(1)

$We have,$ $x^{2} - x + 1 = 0$

$\Rightarrow x = \frac{- 1 (- 1) \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{3} i}{2}$

$∴$ $Let α = \frac{1 + \sqrt{3} i}{2} and β = \frac{1 - \sqrt{3} i}{2}$

$\Rightarrow α = - ω^{2} and β = - ω, where ω = \frac{- 1 + \sqrt{3} i}{2}$

$and ω^{2} = \frac{- 1 - \sqrt{3} i}{2}$

$Now, α^{101} + β^{107} = {(- ω^{2})}^{101} + {(- ω)}^{107}$

$= - ω^{202} + {(- ω)}^{107}$

$= - ω - ω^{2} = - (ω + ω^{2}) = - (- 1)$ $(∵ 1 + ω + ω^{2} = 0)$

$= 1$

Q 20 :

The sum of the roots of the equation $x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{4} (10 - 2^{- x}) = 0,$ is

$\log_{2} 14$
$\log_{2} 12$
$\log_{2} 13$
$\log_{2} 11$

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(4)

$We have, x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{2} (10 - 2^{- x}) = 0$

$\Rightarrow (x + 1) + \log_{2} (\frac{10 \cdot 2^{x} - 1}{2^{x}}) - \log_{2} {(3 + 2^{x})}^{2} = 0$

$\Rightarrow (x + 1) + \log_{2} [\frac{10 \cdot 2^{x} - 1}{{(3 + 2^{x})}^{2}}] - \log_{2} 2^{x} = 0$

$\Rightarrow (x + 1) + \log_{2} [\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}] - x = 0$

$\Rightarrow \log_{2} (\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}) = - 1$

$\Rightarrow (\frac{10 \cdot 2^{x} - 1}{9 + {(2^{x})}^{2} + 6 \cdot 2^{x}}) = \frac{1}{2}$

$\Rightarrow 20 \cdot 2^{x} - 2 = 9 + {(2^{x})}^{2} + 6 \cdot 2^{x} \Rightarrow {(2^{x})}^{2} - 14 \cdot 2^{x} + 11 = 0$

$\Rightarrow 2^{x} = \frac{14 \pm \sqrt{{(14)}^{2} - 4 (11) (1)}}{2} = \frac{14 \pm \sqrt{152}}{2}$

$\Rightarrow 2^{x} = 7 \pm \sqrt{38}$

$\Rightarrow x = \log_{2} (7 + \sqrt{38}) or x = \log_{2} (7 - \sqrt{38})$

$∴$ $Sum of roots = \log_{2} (7 + \sqrt{38}) + \log_{2} (7 - \sqrt{38})$

$= \log_{2} [49 - 38] = \log_{2} 11$

Q 21 :

The least positive value of $' a'$ for which the equation $2 x^{2} + (a - 10) x + \frac{33}{2} = 2 a$ has real roots is _______.

(8)

$Since, the given equation 2 x^{2} + (a - 10) x + \frac{33}{2} = 2 a$ $has real roots.$

$∴$ $D \geq 0 \Rightarrow {(a - 10)}^{2} - 4 (2) (\frac{33}{2} - 2 a) \geq 0$

$\Rightarrow$ $a^{2} + 100 - 20 a - 132 + 16 a \geq 0$

$\Rightarrow$ $a^{2} - 4 a - 32 \geq 0 \Rightarrow (a - 8) (a + 4) \geq 0$

$\Rightarrow$ $a \in (- \infty, - 4] \cup [8, \infty)$

Q 22 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is ________.

(1)

$We have,$ $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$

Taking log both sides and take $\log_{5} x = t$ , we get

$t (16 t^{3} - 68 t) = - 16$

$\Rightarrow$ $16 t^{4} - 68 t^{2} + 16 = 0 \Rightarrow 4 t^{4} - 17 t^{2} + 4 = 0$

$∴$ $t_{1} + t_{2} + t_{3} + t_{4} = 0$

$\Rightarrow$ $\log_{5} x_{1} + \log_{5} x_{2} + \log_{5} x_{3} + \log_{5} x_{4} = 0 \Rightarrow x_{1} x_{2} x_{3} x_{4} = 1$

Q 23 :

$p (x) = x^{4} + a x^{3} + b x^{2} + c x + d$ , where $p (1) = 10$ , $p (2) = 20$ , $p (3) = 30$ , then the value of $\frac{p (12) + p (- 8)}{10} - 1980$ is ________.

(4)

$Let$ $f (x) = p (x) - 10 x$

$∴$ $f (1) = f (2) = f (3) = 0$

$∴$ $p (x) = (x - 1) (x - 2) (x - 3) (x - λ) + 10 x$

$p (12) = 11 \times 10 \times 9 (12 - λ) + 120$

$p (- 8) = (- 9) \times (- 10) \times (- 11) (- 8 - λ) - 80$

$∴$ $p (12) + p (- 8) = 11 \times 10 \times 9 (12 - λ + 8 + λ) + 40$

$= 11 \times 10 \times 9 \times 20 + 40 = 40 \times 496$

$or$ $\frac{p (12) + p (- 8)}{10} = 1984$

$∴$ $\frac{p (12) + p (- 8)}{10} - 1980 = 4$

Q 24 :

Let $α_{1}, α_{2}, \dots, α_{8}$ be roots of the equation $1 + z + z^{2} + \dots + z^{8} = 0$ . On throwing three dice simultaneously, sum of the numbers appearing on them is $n$ . The probability such that $\sum_{i = 1}^{8} {(α_{i})}^{n} = 8$ is $\frac{a}{b}$ (where $a$ and $b$ are co-prime) then the value $| 7 a - b |$ is:

22
16
26
17

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(4)

$' n'$ $should be multiple of 9 for favourable event.$

$So, x_{1} + x_{2} + x_{3} = 9 or 18$

$If x_{1} + x_{2} + x_{3} = 9, 1 \leq x_{i} \leq 6$

$No. of solution of (1) = coefficient of x^{9} in$

${(x + x^{2} + \dots + x^{6})}^{3} = x^{3} {(1 - x^{6})}^{3} {(1 - x)}^{- 3}$

$= coefficient of x^{6} in (1 - 3 x^{6}) {(1 - x)}^{- 3}$

$= {}^{8}{C_{6}} + (- 3) \times 1 = 25$

$And, there is only one solution of$ $x_{1} + x_{2} + x_{3} = 18, 1 \leq x_{i} \leq 6$

$Hence, n (E) = 26, Also n (S) = 6^{3}$

$\Rightarrow P (E) = \frac{13}{108}$

Q 25 :

Let $α$ , $β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}$ , $2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then m is equal to [2026]

–135
–567
135
567

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(4)

$α$ , $β$ are the roots of $x^{2} - 3 x + r = 0$

$∴ α + β = 3 and α β = r$

$\frac{α}{2}, 2 β$ are the roots of $x^{2} + 3 x + r = 0$

$∴ \frac{α}{2} + 2 β = - 3 and α β = r$

Now, $α + β = 3 and α + 4 β = - 6$

$\Rightarrow β = - 3, α = 6 and r = - 18$

Now, roots are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2} = 12 - 3 - 36, 6 + 6 + 9 = - 27, 21$

$∴$ Roots of $x^{2} + 6 x - m = 0$ are –27, 21

$∴$ (–27)(21) = –m $\Rightarrow$ m = (27)(21) $\Rightarrow$ m = 567

Q 26 :

Let $a, b \in C$ . Let $α, β$ be the roots of the equation $x^{2} + a x + b = 0$ . If $β - α = \sqrt{11}$ and $β^{2} - α^{2} = 3 i \sqrt{11}$ , then ${(β^{3} - α^{3})}^{2}$ is equal to : [2026]

160
176
194
187

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(2)

Given, $x^{2} + a x + b = 0$ , $β - α = \sqrt{11}$ and $β^{2} - α^{2} = 3 i \sqrt{11}$

$\Rightarrow (β + α) (β - α) = 3 i \sqrt{11}$

$\Rightarrow β + α = 3 i$

$\Rightarrow β^{2} + α^{2} + 2 α β = - 9$ ... (i)

$∵ β - α = \sqrt{11}$

$\Rightarrow β^{2} + α^{2} - 2 α β = 11$ ... (ii)

From equation (i) and (ii),

$4 α β = - 20 \Rightarrow α β = - 5$

${(β^{3} - α^{3})}^{2} = {(β - α)}^{2} {(β^{2} + α^{2} + α β)}^{2}$

$= {(β - α)}^{2} {({(β + α)}^{2} - α β)}^{2}$

$= {(\sqrt{11})}^{2} {(- 9 + 5)}^{2}$

$= 11 \times 6 = 176$

Q 27 :

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is _____

(36)

$x^{4} - 3 x^{3} - x^{2} - x^{2} + 3 x + 1 = 0$

$(x^{2} - 1) (x^{2} - 3 x - 1) = 0$

$Let the roots of x^{2} - 3 x - 1 = 0 be α and β . Other two roots of the given equation are 1 and - 1 .$

$So, sum of cubes of roots = 1^{3} + {(- 1)}^{3} + α^{3} + β^{3}$

$= {(α + β)}^{3} - 3 α β (α + β) = {(3)}^{3} - 3 (- 1) (3) = 36$

Q 28 :

If $α, β$ are the roots of $x^{2} - 6 x - 2 = 0$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ . Then the value of $\frac{a_{12} - 2 a_{10}}{3 a_{11}}$ is _________

2
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4
3

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(1)

$α^{12} - 6 α^{11} - 2 α^{10} = 0,$ $α^{12} - 2 α^{10} = 6 α^{11},$ $β^{12} - 2 β^{10} = 6 β^{11}$

Topic Question Set

Q 11 :

The set of all real numbers $x$ for which $x^{2} -$ $| x + 2 |$ $+ x > 0,$ is

Q 12 :

Let $α$ and $β$ be the roots of the equation $x^{2} - 6 x - 2 = 0 .$ If $a_{n} = α^{n} - β^{n}$ , for $n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is equal to

Q 13 :

If one root of a quadratic equation is $\frac{1}{1 + \sqrt{3}},$ then the quadratic equation is

Q 14 :

Let $α$ and $β$ be the roots of the equation, $5 x^{2} + 6 x - 2 = 0 .$ If $S_{n} = α^{n} + β^{n},$ $n = 1, 2, 3, \dots,$ then

Q 15 :

Let $a \in R$ and let $α, β$ be the roots of the equation $x^{2} + 60^{\frac{1}{4}} x + a = 0 .$ If $α^{4} + β^{4} = - 30,$ then the product of all possible values of $a$ is _______.

Q 16 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is:

Q 17 :

The solution set of the inequation $\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ is

Q 18 :

Sum of the roots of the equation $| x - 3 |^{2}$ $+$ $| x - 3 |$ $- 2 = 0$ is equal to

Q 19 :

If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0,$ then $α^{101} + β^{107}$ is equal to

Q 20 :

The sum of the roots of the equation $x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{4} (10 - 2^{- x}) = 0,$ is

Q 21 :

The least positive value of $' a'$ for which the equation $2 x^{2} + (a - 10) x + \frac{33}{2} = 2 a$ has real roots is _______.

Q 22 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is ________.

Q 23 :

$p (x) = x^{4} + a x^{3} + b x^{2} + c x + d$ , where $p (1) = 10$ , $p (2) = 20$ , $p (3) = 30$ , then the value of $\frac{p (12) + p (- 8)}{10} - 1980$ is ________.

Q 25 :

Let $α$ , $β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}$ , $2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then m is equal to [2026]

Q 26 :

Let $a, b \in C$ . Let $α, β$ be the roots of the equation $x^{2} + a x + b = 0$ . If $β - α = \sqrt{11}$ and $β^{2} - α^{2} = 3 i \sqrt{11}$ , then ${(β^{3} - α^{3})}^{2}$ is equal to : [2026]

Q 27 :

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is _____

Q 28 :

If $α, β$ are the roots of $x^{2} - 6 x - 2 = 0$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ . Then the value of $\frac{a_{12} - 2 a_{10}}{3 a_{11}}$ is _________

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Topic Question Set

Q 11 : The set of all real numbers x for which x2-|x+2|+x>0, is

Q 12 : Let α and β be the roots of the equation x2-6x-2=0. If an=αn-βn, for n≥1, then the value of a10-2a82a9 is equal to

Q 13 : If one root of a quadratic equation is 11+3, then the quadratic equation is

Q 14 : Let α and β be the roots of the equation, 5x2+6x-2=0. If Sn=αn+βn, n=1,2,3,…, then

Q 15 : Let a∈R and let α, β be the roots of the equation x2+6014x+a=0. If α4+β4=-30, then the product of all possible values of a is _______.

Q 16 : Let α, β be the roots of the equation x2+22x-1=0. The quadratic equation, whose roots are α4+β4 and 110(α6+β6), is:

Q 17 : The solution set of the inequation x2+6x-7|x+4|<0 is

Q 18 : Sum of the roots of the equation |x-3|2+|x-3|-2=0 is equal to

Q 19 : If α, β∈C are the distinct roots of the equation x2-x+1=0, then α101+β107 is equal to

Q 20 : The sum of the roots of the equation x+1-2log2(3+2x)+2log4(10-2-x)=0, is

Q 21 : The least positive value of 'a' for which the equation 2x2+(a-10)x+332=2a has real roots is _______.

Q 22 : The product of all positive real values of x satisfying the equation x(16(log5x)3-68log5x)=5-16 is ________.

Q 23 : p(x)=x4+ax3+bx2+cx+d, where p(1)=10, p(2)=20, p(3)=30, then the value of p(12)+p(-8)10-1980 is ________.

Q 24 : Let α1,α2,…,α8 be roots of the equation 1+z+z2+⋯+z8=0. On throwing three dice simultaneously, sum of the numbers appearing on them is n. The probability such that ∑i=18(αi)n=8 is ab(where a and b are co-prime) then the value |7a-b| is:

Q 25 : Let α, β be the roots of the equation x2–3x+r=0, and α2, 2β be the roots of the equation x2+3x+r=0. If the roots of the equation x2+6x=m are 2α+β+2r and α–2β–r2, then m is equal to [2026]

Q 26 : Let a,b∈C. Let α,β be the roots of the equation x2+ax+b=0. If β–α=11 and β2–α2=3i11, then (β3–α3)2 is equal to : [2026]

Q 27 : The sum of the cubes of all the roots of the equation x4−3x3−2x2+3x+1=0 is _____

Q 28 : If α,β are the roots of x2-6x-2=0. If an=αn-βn for n≥1. Then the value of a12-2a103a11 is _________

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Q 11 :

The set of all real numbers $x$ for which $x^{2} -$ $| x + 2 |$ $+ x > 0,$ is

Q 12 :

Let $α$ and $β$ be the roots of the equation $x^{2} - 6 x - 2 = 0 .$ If $a_{n} = α^{n} - β^{n}$ , for $n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is equal to

Q 13 :

If one root of a quadratic equation is $\frac{1}{1 + \sqrt{3}},$ then the quadratic equation is

Q 14 :

Let $α$ and $β$ be the roots of the equation, $5 x^{2} + 6 x - 2 = 0 .$ If $S_{n} = α^{n} + β^{n},$ $n = 1, 2, 3, \dots,$ then

Q 15 :

Let $a \in R$ and let $α, β$ be the roots of the equation $x^{2} + 60^{\frac{1}{4}} x + a = 0 .$ If $α^{4} + β^{4} = - 30,$ then the product of all possible values of $a$ is _______.

Q 16 :

Let $α, β$ be the roots of the equation $x^{2} + 2 \sqrt{2} x - 1 = 0 .$ The quadratic equation, whose roots are $α^{4} + β^{4}$ and $\frac{1}{10} (α^{6} + β^{6}),$ is:

Q 17 :

The solution set of the inequation $\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ is

Q 18 :

Sum of the roots of the equation $| x - 3 |^{2}$ $+$ $| x - 3 |$ $- 2 = 0$ is equal to

Q 19 :

If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0,$ then $α^{101} + β^{107}$ is equal to

Q 20 :

The sum of the roots of the equation $x + 1 - 2 \log_{2} (3 + 2^{x}) + 2 \log_{4} (10 - 2^{- x}) = 0,$ is

Q 21 :

The least positive value of $' a'$ for which the equation $2 x^{2} + (a - 10) x + \frac{33}{2} = 2 a$ has real roots is _______.

Q 22 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is ________.

Q 23 :

$p (x) = x^{4} + a x^{3} + b x^{2} + c x + d$ , where $p (1) = 10$ , $p (2) = 20$ , $p (3) = 30$ , then the value of $\frac{p (12) + p (- 8)}{10} - 1980$ is ________.

Q 25 :

Let $α$ , $β$ be the roots of the equation $x^{2} - 3 x + r = 0$ , and $\frac{α}{2}$ , $2 β$ be the roots of the equation $x^{2} + 3 x + r = 0$ . If the roots of the equation $x^{2} + 6 x = m$ are $2 α + β + 2 r$ and $α - 2 β - \frac{r}{2}$ , then m is equal to [2026]

Q 26 :

Let $a, b \in C$ . Let $α, β$ be the roots of the equation $x^{2} + a x + b = 0$ . If $β - α = \sqrt{11}$ and $β^{2} - α^{2} = 3 i \sqrt{11}$ , then ${(β^{3} - α^{3})}^{2}$ is equal to : [2026]

Q 27 :

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is _____

Q 28 :

If $α, β$ are the roots of $x^{2} - 6 x - 2 = 0$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ . Then the value of $\frac{a_{12} - 2 a_{10}}{3 a_{11}}$ is _________