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

Q 1 :

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b},$ is [2024]

$2 x^{2} + 11 x + 12 = 0$
$x^{2} + 8 x + 12 = 0$
$4 x^{2} + 14 x + 12 = 0$
$x^{2} + 10 x + 16 = 0$

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(2)

2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$

$∴$ Sum of roots = $\frac{- b}{a} = 8 \Rightarrow b = - 8 a$

Product of roots = $\frac{1}{a} = 12 \Rightarrow a = \frac{1}{12}$

$∴$ $b = \frac{- 2}{3} .$ Now, $\frac{1}{2 a + b} = \frac{1}{2 \times \frac{1}{12} - \frac{2}{3}} = - 2$

and $\frac{1}{6 a + b} = \frac{1}{6 \times \frac{1}{12} - \frac{2}{3}} = - 61$

$∴$ Required equation is $(x + 2) (x + 6) = 0$

i.e., $x^{2} + 8 x + 12 = 0$

Q 2 :

Let $α, β$ be roots of $x^{2} + \sqrt{2} x - 8 = 0 .$ If $U_{n} = α^{n} + β^{n},$ then $\frac{U_{10} + \sqrt{2} U_{9}}{2 U_{8}}$ is equal to __________ . [2024]

(4)

Given that $α, β$ are roots of $x^{2} + \sqrt{2} x - 8 = 0$

$\Rightarrow α^{2} + \sqrt{2} α - 8 = 0$ and $β^{2} + \sqrt{2} β - 8 = 0$

$\Rightarrow α^{2} + \sqrt{2} α = 8$ and $β^{2} + \sqrt{2} β = 8$ ...(i)

Now, $\frac{U_{10} + \sqrt{2} U_{9}}{2 U_{8}} = \frac{α^{10} + β^{10} + \sqrt{2} α^{9} + \sqrt{2} β^{9}}{2 (α^{8} + β^{8})}$

$= \frac{α^{9} (α + \sqrt{2}) + β^{9} (β + \sqrt{2})}{2 (α^{8} + β^{8})}$

$= \frac{α^{9} (\frac{8}{α}) + β^{9} (\frac{8}{β})}{2 (α^{8} + β^{8})}$ [Using (i)]

$= \frac{8 (α^{8} + β^{8})}{2 (α^{8} + β^{8})} = 4$

Q 3 :

Let $α, β \in N$ be roots of the equation $x^{2} - 70 x + λ = 0,$ where $\frac{λ}{2}, \frac{λ}{3} \notin N .$ If $λ$ assumes the minimum possible value, then $\frac{(\sqrt{α - 1} + \sqrt{β - 1}) (λ + 35)}{| α - β |}$ is equal to _______ . [2024]

(60)

Given $α$ and $β$ are roots of $x^{2} - 70 x + λ = 0$

$\Rightarrow α + β = 70$ and $α β = λ$

We need to minimise $λ$

Let $f = α β = α (70 - α) .$

So, $f (α) = α (70 - α)$

0 and 70 are roots of $f .$

$f$ has a minimum value at $α = 0 or 70$

Now, $\frac{λ}{2}, \frac{λ}{3} \notin N$

It means $α$ and $β$ should not be multiples of 2 and 3.

$\Rightarrow For α = 1, β = 69$ which is a multiple of 3 similarly we can't take $α = 2, 3, 4 .$

So, for $α = 5, β = 65$

$\Rightarrow λ = α β = 325$ is the minimum value of $λ$ satisfying all the conditions.

Now, $\frac{(\sqrt{α - 1} + \sqrt{β - 1}) (λ + 35)}{| α - β |} = \frac{(\sqrt{4} + \sqrt{64}) (360)}{| - 60 |} = 60$

Q 4 :

Let $P_{n} = α^{n} + β^{n}, n \in N$ . If $P_{10} = 123, P_{9} = 76, P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is: [2025]

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

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(1)

Since, $P_{10} = 123 \Rightarrow α^{10} + β^{10} = 123$

Since, $P_{9} = 76 \Rightarrow α^{9} + β^{9} = 76$

Since, $P_{8} = 47 \Rightarrow α^{8} + β^{8} = 47$

Since, $P_{1} = 1 \Rightarrow α + β = 1$

$∵ P_{1} \cdot P_{9} = P_{10} + α β P_{8}$

$\Rightarrow 1 \times 76 = 123 + α β (47)$

$\Rightarrow α β = - 1$

Also, $α + β = 1$

Clearly, $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{1}{- 1} = - 1, \frac{1}{α β} = - 1$

Thus, the required quadratic equation is $x^{2} + x - 1 = 0$ .

Q 5 :

Let $α$ and $β$ be the roots of $x^{2} + \sqrt{3} x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$ is equal to [2025]

4
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5
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(3)

We have, $x^{2} + 3 x - 1 = 0$ ... (i)

$\Rightarrow x^{2} - 1 = - 3 x$

$γ$ and $δ$ are the roots of equation (i)

Now, $Q_{n} = γ^{n} + δ^{n}$

$Q_{25} - Q_{23} = (γ^{25} - γ^{23}) + (δ^{25} - δ^{23})$

$= γ^{23} (γ^{2} - 1) + δ^{23} (δ^{2} - 1)$

$= γ^{23} (- 3 γ) + δ^{23} (- 3 δ) = - 3 [γ^{24} + δ^{24}] = - 3 Q_{24}$

$\Rightarrow \frac{Q_{25} - Q_{23}}{Q_{24}} = (- 3)$

Similarly, $x^{2} + \sqrt{3} x - 16 = 0$ ... (ii)

$\Rightarrow x^{2} + \sqrt{3} x = 16$

$α$ and $β$ are the roots of equation (ii).

Now, $P_{n} = α^{n} + β^{n}$

$\Rightarrow P_{25} + \sqrt{3} P_{24} = (α^{25} + \sqrt{3} α^{24}) + (β^{25} + \sqrt{3} β^{24})$

$= α^{23} (α^{2} + \sqrt{3} α) + β^{23} (β^{2} + \sqrt{3} β)$

$= α^{23} (16) + 16 β^{23}$

$\Rightarrow \frac{P_{25} + \sqrt{3} P_{24}}{2 \cdot P_{23}} = \frac{16 (α^{23} + β^{23})}{2 (α^{23} + β^{23})} = 8$

Now, $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{(Q_{25} - Q_{23})}{Q_{24}} = 8 + (- 3) = 5$ .

Q 6 :

Let $z \in C$ be such that $\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is [2025]

–19 + 2i
19 + 2i
19 – 2i
–19 –2i

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(4)

We have, $\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i$

$\Rightarrow z^{2} + 3 i = z (2 + 3 i) + (2 + 3 i) (- 2 + i)$

$\Rightarrow z^{2} + 3 i = z (2 + 3 i) - 7 - 4 i$

$\Rightarrow z^{2} - z (2 + 3 i) + 7 + 7 i = 0$

It is a quadratic equation in z

$∴$ Sum of roots = $z_{1} + z_{2} = 2 + 3 i$

Product of roots = $z_{1} z_{2} = 7 + 7 i$

Now, $z_{1}^{2} + z_{2}^{2} = {(z_{1} + z_{2})}^{2} - 2 z_{1} z_{2}$

= ${(2 + 3 i)}^{2} - 2 (7 + 7 i)$

= 4 – 9 + 12i – 14 –14i

= –19 – 2i.

Q 7 :

If the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ has equal roots, where a + c = 15 and $b = \frac{36}{5}$ , then $a^{2} + c^{2}$ is equal to __________. [2025]

(117)

Since, $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ has equal roots

Put x = 1 in given equation.

$∴$ ab – ac + bc – ab + ac – bc = 0

$∴$ Both roots of the equation is 1.

$∴ 1 + 1 = \frac{- b (c - a)}{a (b - c)}$

$\Rightarrow 2 a b - 2 a c = - b c + a b$

$\Rightarrow 2 a c = b c + a b \Rightarrow 2 a c = b (a + c)$

$\Rightarrow 2 a c = \frac{36}{5} \times 15 = 108$

Also, a + c = 15 [Given]

$∴ a^{2} + c^{2} + 2 a c = 225$

$\Rightarrow a^{2} + c^{2} = 225 - 108 \Rightarrow a^{2} + c^{2} = 117$ .

Q 8 :

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ , where $i = \sqrt{- 1}$ , then $16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$ is equal to [2025]

441
409
312
398

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(1)

We have, $2 z^{2} - 3 z - 2 i = 0$ ... (i)

$\Rightarrow 2 (z - \frac{i}{z}) = 3$

$\Rightarrow α - \frac{i}{α} = \frac{3}{2}$ [ $∵ α$ is a root of equation (i)]

$\Rightarrow α^{2} - \frac{1}{α^{2}} - 2 i = \frac{9}{4} \Rightarrow α^{2} - \frac{1}{α^{2}} = \frac{9}{4} + 2 i$

$\Rightarrow α^{4} + \frac{1}{α^{4}} - 2 = \frac{81}{16} - 4 + 9 i \Rightarrow α^{4} + \frac{1}{α^{4}} = \frac{49}{16} + 9 i$

Similarly, $β^{4} + \frac{1}{β^{4}} = \frac{49}{16} + 9 i$

$\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}} = \frac{α^{15} (α^{4} + \frac{1}{α^{4}}) + β^{15} (β^{4} + \frac{1}{β^{4}})}{α^{15} + β^{15}}$

$= \frac{(\frac{49}{16} + 9 i) (α^{15} + β^{15})}{α^{15} + β^{15}} = \frac{49}{16} + 9 i$

$∴ 16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$

$= 16 \times \frac{49}{16} \times 9 = 441$ .

Q 9 :

Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2, satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

1
6
9
7

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(2)

We have, $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$

$\Rightarrow f (x) f (\frac{1}{x}) - f (x) = f (\frac{1}{x})$

$\Rightarrow f (x) = \frac{f (\frac{1}{x})}{f (\frac{1}{x}) - 1}$ ... (i)

Also, $f (x) f (\frac{1}{x}) - f (\frac{1}{x}) = f (x)$

$\Rightarrow f (\frac{1}{x}) = \frac{f (x)}{f (x) - 1}$ ...(ii)

On multiplying (i) and (ii), we get

$f (x) f (\frac{1}{x}) = \frac{f (\frac{1}{x}) f (x)}{{f (\frac{1}{x}) - 1} {f (x) - 1}}$

$\Rightarrow (f (\frac{1}{x}) - 1) (f (x) - 1) = 1$ ... (iii)

Since, f(x) is polynomial function, so f(x) – 1 and $f (\frac{1}{x}) - 1$ are reciprocal of each other. Also x and $\frac{1}{x}$ are reciprocal of each other.

So, (iii) hold only if, $f (x) - 1 = \pm x^{n}, n \in R$

$∴ f (x) = 1 \pm x^{n}$

As, f(x) is a polynomial of degree 2 and range $\in (- \infty, 1)$

$∴ f (x) = 1 - x^{2}$

Now, $f (K) = - 2 K \Rightarrow 1 - k^{2} = - 2 K \Rightarrow K^{2} - 2 K - 1 = 0$

Let its roots i.e., possible values of K be $α$ and $β$

$∴ α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = 4 - 2 (- 1) = 6$ .

Q 10 :

If $α$ is a root of the equation $x^{2} + x + 1 = 0$ and $\sum_{k = 1}^{n} {(α^{k} + \frac{1}{α^{k}})}^{2} = 20$ , then n is equal to __________. [2025]

(11)

We have, $α$ is root of equation $x^{2} + x + 1 = 0$ .

Let $α = ω$ .

Then, $\sum_{k = 1}^{n} {(ω^{k} + \frac{1}{ω^{k}})}^{2} = \sum_{k = 1}^{n} (ω^{2 k} + \frac{1}{ω^{2 k}} + 2)$

$= \sum_{k = 1}^{n} (ω^{2 k} + ω^{k} + 2) (∵ ω^{3 k} = 1)$

But $\sum_{k = 1}^{n} (ω^{2 k} + ω^{k} + 2) = 20$

$\Rightarrow (ω^{2} + ω^{4} + . . . + ω^{2 n}) + (ω + ω^{2} + . . . + ω^{n}) + 2 n = 20$ .

If n = 3m, m $\in$ $I$

Then, 0 + 0 + 2n = 20

$\Rightarrow n = 10$ (not satisfy as m is integer).

If n = 3m + 1, then $ω^{2} + ω + 2 n = 20$ .

$\Rightarrow n = \frac{21}{2}$ (not possible)

If n = 3m + 2, then $(ω^{2} + ω^{4}) + (ω + ω^{2}) + 2 n = 20$ .

$\Rightarrow 2 (ω^{2} + ω + n) = 20$ .

$\Rightarrow n = 10 + 1 = 11$ .

$∴$ For n = 11 satisfies n = 3m + 2.

Q 11 :

The sum of all the roots of the equation $| x^{2} - 8 x + 15 |$ $- 2 x + 7 = 0$ is [2023]

$11 + \sqrt{3}$
$11 - \sqrt{3}$
$9 + \sqrt{3}$
$9 - \sqrt{3}$

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(3)

Given equation is $| x^{2} - 8 x + 15 |$ $- 2 x + 7 = 0$

$x^{2} - 8 x + 15 \geq 0$

$\Rightarrow (x - 3) (x - 5) \geq 0 \Rightarrow x \leq 3 or x \geq 5$

$When x \leq 3 or x \geq 5$

$x^{2} - 8 x + 15 - 2 x + 7 = 0 \Rightarrow x^{2} - 10 x + 22 = 0$

$∴$ $x = 5 + \sqrt{3}$ $[5 - \sqrt{3} is rejected]$

$When 3 < x < 5$

$- (x^{2} - 8 x + 15) - 2 x + 7 = 0 \Rightarrow x^{2} - 8 x + 15 + 2 x - 7 = 0$

$\Rightarrow x^{2} - 6 x + 8 = 0 \Rightarrow (x - 4) (x - 2) = 0$

$\Rightarrow x = 4 [2 does not lie between 3 and 5, so rejected]$

$Sum of roots = 5 + \sqrt{3} + 4 = 9 + \sqrt{3}$

Q 12 :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ is [2023]

$x^{2} - 19 x + 90 = 0$
$x^{2} - 20 x + 99 = 0$
$x^{2} - 18 x + 80 = 0$
$x^{2} - 22 x + 120 = 0$

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(2)

Let A(2, 3), B(1, 2), and C(3, 1) be the vertices of a triangle.

$Now, B C = \sqrt{{(3 - 1)}^{2} + {(1 - 2)}^{2}} and A C = \sqrt{{(3 - 2)}^{2} + {(1 - 3)}^{2}}$

$∴ ∆ A B C is an isosceles triangle.$

$∴ The median and altitude passing through (α, β) are the same.$

$Now, slope of A B = 1$

$∴ slope of C D = - 1$

$∴ Equation of line passing through (a, b) and having slope - 1 is given by a + b = c, where c is some constant.$

$Now, line a + b = c will pass through the midpoint of A B as ∆ A B C is isosceles.$

$\Rightarrow \frac{3}{2} + \frac{5}{2} = c = 4 \Rightarrow α + β = 4 ...(i)$

$Now, we need to find the equation where roots are α + 4 β and 4 α + β .$

$Sum of roots = (α + 4 β) + (4 α + β) = 5 α + 5 β = 5 (α + β)$

$= 5 \times 4 [Using (i)]$

$= 20$

$From the options, we can see that the equation x^{2} - 20 x + 99 = 0 has sum of roots as 20 .$

Q 13 :

Let $p, q \in R$ and ${(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q), i = \sqrt{- 1} .$ The $p + q + q^{2}$ and $p - q + q^{2}$ are roots of the equation [2023]

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

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(1)

$Given: {(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q)$

$\Rightarrow {(2 e^{- i π / 3})}^{200} = 2^{199} (p + i q)$

$\Rightarrow 2^{200} {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{200} = 2^{199} (p + i q)$

$2 (\cos \frac{200 π}{3} - i \sin \frac{200 π}{3}) = p + i q$

$∴ p = - 1, q = - \sqrt{3}$

$\Rightarrow p + q + q^{2} = - 1 - \sqrt{3} + 3 = 2 - \sqrt{3}$

$and p - q + q^{2} = - 1 + \sqrt{3} + 3 = 2 + \sqrt{3}$

$Sum of roots = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$

$Product of roots = (2 + \sqrt{3}) (2 - \sqrt{3}) = 4 - 3 = 1$

$Required quadratic equation is x^{2} - 4 x + 1 = 0$

Q 14 :

Let $λ \neq 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0 .$ Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation [2023]

$7 x^{2} + 245 x - 250 = 0$
$49 x^{2} + 245 x + 250 = 0$
$49 x^{2} - 245 x + 250 = 0$
$7 x^{2} - 245 x + 250 = 0$

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(3)

$Root α will satisfy the equations,$

$14 α^{2} - 31 α + 3 λ = 0$ $...(i)$

and $35 α^{2} - 53 α + 4 λ = 0$ $...(ii)$

$Now, equation (i) \times 5 - (ii) \times 2 \Rightarrow 49 α - 7 λ = 0 \Rightarrow α = \frac{λ}{7}$

$Put α = \frac{λ}{7} in equation (i): 14 {(\frac{λ}{7})}^{2} - 31 (\frac{λ}{7}) + 3 λ = 0$

$\Rightarrow λ = 0, 5$

$Since λ \neq 0, so λ = 5$

$∴$ $α = \frac{5}{7} \Rightarrow 7 α - 5 = 0$

$So, other root β = \frac{3}{2} \Rightarrow γ = \frac{4}{5}$

$Then, \frac{3 α}{β} = \frac{3 \times \frac{5}{7}}{\frac{3}{2}} = \frac{30}{21} = \frac{10}{7}, and \frac{4 α}{γ} = \frac{4 \times \frac{5}{7}}{\frac{4}{5}} = \frac{25}{7}$

$So, equation can be written as:$

$x^{2} - (\frac{10}{7} + \frac{25}{7}) x + \frac{10}{7} \times \frac{25}{7} = 0$

$\Rightarrow x^{2} - 5 x + \frac{250}{49} = 0 \Rightarrow 49 x^{2} - 245 x + 250 = 0$

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 __________ . [2023]

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

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

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 $α_{1}, α_{2}, \dots, α_{7}$ be the roots of the equation $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$ and $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$ . Then $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6}$ is equal to ______. [2023]

(9)

Given equation is, $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$

$\Rightarrow x (x^{6} + 3 x^{4} - 13 x^{2} - 15) = 0$

Here, $x = 0$ is one of the roots. Replacing $x^{2} = t$

So, $t^{3} + 3 t^{2} - 13 t - 15 = 0 \Rightarrow (t - 3) (t^{2} + 6 t + 5) = 0$

So, $t = 3, t = - 1, t = - 5$

Now, $x^{2} = 3, x^{2} = - 1, x^{2} = - 5$

$\Rightarrow x = \pm \sqrt{3}, x = \pm i, x = \pm \sqrt{5} i$

From the given condition, $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$

We can say that $| α_{7} |$ $= 0$ and $| α_{1} |$ $= \sqrt{5} =$ $| α_{2} |$

and $| α_{4} |$ $= \sqrt{3} =$ $| α_{3} |$ , and $| α_{5} |$ $= 1 =$ $| α_{6} |$

So, we have,

$α_{5} = i, α_{6} = - i, α_{3} = \sqrt{3}, α_{4} = - \sqrt{3}, α_{2} = \sqrt{5} i, α_{1} = - \sqrt{5} i$

So, $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6} = 1 - (- 3) + 5 = 9$

Q 17 :

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2 β}}$ then $β$ is equal to _______ . [2023]

(13)

$x^{2} - 5 a x + 1 = 0$ ...(i)

$Here, a_{1} = 1, b_{1} = - 5 a, c_{1} = 1$

$x^{2} - a x - 5 = 0$ ...(ii)

$a_{2} = 1, b_{2} = - a, c_{2} = - 5$

$∵ Equation (i) and (ii) have one common real root, so$

${[1 \times 1 - (- 5) (1)]}^{2} = [(- 5 a) (- 5) - (- a) (1)] (- a + 5 a)$

$\Rightarrow 36 = (25 a + a) (4 a) \Rightarrow 36 = 26 a \times 4 a$

$∴$ $a^{2} = \frac{36}{26 \times 4} \Rightarrow a = \pm \frac{3}{\sqrt{26}}$

Since, the value of the root is given as $\frac{3}{\sqrt{2 β}}$ .

So, $β = 13$

Topic Question Set

Q 1 :

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b},$ is [2024]

Q 2 :

Let $α, β$ be roots of $x^{2} + \sqrt{2} x - 8 = 0 .$ If $U_{n} = α^{n} + β^{n},$ then $\frac{U_{10} + \sqrt{2} U_{9}}{2 U_{8}}$ is equal to __________ . [2024]

Q 3 :

Let $α, β \in N$ be roots of the equation $x^{2} - 70 x + λ = 0,$ where $\frac{λ}{2}, \frac{λ}{3} \notin N .$ If $λ$ assumes the minimum possible value, then $\frac{(\sqrt{α - 1} + \sqrt{β - 1}) (λ + 35)}{| α - β |}$ is equal to _______ . [2024]

Q 4 :

Let $P_{n} = α^{n} + β^{n}, n \in N$ . If $P_{10} = 123, P_{9} = 76, P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is: [2025]

Q 5 :

Let $α$ and $β$ be the roots of $x^{2} + \sqrt{3} x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$ is equal to [2025]

Q 6 :

Let $z \in C$ be such that $\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is [2025]

Q 7 :

If the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ has equal roots, where a + c = 15 and $b = \frac{36}{5}$ , then $a^{2} + c^{2}$ is equal to __________. [2025]

Q 8 :

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ , where $i = \sqrt{- 1}$ , then $16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$ is equal to [2025]

Q 9 :

Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2, satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

Q 10 :

If $α$ is a root of the equation $x^{2} + x + 1 = 0$ and $\sum_{k = 1}^{n} {(α^{k} + \frac{1}{α^{k}})}^{2} = 20$ , then n is equal to __________. [2025]

Q 11 :

The sum of all the roots of the equation $| x^{2} - 8 x + 15 |$ $- 2 x + 7 = 0$ is [2023]

Q 12 :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ is [2023]

Q 13 :

Let $p, q \in R$ and ${(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q), i = \sqrt{- 1} .$ The $p + q + q^{2}$ and $p - q + q^{2}$ are roots of the equation [2023]

Q 14 :

Let $λ \neq 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0 .$ Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation [2023]

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 __________ . [2023]

Q 16 :

Let $α_{1}, α_{2}, \dots, α_{7}$ be the roots of the equation $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$ and $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$ . Then $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6}$ is equal to ______. [2023]

Q 17 :

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2 β}}$ then $β$ is equal to _______ . [2023]

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Topic Question Set

Q 1 : If 2 and 6 are the roots of the equation ax2+bx+1=0, then the quadratic equation, whose roots are 12a+b and 16a+b, is [2024]

Q 2 : Let α,β be roots of x2+2x-8=0. If Un=αn+βn, then U10+2U92U8 is equal to __________ . [2024]

Q 3 : Let α,β∈N be roots of the equation x2-70x+λ=0, where λ2,λ3∉N. If λ assumes the minimum possible value, then (α-1+β-1)(λ+35)|α-β| is equal to _______ . [2024]

Q 4 : Let Pn=αn+βn, n∈N. If P10=123, P9=76, P8=47 and P1=1, then the quadratic equation having roots 1α and 1β is: [2025]

Q 5 : Let α and β be the roots of x2+3x–16=0, and γ and δ be the roots of x2+3x–1=0. If Pn=αn+βn and Qn=γn+δn, then P25+3P242P23+Q25-Q23Q24 is equal to [2025]

Q 6 : Let z∈C be such that z2+3iz–2+i=2+3i. Then the sum of all possible values of z2 is [2025]

Q 7 : If the equation a(b–c)x2+b(c–a)x+c(a–b)=0 has equal roots, where a + c = 15 and b=365, then a2+c2 is equal to __________. [2025]

Q 8 : If α and β are the roots of the equation 2z2–3z–2i=0, where i=–1, then 16·Re(α19+β19+α11+β11α15+β15)·Im(α19+β19+α11+β11α15+β15) is equal to [2025]

Q 9 : Let f : R–{0}→(–∞,1) be a polynomial of degree 2, satisfying f(x)f(1x)=f(x)+f(1x). If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

Q 10 : If α is a root of the equation x2+x+1=0 and ∑k=1n(αk+1αk)2=20, then n is equal to __________. [2025]

Q 11 : The sum of all the roots of the equation |x2-8x+15|-2x+7=0 is [2023]

Q 12 : If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is (α,β), then the quadratic equation whose roots are α+4β and 4α+β is [2023]

Q 13 : Let p,q∈R and (1-3i)200=2199(p+iq), i=-1. The p+q+q2 and p-q+q2 are roots of the equation [2023]

Q 14 : Let λ≠0 be a real number. Let α,β be the roots of the equation 14x2-31x+3λ=0 and α,γ be the roots of the equation 35x2-53x+4λ=0. Then 3αβ and 4αγ are the roots of the equation [2023]

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 __________ . [2023]

Q 16 : Let α1,α2,…,α7 be the roots of the equation x7+3x5-13x3-15x=0 and |α1|≥|α2|≥…≥|α7|. Then α1α2-α3α4+α5α6 is equal to ______. [2023]

Q 17 : If the value of real number a>0 for which x2-5ax+1=0 and x2-ax-5=0 have a common real root is 32β then β is equal to _______ . [2023]

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Q 1 :

If 2 and 6 are the roots of the equation $a x^{2} + b x + 1 = 0$ , then the quadratic equation, whose roots are $\frac{1}{2 a + b}$ and $\frac{1}{6 a + b},$ is [2024]

Q 2 :

Let $α, β$ be roots of $x^{2} + \sqrt{2} x - 8 = 0 .$ If $U_{n} = α^{n} + β^{n},$ then $\frac{U_{10} + \sqrt{2} U_{9}}{2 U_{8}}$ is equal to __________ . [2024]

Q 3 :

Let $α, β \in N$ be roots of the equation $x^{2} - 70 x + λ = 0,$ where $\frac{λ}{2}, \frac{λ}{3} \notin N .$ If $λ$ assumes the minimum possible value, then $\frac{(\sqrt{α - 1} + \sqrt{β - 1}) (λ + 35)}{| α - β |}$ is equal to _______ . [2024]

Q 4 :

Let $P_{n} = α^{n} + β^{n}, n \in N$ . If $P_{10} = 123, P_{9} = 76, P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is: [2025]

Q 5 :

Let $α$ and $β$ be the roots of $x^{2} + \sqrt{3} x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P_{25} + \sqrt{3} P_{24}}{2 P_{23}} + \frac{Q_{25} - Q_{23}}{Q_{24}}$ is equal to [2025]

Q 6 :

Let $z \in C$ be such that $\frac{z^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is [2025]

Q 7 :

If the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ has equal roots, where a + c = 15 and $b = \frac{36}{5}$ , then $a^{2} + c^{2}$ is equal to __________. [2025]

Q 8 :

If $α$ and $β$ are the roots of the equation $2 z^{2} - 3 z - 2 i = 0$ , where $i = \sqrt{- 1}$ , then $16 \cdot Re (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}}) \cdot Im (\frac{α^{19} + β^{19} + α^{11} + β^{11}}{α^{15} + β^{15}})$ is equal to [2025]

Q 9 :

Let $f : R - {0} \to (- \infty, 1)$ be a polynomial of degree 2, satisfying $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})$ . If f(K) = –2K, then the sum of squares of all possible values of K is : [2025]

Q 10 :

If $α$ is a root of the equation $x^{2} + x + 1 = 0$ and $\sum_{k = 1}^{n} {(α^{k} + \frac{1}{α^{k}})}^{2} = 20$ , then n is equal to __________. [2025]

Q 11 :

The sum of all the roots of the equation $| x^{2} - 8 x + 15 |$ $- 2 x + 7 = 0$ is [2023]

Q 12 :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $(α, β)$ , then the quadratic equation whose roots are $α + 4 β$ and $4 α + β$ is [2023]

Q 13 :

Let $p, q \in R$ and ${(1 - \sqrt{3} i)}^{200} = 2^{199} (p + i q), i = \sqrt{- 1} .$ The $p + q + q^{2}$ and $p - q + q^{2}$ are roots of the equation [2023]

Q 14 :

Let $λ \neq 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0 .$ Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation [2023]

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 __________ . [2023]

Q 16 :

Let $α_{1}, α_{2}, \dots, α_{7}$ be the roots of the equation $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$ and $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$ . Then $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6}$ is equal to ______. [2023]

Q 17 :

If the value of real number $a > 0$ for which $x^{2} - 5 a x + 1 = 0$ and $x^{2} - a x - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2 β}}$ then $β$ is equal to _______ . [2023]