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]

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(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]

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(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
7
5
3

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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 :
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

(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 8 :
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

(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 9 :
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)

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

$\Rightarrow α - \frac{1}{α} = \frac{3}{2}$ [ $∵ α$ 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 10 :
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, $f (x) f (\frac{1}{x}) - f (\frac{1}{x}) = f (x)$

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

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 $β$

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

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]

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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]

0%

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 :
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 8 :
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 9 :
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 10 :
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]

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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]

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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]

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

Q 8 : Let z∈C be such that z2+3iz–2+i=2+3i. Then the sum of all possible values of z2 is [2025]

Q 9 : 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 10 : 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]

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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 :
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 8 :
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 9 :
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 10 :
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]