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

Q 1 :
The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

$1 + \log_{8} (8)$
$\log_{8} (6)$
$1 + \log_{8} (6)$
$\log_{8} (4)$

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(C)

We have, $8^{2 x} - 16 \cdot 8^{x} + 48 = 0$

Let $8^{x} = t$

$\Rightarrow t^{2} - 16 t + 48 = 0 \Rightarrow t^{2} - 12 t - 4 t + 48 = 0$

$\Rightarrow (t - 12) (t - 4) = 0 \Rightarrow t = 4 or t = 12$

$\Rightarrow 8^{x} = 4 or 8^{x} = 12$

$\Rightarrow x = \log_{8} 4 or x = \log_{8} 12$

Sum of solutions $= \log_{8} 4 + \log_{8} 12 = \log_{8} 48 = \log_{8} (6 \cdot 8)$

$= \log_{8} 8 + \log_{8} 6 = 1 + \log_{8} 6$

Q 2 :
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: [2024]

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

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(C)

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 3 :
Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

$11 \sqrt{2} P_{9}$
$10 \sqrt{2} P_{9}$
$10 \sqrt{3} P_{9}$
$11 \sqrt{3} P_{9}$

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(C)

We have, $x^{2} - \sqrt{2} x - \sqrt{3} = 0$ has two roots $α$ and $β$

So for $S_{n} = α^{n} \neq β^{n},$ by Newton's theorem, we have $S_{n + 2} - \sqrt{2} S_{n + 1} - \sqrt{3} S_{n} = 0$

$\Rightarrow (α^{n + 2} - β^{n + 2}) - \sqrt{2} (α^{n + 1} - β^{n + 1}) - \sqrt{3} (α^{n} - β^{n}) = 0$

$\Rightarrow P_{n + 2} - \sqrt{2} P_{n + 1} - \sqrt{3} P_{n} = 0 [∵ P_{n} = α^{n} - β^{n}]$

For $n = 9,$ we have

$P_{12} - \sqrt{2} P_{11} - \sqrt{3} P_{10} = 0$ ...(i)

$\Rightarrow P_{12} = \sqrt{2} P_{11} + \sqrt{3} P_{10}$

For $n = 9,$ we have

$P_{11} - \sqrt{2} P_{10} - \sqrt{3} P_{9} = 0$

$\Rightarrow P_{11} = \sqrt{2} P_{10} + \sqrt{3} P_{9}$

So, $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$

$= (11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) (\sqrt{2} P_{10} + \sqrt{3} P_{9}) - 11 (\sqrt{2} P_{11} + \sqrt{3} P_{10})$

$= 11 \sqrt{3} P_{10} - 10 \sqrt{2} P_{10} + 22 P_{10} + 10 \sqrt{2} P_{10} - 11 \sqrt{3} P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 11 \sqrt{2} (\sqrt{2} P_{10} + \sqrt{3} P_{9})$

$= 22 P_{10} + 11 \sqrt{6} P_{9} + 10 \sqrt{3} P_{9} - 22 P_{10} - 11 \sqrt{6} P_{9} = 10 \sqrt{3} P_{9}$

Q 4 :
Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

$2 S_{12} = S_{11} + S_{10}$
$2 S_{11} = S_{12} + S_{10}$
$S_{11} = S_{10} + S_{12}$
$S_{12} = S_{11} + S_{10}$

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(D)

Given, $α$ and $β$ are the roots of $x^{2} - x - 1 = 0$

$\Rightarrow α^{2} - α - 1 = 0$ and $β^{2} - β - 1 = 0$

$\Rightarrow α^{2} = 1 + α$ and $β^{2} = 1 + β$ (i)

Now, $S_{10} = 2023 α^{10} + 2024 β^{10} [∵ S_{n} = 2023 α^{n} + 2024 β^{n}]$

$S_{11} = 2023 α^{11} + 2024 β^{11}$

$\Rightarrow S_{10} + S_{11} = 2023 α^{10} (1 + α) + 2024 β^{10} (1 + β)$

$= 2023 α^{10} \cdot α^{2} + 2024 β^{10} \cdot β^{2} [From (i)]$

$= 2023 α^{12} + 2024 β^{12} = S_{12}$

$∴ S_{12} = S_{10} + S_{11}$

Q 5 :
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

0%

(221)

We have, $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are solution of $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$

$\Rightarrow 4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9$

$= 4 (x - x_{1}) (x - x_{2}) (x - x_{3}) (x - x_{4})$ ...(i)

Let us substitute $x = 2 i$ in (i)

$∴ 64 - 64 i + 68 - 24 i + 9$

$= 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4})$

$\Rightarrow 4 (2 i - x_{1}) (2 i - x_{2}) (2 i - x_{3}) (2 i - x_{4}) = 141 - 88 i$ ...(ii)

Substitute $x = - 2 i$ in (i), we get

$4 (2 i + x_{1}) (2 i + x_{2}) (2 i + x_{3}) (2 i + x_{4}) = 141 + 88 i$ ...(iii)

Multiplying (ii) and (iii), we get

$(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{{(141)}^{2} + 88^{2}}{16}$

$\Rightarrow \frac{125 m}{16} = \frac{27625}{16} \Rightarrow m = 221$

Q 6 :
If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

0%

(5)

As $α$ satisfies $x^{2} + x + 1 = 0 \Rightarrow α = ω or ω^{2}$

Now, ${(1 + α)}^{7} = {(1 + ω)}^{7} = A + B ω + C ω^{2}$

$\Rightarrow - ω^{2} = A + B ω + C ω^{2} [∵ 1 + ω + ω^{2} = 0 and ω^{3} = 1]$

$\Rightarrow A + B ω + (C + 1) ω^{2} = 0$

As $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ & $ω^{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

$\Rightarrow A + B (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) + (C + 1) (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) = 0$

$\Rightarrow A - \frac{B}{2} - \frac{C + 1}{2} = 0 and \frac{\sqrt{3}}{2} B - \frac{\sqrt{3}}{2} (C + 1) = 0$

$\Rightarrow 2 A - B - C - 1 = 0 and B - C = 1$

On solving, we get A = B and B - C = 1

$∴$ $5 (3 A - 2 B - C) = 5 (3 B - 2 B - C) = 5 (B - C) = 5 \times 1 = 5$

Q 7 :
Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

0%

(13)

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

As $α, β$ are the roots of equation so, it satisfy the equation.

$α^{2} - α + 2 = 0 \Rightarrow α^{2} = α - 2$

$\Rightarrow α^{4} = {(α - 2)}^{2}$ [Squaring both sides]

$\Rightarrow α^{4} = 4 + α^{2} - 4 α = 4 + α - 2 - 4 α = 2 - 3 α$

and $α^{6} = α^{2} \cdot α^{4} = (α - 2) (2 - 3 α) = 2 α - 3 α^{2} - 4 + 6 α$

$= 8 α - 4 - 3 (α - 2) = 8 α - 4 - 3 α + 6 = 5 α + 2$

Similarly, $β^{4} = 2 - 3 β$

Now, $α^{6} + α^{4} + β^{4} - 5 α^{2}$

$= 5 α + 2 + 2 - 3 α + 2 - 3 β - 5 (α - 2)$

$= 16 - 3 (α + β) = 16 - 3 (1) = 13$

Q 8 :
Let $α, β$ be the roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ such that $Im (α) > Im (β) .$ Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b), i = \sqrt{- 1} .$ Then $n + a + b$ is equal to ______ . [2024]

0%

(49)

Roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ are

$x = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} \Rightarrow x = \frac{\sqrt{6} \pm \sqrt{6 i}}{2}$

$∵$ $Im (α) > Im (β)$

$∴$ $α = \frac{\sqrt{6} + \sqrt{6 i}}{2}, β = \frac{\sqrt{6} - \sqrt{6 i}}{2}$

$\Rightarrow α^{2} = \frac{2 \cdot \sqrt{6} \cdot \sqrt{6} \cdot i}{4}, β^{2} = \frac{- 2 \cdot \sqrt{6} \cdot \sqrt{6 i}}{4}$

$α^{2} = 3 i, β^{2} = - 3 i \Rightarrow \frac{α}{β} = i$

Now, $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b)$

$\Rightarrow α^{98} (\frac{α}{β} + 1) = 3^{n} (a + i b) \Rightarrow {(3 i)}^{49} (i + 1) = 3^{n} (a + i b)$

$\Rightarrow 3^{49} \cdot i (i + 1) = 3^{n} (a + i b) \Rightarrow 3^{49} \cdot (- 1 + i) = 3^{n} (a + i b)$

$\Rightarrow n = 49, a = - 1, b = 1$ $∴$ $n + a + b = 49$

Q 9 :
Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

0%

(36)

We have, $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0$

$\Rightarrow a^{2} x^{2} + b^{2} x^{2} - 2 a b x - 2 b c x + b^{2} + c^{2} = 0$

$\Rightarrow a^{2} x^{2} - 2 a b x + b^{2} + b^{2} x^{2} - 2 b c x + c^{2} = 0$

$\Rightarrow {(a x - b)}^{2} + {(b x - c)}^{2} = 0$

$\Rightarrow a x - b = 0, b x - c = 0$

$\Rightarrow a x = b, b x = c \Rightarrow a x^{2} = b x, b x^{2} = c x$

Since, $a + b > c, b + c > a and c + a > b$

$a + a x > b x, a x + b x > a and b x + a > a x$

$\Rightarrow a + a x > a x^{2}, a x + a x^{2} > a and a x^{2} + a > a x$

$\Rightarrow x^{2} - x - 1 < 0, x^{2} + x - 1 > 0 and x^{2} - x + 1 > 0$

$\Rightarrow \frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2} \Rightarrow x < \frac{- 1 - \sqrt{5}}{2} or x > \frac{- 1 + \sqrt{5}}{2}$

$\Rightarrow \frac{\sqrt{5} - 1}{2} < x < \frac{\sqrt{5} + 1}{2} \Rightarrow α = \frac{\sqrt{5} - 1}{2}, β = \frac{\sqrt{5} + 1}{2}$

$∴$ $12 (α^{2} + β^{2}) = 36$

Topic Question Set

Q 1 :
The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

Q 2 :
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: [2024]

Q 3 :
Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

Q 4 :
Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

Q 5 :
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

0%

Q 6 :
If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

0%

Q 7 :
Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

0%

Q 8 :
Let $α, β$ be the roots of the equation $x^{2} - \sqrt{6} x + 3 = 0$ such that $Im (α) > Im (β) .$ Let $a, b$ be integers not divisible by 3 and $n$ be a natural number such that $\frac{α^{99}}{β} + α^{98} = 3^{n} (a + i b), i = \sqrt{- 1} .$ Then $n + a + b$ is equal to ______ . [2024]

0%

Q 9 :
Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]

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Topic Question Set

Q 1 : The sum of all the solutions of the equation (8)2x-16·(8)x+48=0 is : [2024]

Q 2 : 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: [2024]

Q 3 : Let α,β:α>β, be the roots of the equation x2-2x-3=0. Let Pn=αn-βn, n∈N. Then (113-102)P10+(112+10)P11-11P12 is equal to [2024]

Q 4 : Let α,β are the roots of the equation, x2-x-1=0 and Sn=2023αn+2024βn, then : [2024]

Q 5 : Let x1,x2,x3,x4 be the solutions of the equation 4x4+8x3-17x2-12x+9=0 and (4+x12)(4+x22)(4+x32)(4+x42)=12516m. Then the value of m is _______ . [2024]

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Q 6 : If α satisfies the equation x2+x+1=0 and (1+α)7=A+Bα+Cα2, A,B,C≥0, then 5(3A-2B-C) is equal to _______ . [2024]

0%

Q 7 : Let α,β be the roots of the equation x2-x+2=0 with Im(α)>Im(β). Then α6+α4+β4-5α2 is equal to ____ . [2024]

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Q 8 : Let α,β be the roots of the equation x2-6x+3=0 such that Im(α)>Im(β). Let a,b be integers not divisible by 3 and n be a natural number such that α99β+α98=3n(a+ib), i=-1. Then n+a+b is equal to ______ . [2024]

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Q 9 : Let a,b,c be the lengths of three sides of a triangle satisfying the condition (a2+b2)x2-2b(a+c)x+(b2+c2)=0. If the set of all possible values of x is the interval, (α,β) then 12(α2+β2) is equal to _____ . [2024]

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Q 1 :
The sum of all the solutions of the equation ${(8)}^{2 x} - 16 \cdot {(8)}^{x} + 48 = 0$ is : [2024]

Q 2 :
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: [2024]

Q 3 :
Let $α, β : α > β,$ be the roots of the equation $x^{2} - \sqrt{2} x - \sqrt{3} = 0 .$ Let $P_{n} = α^{n} - β^{n}, n \in N .$ Then $(11 \sqrt{3} - 10 \sqrt{2}) P_{10} + (11 \sqrt{2} + 10) P_{11} - 11 P_{12}$ is equal to [2024]

Q 4 :
Let $α, β$ are the roots of the equation, $x^{2} - x - 1 = 0$ and $S_{n} = 2023 α^{n} + 2024 β^{n},$ then : [2024]

Q 5 :
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the solutions of the equation $4 x^{4} + 8 x^{3} - 17 x^{2} - 12 x + 9 = 0$ and $(4 + x_{1}^{2}) (4 + x_{2}^{2}) (4 + x_{3}^{2}) (4 + x_{4}^{2}) = \frac{125}{16} m .$ Then the value of $m$ is _______ . [2024]

Q 6 :
If $α$ satisfies the equation $x^{2} + x + 1 = 0$ and ${(1 + α)}^{7} = A + B α + C α^{2}, A, B, C \geq 0,$ then $5 (3 A - 2 B - C)$ is equal to _______ . [2024]

Q 7 :
Let $α, β$ be the roots of the equation $x^{2} - x + 2 = 0$ with $Im (α) > Im (β) .$ Then $α^{6} + α^{4} + β^{4} - 5 α^{2}$ is equal to ____ . [2024]

Q 9 :
Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0 .$ If the set of all possible values of $x$ is the interval, $(α, β)$ then $12 (α^{2} + β^{2})$ is equal to _____ . [2024]