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

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4

(B)

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

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]

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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 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]