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

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.

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

$∴ ∆ A B C is an isosceles triangle.$

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

$Now, slope of A B = 1$

$∴ slope of C D = - 1$

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

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

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

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

[IMAGE 11]---------

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

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

$= 20$

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

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]

0%

(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]

0%

(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]

0%

(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 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]

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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]

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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 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]

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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]

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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 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]