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

Q 11 :
The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

(2)

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

$\Rightarrow | x - 2 |^{2} + 2 | x - 2 | - | x - 2 | - 2 = 0$

$\Rightarrow (| x - 2 | + 2) (| x - 2 | - 1) = 0$

$\Rightarrow | (x - 2) | = 1$ [ $∵ | x - 2 | \neq - 2$ ]

$\Rightarrow x = 2 \pm 1 \Rightarrow x = 3, 1$

Sum of square of roots = 9 + 1 = 10

Now, we have $x^{2} - 2 | x - 3 | - 5 = 0$

Case I : When x – 3 > 0

$\Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x = 1$

but x > 3 $\Rightarrow x \neq 1$

Case II : When x – 3 < 0

$\Rightarrow x^{2} + 2 x - 11 = 0$

Discriminant, D = 4 + 44 = 48 > 0

$x = \frac{- 2 \pm \sqrt{48}}{2} = \frac{- 2 \pm 4 \sqrt{3}}{2} = - 1 \pm 2 \sqrt{3}$

SInce, i < 3, so both roots are valid.

Sum of squares of roots = ${(- 1 + 2 \sqrt{3})}^{2} + {(- 1 - 2 \sqrt{3})}^{2}$

$= 1 + 12 - 4 \sqrt{3} + 1 + 12 + 4 \sqrt{3} = 26$

$∴$ Required sum = 10 + 26 = 36.

Q 12 :
If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

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Let $x_{1}$ and $x_{2}$ be the roots of ${(1 - a)}^{2} + 2 (a - 3) x + 9 = 0$ .

Since, $x_{1} x_{2} > 0$

$∴ x_{1} + x_{2} > 0 and x_{1} x_{2} > 0$

$\Rightarrow \frac{- 2 (a - 3)}{1 - a} > 0 \Rightarrow \frac{a - 3}{1 - a} < 0$

$\Rightarrow a \in (- \infty, 1) \cup (3, \infty)$

Also, $x_{1} x_{2} > 0 \Rightarrow \frac{9}{1 - a} > 0 \Rightarrow 1 - a > 0 \Rightarrow a < 1$

On combining both conditions, we get $a \in (- \infty, 1)$

Now, for real roots, discriminaqnt must be non negative.

i.e., $(2 {(a - 3))}^{2} - 4 (1 - a) 9 \geq 0$

$\Rightarrow 4 (a^{2} - 6 a + 9) - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} - 24 a + 36 - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} + 12 a \geq 0 \Rightarrow a (a + 3) \geq 0$

$\Rightarrow a \in (- \infty, - 3] \cup [0, 1)$

Combining all the conditions, we get $a \in (- \infty, - 3] \cup [0, 1)$

$∴ α = 3, β = 0 and γ = 1$

$∴ 2 α + β + γ = 2 \times 3 + 0 + 1 = 7$

Topic Question Set

Q 11 :
The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

Q 12 :
If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

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Topic Question Set

Q 11 : The sum of the squares of the roots of |x–2|2+|x–2|–2=0 and the squares of the roots of x2–2|x–3|–5=0 is [2025]

Q 12 : If the set of all a∈R–{1}, for which the roots of the equation (1–a)x2+2(a–3)x+9=0 are positive is (–∞,–α]∪[β,γ), then 2α+β+γ is equal to __________. [2025]

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Q 11 :
The sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

Q 12 :
If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]