Quadratic Equations Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

When mistake is done in first degree term the roots of the equation are - 9 and -1.

∴ Equation is (x + 1) (x + 9) = $x^2$ + 10x + 9 ......(i)

When mistake is done in constant term, the roots of equation are 8 and 2.

∴ Equation is (x -2) (x - 8) = $x^2$ - 10x + 16 ......(ii)

∴ Required equation from equations (i) and (ii), we get

$x^2$ - 10x + 9

Answer: (d)

Quadratic equations $7x^2$ - 50x + k = 0

Here, a = 7, b = -50 and c = k

Since, α + β = ${-b}/a$

∴ α + β = ${50}/7$

⇒ β = $1/7$ (∵ α = 7, given)

and α β = $c/a or 7 × k/7 = k/7$ ⇒ k = 7

Answer: (a)

Let the roots of equation,

$ax^2 + bx + c = 0$ are - α and $- 1/{α}$.

∴ (- α)$(- 1/{α}) = c/a$

⇒ 1 = $c/a$ ⇒ c = a

Answer: (b)

Given equation,

$√{{2x}/{3 - x}} - √{{3 - x}/{2x}} = 3/2$

Let $√{{2x}/{3 - x}}$ = a

∴ a - $1/a = 3/2$

⇒ $2(a^2 - 1)$ = 3a

⇒ $2a^2$ - 3a - 2 = 0

⇒ $2a^2$ - 4a + a - 2 = 0

⇒ 2a(a - 2) + 1(a - 2) = 0

⇒ (2a + 1) (a - 2) = 0

If a - 2 = 0

Now, put a = 2

⇒ $√{{2x}/{3 - x}}$ = 2

Squaring both sides, then we get

⇒ 2x = 4(3 - x)

⇒ 6x = 12 ⇒ x = 2

If 2a + 1 = 0,

⇒ a = -$1/2, a ≠ {-1}/2$

x = 2 is the root of equation.

Answer: (b)

Given equation,

$2x^2$ - 3x - 4 = 0

For a reciprocal roots, we replace x by $1/x$, we get

$2(1/x)^2 - 3(1/x) - 4 = 0$

⇒ $-4x^2$ - 3x + 2 = 0

⇒ $4x^2$ + 3x - 2 = 0

Answer: (b)

We know that two equation

$a_1 x^2 + b_1 x + c_1$ = 0

$a_2 x^2 + b_2 + c_2$ = 0

have common root when

$(c_1 a_2 - a_1 c_2)^2 = (b_1 c_2 - c_1 b_2) (a_1 b_2 - b_1 a_2)$

So, for $x^2 + 5x + 6 = 0 and x^2 + kx = 1 = 0$

we have $(5)^2$ = (5 - 6x) (x - 5)

⇒ 25 = $-6x^2$ + 35x - 25

⇒ $6x^2$ - 35x + 50 = 0

⇒ x = $5/2 or {10}/3$