Quadratic Equations Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

The equation will have equal roots, if

$B^2$ - 4AC = 0

∴ $(2nc)^2 - 4(1 + n^2) (c^2 - a^2)$ = 0

⇒ $4n^2 c^2 - 4 (c^2 - n^2 c^2 - a^2 - n^2 a^2)$ = 0

⇒ $-4c^2 + 4a^2 + 4n^2 a^2$ = 0

⇒ $c^2 = a^2 (1 + n^2)$

Answer: (b)

Given equation

$x^2$ - 15x + r = 0

Sum of roots = 15

p + q = 15 .....(i)

and p - q = 1 ......(ii)

From equation (i) and (ii) we have

p = 8, q = 7

Now, $p^2$ - 15 p + r = 0

$(8)^2$ - 15 (8) + 4 = 0

∴ r = 56

Answer: (a)

Let the roots of equation are α and β

$x^2 - 3kx + 2k^2 - 1 = 0$

∴ α β = $2k^2$ - 1

But given, - α β = 7

∴ $2k^2$ - 1= 7

⇒ $2k^2 = 8 ⇒ k^2$ = 4

k ≠ 2

On putting k = ± 2 in the equation, then we get

$x^2$ ± 6x + 7 = 0

D = $√{b^2 - 4ac} = √{(6)^2 - 4 × 7} = √{36 - 28} = 2 √2$

D = $2√2$, so roots of equation are irrational.

Answer: (a)

$x^2 - 6x - 27 > 0 ; x^2 - 9x + 3x - 27 > 0$

(x - 9) (x + 3) > 0 ∴ x = (- ∞ , 3) ∪ (9, ")

Answer: (c)

Given, $1/{a + b + x} = 1/a + 1/b + 1/x$

∴ $1/{a + b + x} - 1/x = 1/a + 1/b ⇒ {-(a + b)}/{(a + b + x)x} ={(a + b)}/{ab}$

⇒ $x^2$ + (a + b)x + ab = 0 ⇒ (x + a) (x + b) = 0

Hence, x = -a, -b.