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

Answer: (a)

Given, $log_{10} (x^2 - 6x + 45) = 2$

⇒ $(x^2 - 6x + 45) = 10^2 = 100$

⇒ $x^2$ - 6x - 55 = 0

⇒ $x^2$ - 11x + 5x - 55 = 0

⇒ x(x - 11) + 5(x - 11) = 0

⇒ (x + 5) (x - 11) = 0

∴ x = 11, -5.

Answer: (b)

Let the roots of the equation be α and β

∴ α + β = $5/k$ and α β = $6/k$

Given, ${α}/{β} = 2/3$

⇒ α = $2/3$ β

∴ $2/3 β + β = 5/k and 2/3 β^2 = 6/k$

⇒ $5/3 β = 5/k and β^2 = 9/k$

β = $3/k and β^2 = 9/k$

⇒ $9/{k^2} = 9/k$

⇒ k = 1 and k ≠ 0. It is not satisfy the given condition.

Answer: (d)

α and β are the roots of the equation $ax^2$ + bx + c = 0

∴ α + β = - $b/a and α β = c/a$

∴ $(1/{α^2} - 1/{β^2})^2 = ({β^2 - α^2}/{α^2 β^2})^2$

= ${(α + β)^2 [(α + β)^2 - 4 α β]}/{(α^2 β^2)^2}$

= ${{b^2}/{a^2}({b^2}/{a^2} - {4c}/a)}/{({c^2}/{a^2})^2}$

= ${b^2}/{c^4} (b^2 - 4ac)$

Answer: (a)

Let $√{x/{1 - x}}$ = y

∴ $y + 1/y = {13}/6 ⇒ (y^2 + 1)6 = 13y$

⇒ $6y^2 - 13y + 6 = 0⇒6y^2 - 9y - 4y + 6 = 0$

⇒ 3y(2y - 3) -2(2y - 3) = 0

⇒ (3y - 2) (2y - 3) = 0

∴ $y = 2/3 and 3/2$

When, we put y = $2/3 ⇒ x/{1 - x} = 4/9$

⇒ 9x = 4 - 4x ⇒ x = $4/{13}$

When we put y = $3/2$

⇒ $x/{1 - x} = 9/4 ⇒ 4x = 9 - 9x$

∴ x = $9/{13}$

Answer: (d)

Given equation, $2x^2$ - 7x + 3 = 0

∴ $2x^2$ - 6x - x + 3 = 0

⇒ 2x(x - 3) - 1(x - 3) = 0

⇒ (2x - 1) (x - 3) = 0

Both equation have a common root.

So, we put x = $1/2 ⇒ 4(1/2)^2 + a(1/2)$ - 3 = 0

⇒ 1 + $a/2 - 3 = 0 ⇒ a/2$ = 2 ⇒ a = 4

Again, we put x = 3

$4(3)^2 + a(3) - 3 = 0$

⇒ 36 + 3a - 3 = 0 ⇒ a = - 11

a = - 11 or 4.