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

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: (c)

Given that, f(x) is a polynomial with constant term 10 and all coefficient are integer. Let $k, k_1, k_2, ...... k_{n - 1}$ be roots of nth degree polynomial.

Now (x - k) is a factor of f(x), where k is an integer.

Product of roots = ${\text"Constant term"}/{\text"Coefficient of" x^n}$

i.e., $k. k_1. k_2......k_{n - 1} = {10}/1$ Coefficient of $x^n$ is 1

⇒ $k. k_1. k_2....k_{n -1}$ = 10 = 5.2.1....1.

Therefore, the possible value of k is 5.

Answer: (b)

Let, the roots of the quadratic equation

$x^2 + px + q$ = 0 is (α, β).

Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct.

i.e, Product of roots

q = α. β = 6 × 2 = 12........(i)

and B starts with a wrong value of q and gets the roots as 2 and - 9. But this time p is correct.

i.e, Sum of roots p = α + β = -9 + 2 = -7.....(ii)

$(α = β)^2 = (α + β)^2$ - 4 α β

= $(- 7)^2$ - 4.12 = 49 - 48 = 1

[From equations (i) and (ii)]

⇒ α - β = 1 ........(iii)

From equations (ii) and (iii),

α = -3 and β = -4

which is correct roots.