Practice Quadratic equations - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

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.

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.

Let &alpha, β be the roots of the equation $lx^2$ + mx + m = 0

Given ${α}/{β} = p/q$

Now α + β (sum of roots) = ${- m}/l$

and α β (product of roots) = $m/l$

Consider $√{p/q} + √{q/p} + √{m/l}$

Using (1)

= $√{{α}/{β}} = √{{β}/{α}} + √{m/l}$

= ${α + β}/{√{α β}} + √{m/l}$

= ${- m/l}/{√{m/l}} + √{m/l} = - √{m/l} + √{m/l}$ = 0

∴ option (d) is correct.

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

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.

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

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.

Let the roots of equation be α and 2 α

Sum of root

- (2 α + α) = ${3a - 1}/{a^2 - 5a + 3}$

-3 α = ${3a - 1}/{a^2 - 5a + 3}$....

Squaring Both the side

$9α^2 = {(3a - 1)^2}/{(a^2 - 5a + 3)^2}$ ......(i)

Product of root

2 α × α = $2/{a^2 - 5a + 3}$

$α^2 = 1/{a^2 - 5a + 3}$

Dividing equation (i) from (ii)

${9 α^2}/{α^2} = {(3a - 1)^2}/{(a^2 - 5a + 3)^2} × {a^2 - 5a + 3}/1$

9 = ${(3a - 1)^2}/{(a^2 - 5a + 3)}$

$9a^2 - 45a + 27 = 9a^2 + 1 - 6a$

26 = 39a

a = $2/3$

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.

Note: Let the quardatic equation be $ax^2$ + b + c = 0.

To find roots of this equation quickly, we find two factors of 'b' such that their sum is equal to b and their product is equal to the product of the coefficient of x2 and the constant term 'c'.

Let two such factors be α and β.

The α + β = b and α β = ca

In the second step, we divide these factors by the coefficient of $x^2$ ,

ie be 'a'.

In the next step, we change the signs of the outcome. These are the

roots of the equation.

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

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 = $9/{13}$

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.

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.

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.

Here m and n are the roots of the equation

$x^2$ + ax + b = 0.

m + n = - a .......(i)

mn = b ......(ii)

Also m $m^2 and n^2$ are the roots of the equation of

$x^2$ - cx + d = 0.

$m^2 + n^2$ = c...(iii)

$m^2 n^2$ = d .....(iv)

by squaring Eq. (i) both sides, we get

$m^2 + n^2 + 2mn = a^2$ [from Eqs. (i) and (ii)]

⇒ c + 2b = $a^2 ⇒ c = a^2 - 2b$

⇒ 2b - $a^2$ = -c

Therefore, Statement 1 is incorrect.

From Eq. (ii)

$m^2 n^2 = b^2 ⇒ b^2$ = d

Therefore, Statement 2 is correct.

Consider $ax^2 - (a^2 + 1)$ x + a .......(1)

⇒ $ax^2 - a^2 x$ - x + a

⇒ $a^x$ (x - a) - 1 (x - a)

= (x - a) (ax - 1)

Given p and q are two linear factors of (1)

∴ p = x + a and q = ax - 1

⇒ p + q = x - a + ax - 1

= x(a + 1) - 1 (a + 1)

= (x - 1) (a + 1)

∴ Option (d) is correct.

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 = $5/2 or {10}/3$

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

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