Practice Quadratic equations - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What is one of the roots of the equation $√{{2x}/{3 - x}} - √{{3 - x}/{2x}} = 3/2$
(a)
(b)
(c)
(d)
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.
Q-2) If f(x) is a polynomial with all coefficients are integers and constant term 10 having a factor (x - k), where k is an integer, then what is the possible value of k?
(a)
(b)
(c)
(d)
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.
Q-3) If the roots of the equation $lx^2$ + mx + m = 0 are in the ratio p : q, then $√{p/q} + √{q/p} + √{m/l}$ is equal to
(a)
(b)
(c)
(d)
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.
Q-4) If p and q are the roots of the equation $x^2$ - 15x + r = 0 and p - q = 1, then what is the value of r ?
(a)
(b)
(c)
(d)
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
Q-5) For what value of k, will the roots of the equation $kx^2$ - 5x + 6 = 0 be in the ratio of 2 : 3?
(a)
(b)
(c)
(d)
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.
Q-6) When the roots of the quadratic equation $ax^2$ + bx + c = 0 are negative of reciprocals of each other, then which one of the following is correct?
(a)
(b)
(c)
(d)
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
Q-7) What are the roots of the equation, $(a + b + x)^{-1} = a^{-1} + b^{-1} + x^{-1} ?$
(a)
(b)
(c)
(d)
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.
Q-8) If one root of $(a^2 - 5a + 3)x^2 + (3a - 1)$ x + 2 = 0 is twice the other, then what is the value of 'a' ?
(a)
(b)
(c)
(d)
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 = $2/3$
Q-9) What are the roots of the equation $log_{10} (x^2 - 6x + 45)$ = 2?
(a)
(b)
(c)
(d)
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.
Q-10) Directions :
In each of these questions, two equations numbered I and II are given. You have to solve both the equations and –- if x < y
- if x ≤. y
- if x > y
- if x ≥ y
- if x = y or the relationship cannot be established.
I. $x^2$ + 15x + 56 = 0
II. $y^2$ – 23y + 132 = 0
(a)
(b)
(c)
(d)
(e)
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.
Q-11) If one of the roots of quadratic equation $7x^2$ - 50x + k = 0 is 7, then what is the value of k ?
(a)
(b)
(c)
(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
Q-12) What is one of the value of x in the equation $√{x/{1 - x}} + √{{1 - x}/x} = {13}/6$
(a)
(b)
(c)
(d)
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 = $9/{13}$
Q-13) Two students A and B solve an equation of the form $x^2$ + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and -9. What are the correct roots of the equation?
(a)
(b)
(c)
(d)
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.
Q-14) If the equations, $2x^2 - 7x + 3 = 0 and 4x^2$ + ax - 3 = 0 have a common root, then what is the value of a ?
(a)
(b)
(c)
(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.
Q-15) If the product of the roots of $x^2 - 3kx + 2k^2$ - 1 = 0 is 7 for a fixed k, then what is the nature of roots?
(a)
(b)
(c)
(d)
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.
Q-16) If m and n are the roots of the equation $x^2$ + ax + b = 0 and $m^2, n^2$ are the roots of the equation $x^2$ - cx + d = 0, then which of the following is/are correct ?
1. 2b - $a^2$ = c
2. $b^2$ = d
Select the correct answer using the codes given below:
(a)
(b)
(c)
(d)
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.
Q-17) If the linear factors of $ax^2 - (a^2 + 1)$ x + a are p and q then p + q is equal to
(a)
(b)
(c)
(d)
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.
Q-18) If the equations $x^2 + 5x + 6 = 0 and x^2$ + kx + 1 = 0 have a common root, then what is the value of k?
(a)
(b)
(c)
(d)
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 = $5/2 or {10}/3$
Q-19) Which one of the following is the quadratic equation whose roots are reciprocal to the roots of the quadratic equation $2x^2$ - 3x - 4 = 0 ?
(a)
(b)
(c)
(d)
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
Q-20) In solving a problem, one student makes a mistake in the coefficient of the first degree term and obtains - 9 and - 1 for the roots. Another student makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct equation was
(a)
(b)
(c)
(d)
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