Quadratic Equations Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts
QUADRATIC EQUATIONS PRACTICE TEST [1 - EXERCISES]
Quadratic Equations Model Questions Set 1
Question : 1
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) Both 1 and 2
b) Only 2
c) Neither 1 nor 2
d) Only 1
Answer »Answer: (b)
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.
Question : 2
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) $5/2 or - {10}/3$
b) $5/2 or {10}/3$
c) $- 5/2 or {10}/3$
d) $- 5/2 or {10}/3$
Answer »Answer: (b)
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 = $5/2 or {10}/3$
Question : 3
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) $3x^2$ - 4x - 2 = 0
b) $4x^2$ + 3x - 2 = 0
c) $4x^2$ - 2x - 3 = 0
d) $3x^2$ - 2x - 4 = 0
Answer »Answer: (b)
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
Question : 4
What is one of the roots of the equation $√{{2x}/{3 - x}} - √{{3 - x}/{2x}} = 3/2$
a) 3
b) 2
c) 4
d) 1
Answer »Answer: (b)
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.
Question : 5
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) if x > y
b) if x ≤ y
c) if x ≥ y
d) if x < y
e) if x = y or the relationship cannot be established.
Answer »Answer: (d)
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.
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Quadratic Equations Model Questions Set 1 Online Quiz
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