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

#### QUADRATIC EQUATIONS PRACTICE TEST [1 - EXERCISES]

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

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\$

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

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

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 –

1. if x < y
2. if x ≤. y
3. if x > y
4. if x ≥ y
5. 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.

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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