# CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations

**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

If roots of an equation ax^{2} + bx + c = 0 are positive, then which of the following is correct?

- A.
Signs of a and b should be alike

- B.
Signs of b and c should be alike

- C.
Signs of a and c should be alike

- D.
None of the above

Answer: Option C

**Explanation** :

Product of roots of a quadratic equation = c/a

When roots are positive their product will also be positive.

∴ c/a should be positive. This is possible only if a and c both have the same sign.

∴ If roots of an equation ax^{2} + bx + c are positive, then signs of a and c should be alike.

Hence, option (c).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

What will be the nature of roots of the equation 3x^{2} + 7x + 2 = 0.

- A.
Real and unequal

- B.
Real and equal

- C.
Imaginary

- D.
None of these

Answer: Option A

**Explanation** :

To find the nature of the roots of a quadratic equation, we need to look at the discriminant of the equation.

∴ D = b^{2} – 4ac

⇒ D = (-7)^{2} – 4 × 3 × 2 = 25

Since, discriminant is positive (i.e. > 0), roots will be real and unequal.

Hence, option (a).

Workspace:

**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

For what value of a, are the roots of the quadratic equation: ax^{2} – 2(a – 2)x + 1 = 0 real and equal?

- A.
a = 1 only

- B.
a = 4 only

- C.
a = 1 or 4

- D.
None of these

Answer: Option C

**Explanation** :

Roots of a quadratic equation are real and equal when discriminant is 0.

⇒ D = b^{2} – 4ac = 0

Since, the roots of the equation ax^{2} – 2(a – 2) x + 1 = 0 are real and equal.

∴ [-2(a – 2)]^{2} – 4a = 0

⇒ 4(a^{2} – 4a + 4) – 4a = 0

⇒ a^{2} – 4a + 4 – a = 0

⇒ a^{2} – 5a + 4 = 0

⇒ (a - 4)(a - 1) = 0

⇒ a = 1 or 4.

Hence, option (c).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

For what value of a will the roots of the quadratic: x^{2} + ax + 1 = 0 are non-real?

- A.
-4 < a < 4

- B.
-2 < a < 2

- C.
a > -2

- D.
a < 2

Answer: Option B

**Explanation** :

Roots of a quadratic equation are non-real when discriminant is < 0.

i.e., D = b^{2} – 4ac < 0

Since the roots of x^{2} + ax + 1 = 0 are non-real.

∴ D = (a^{2}) – 4 < 0

⇒ a^{2} – 4 < 0

⇒ a^{2} < 4

⇒ |a| < 2

⇒ -2 < a < 2

Hence, option (b).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

If bx^{2} + cx + a = 0 has real and different roots, then

- A.
c

^{2}- 4ba = 0 - B.
c

^{2}- 4ba > 0 - C.
c

^{2}- 4ba < 0 - D.
c

^{2}- 4ba ≤ 0

Answer: Option B

**Explanation** :

Roots of a quadratic equation (ax^{2} + bx + c = 0) are real and unequal when discriminant is > 0.

i.e., D = b^{2} – 4ac > 0

Since roots of bx^{2} + cx + a = 0 are real.

∴ D = c^{2} – 4ba > 0

Hence, option (b).

Workspace:

**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

If the roots of the equation x^{2} – px + q = 0 differ by 1, then which of the following is true?

- A.
q

^{2}= 4(q + 1) - B.
p

^{2}= 4q + 1 - C.
q

^{2}= p + 4 - D.
p

^{2}= 4(q + 2)

Answer: Option B

**Explanation** :

Let the roots be α and α + 1.

Then α + α + 1 = p

⇒ α = (p – 1)/2 …(1)

And, α(α + 1) = q

⇒ α^{2} + α = q …(2)

Putting the value of α from (1) in (2).

${\left(\frac{p-1}{2}\right)}^{2}$ + $\left(\frac{p-1}{2}\right)$ = q

⇒ $\frac{{p}^{2}+1\u20132p}{4}$ + $\left(\frac{p-1}{2}\right)$ = q

⇒ p^{2} + 1 – 2p + 2p - 2 = 4q

⇒ p^{2} = 4q + 1.

Hence, option (b).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

The roots of the equation x^{2} + 2√3x + 4 = 0 are

- A.
real and equal

- B.
rational and equal

- C.
rational and unequal

- D.
imaginary

Answer: Option D

**Explanation** :

To find the nature of roots of a quadratic equation we need to look at its discriminant.

D = b^{2} – 4ac = (2√3)^{2} – 4(1)(4) = -4 < 0.

So, the roots are imaginary.

Hence, option (d).

Workspace:

**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

If one root of x^{2} + ax + 6 = 0 is 3, while the equation x^{2} + ax + b = 0 has equal roots, then the value of b is

- A.
25/4

- B.
4/25

- C.
4

- D.
1/4

Answer: Option A

**Explanation** :

Given x^{2} + ax + 6 = 0

Since, x = 3 is the one root of the equation, therefore x = 3 will satisfy this equation

∴ 9 + 3a + 6 = 0

⇒ a = -5

Other quadratic equation becomes x^{2} – 5x + b = 0 (By putting value of a)

Its roots are equal, hence, D = b^{2} - 4ac = 0

⇒ b^{2} = 4ac.

⇒ 25 = 4b or b = 25/4

Hence, option (a).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

If the roots of the quadratic equation 2x^{2} – 4x + a = 0 are real and unequal, then which one of the following is correct?

- A.
a = 2

- B.
a < 2

- C.
a ≤ 2

- D.
None of these

Answer: Option B

**Explanation** :

The given equation is 2x^{2} – 4x + a = 0

We have, a = 2, b = -4, c = a

For roots of a quadratic equation to be real and unequal D = b^{2} – 4ac > 0

⇒ 16 – 8a > 0

⇒ a < 2

Hence, option (b).

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

How many real values of x are there for which (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} + (x - 4)^{2} = 0

Answer: 0

**Explanation** :

We know square of any real number is always greater than or equal to zero.

We have (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} + (x - 4)^{2}

Each of these terms is a square hence greater than or equal to zero.

Sum of these numbers will be equal to zero only when all of them are equal to zero.

But,

(x - 1) = 0 when x = 1

(x - 2) = 0 when x = 2

(x - 3) = 0 when x = 3

(x - 4) = 0 when x = 4

Hence, there is no value of x for which all of them are equal to zero at the same time.

∴ There is no value of x for which (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} + (x - 4)^{2} = 0

**Alternately**,

Given, (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} + (x - 4)^{2} = 0

⇒ (x^{2} - 2x + 1) + (x^{2} - 4x + 4) + (x^{2} - 6x + 9) + (x^{2} - 8x + 16) = 0

⇒ 4x^{2} - 20x + 30 = 0

⇒ 2x^{2} - 10x + 15 = 0

D = (-10)^{2} - 4 × 2 × 15 = -20 < 0

Hence, there is no real root for (x - 1)^{2} + (x - 2)^{2} + (x - 3)^{2} + (x - 4)^{2} = 0

Hence, 0.

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**CRE 2 - Discriminant and Roots of Quadratic Equation | Algebra - Quadratic Equations**

Find the nature of the roots of the equation: x^{2} + 5(p - 2)x - 25p = 0.

- A.
Real and Equal

- B.
Real and Unequal

- C.
Complex and conjugate

- D.
Cannot be determined

Answer: Option B

**Explanation** :

Given, x^{2} + 5(p - 2)x - 25p = 0.

D = (5(p - 2))^{2} - 4 × 1 × -25p

⇒ D = 25(p^{2} - 4p + 4) + 4 × 1 × 25p

⇒ D = 25[p^{2} - 4p + 4 + 4p]

⇒ D = 25[p^{2} + 4]

Here, p^{2} + 4 is always greater than 0, hence D > 0

Since, D > 0, the roots will be real and unequal

Hence, option (b).

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