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

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 ax² + bx + c are positive, then signs of a and c should be alike.

Hence, option (c).

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 = b² – 4ac

⇒ D = (-7)² – 4 × 3 × 2 = 25

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

Hence, option (a).

Answer: Option C

Explanation :

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

⇒ D = b² – 4ac = 0

Since, the roots of the equation ax² – 2(a – 2) x + 1 = 0 are real and equal.

∴ [-2(a – 2)]² – 4a = 0

⇒ 4(a² – 4a + 4) – 4a = 0

⇒ a² – 4a + 4 – a = 0

⇒ a² – 5a + 4 = 0

⇒ (a - 4)(a - 1) = 0

⇒ a = 1 or 4.

Hence, option (c).

Answer: Option B

Explanation :

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

i.e., D = b² – 4ac < 0

Since the roots of x² + ax + 1 = 0 are non-real.

∴ D = (a²) – 4 < 0 

⇒ a² – 4 < 0

⇒ a² < 4 

⇒ |a| < 2 

⇒ -2 < a < 2

Hence, option (b).

Answer: Option B

Explanation :

Roots of a quadratic equation (ax² + bx + c = 0) are real and unequal when discriminant is > 0.

i.e., D = b² – 4ac > 0

Since roots of bx² + cx + a = 0 are real.

∴ D = c² – 4ba > 0

Hence, option (b).

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 

⇒ α² + 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

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

⇒ p² + 1 – 2p + 2p - 2 = 4q

⇒ p² = 4q + 1.

Hence, option (b).

Answer: Option D

Explanation :

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

D = b² – 4ac = (2√3)² – 4(1)(4) = -4 < 0. 

So, the roots are imaginary.

Hence, option (d).

Answer: Option A

Explanation :

Given x² + 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 x² – 5x + b = 0 (By putting value of a)

Its roots are equal, hence, D = b² - 4ac = 0

⇒ b² = 4ac.

⇒ 25 = 4b or b = 25/4

Hence, option (a).

Answer: Option B

Explanation :

The given equation is 2x² – 4x + a = 0

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

For roots of a quadratic equation to be real and unequal D = b² – 4ac > 0

⇒ 16 – 8a > 0

⇒ a < 2

Hence, option (b).

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

⇒ (x² - 2x + 1) + (x² - 4x + 4) + (x² - 6x + 9) + (x² - 8x + 16) = 0

⇒ 4x² - 20x + 30 = 0

⇒ 2x² - 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.

Answer: Option B

Explanation :

Given, x² + 5(p - 2)x - 25p = 0.

D = (5(p - 2))² - 4 × 1 × -25p

⇒ D = 25(p² - 4p + 4) + 4 × 1 × 25p

⇒ D = 25[p² - 4p + 4 + 4p]

⇒ D = 25[p² + 4]

Here, p² + 4 is always greater than 0, hence D > 0

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

Hence, option (b).