Quadratic Equations Questions and Answers

Explanation :

First, we have to find the discriminant of the Quadratic Equation :
Discriminant = b^{2}-4ac

Here, a = 1,b = 5 and c = 6
So, Discriminant = 1 > 0
So, the roots are real and have two distinct roots.
Hence Option (D) is correct.

To calculate positive roots of Eqn. f(a) =  a^{4}+ a^{2}+1= 0

Since all the signs are positive, therefore none of the roots will be positive.

To calculate the negative roots of Eqn. f(a)= a^{4}+ a^{2}+1= 0

f(-a)= (- a)^{4} (- a)^{2}+1= 0

Again, as all the signs are same, hence no roots of this eqn. will be negative and also none of the roots will be equal to 0

Therefore Option D is the correct one.

Since f(0)<0 and f(1)>0, therefore,the root of the equation will be somewhere between 0 and 1

Now for all the negative values of p<0, the above equation will give negative values as p11 will eventually convert to a significant positive number. 

Therefore it will not have any real root for p>1 or p<0 

Hence the total number of roots= 1 

Thus option B is the correct one. 

α+β=p+6
αβ=2(2p-1)
According to the question,
p+6 = \frac{2(2p-1)}{2}

=> p = 7.

Option A – has no real roots

Option B is independent of term x²

Option C is also independent of term x² for k= -1

Option D implies

x² =1

x= ±1

Hence answer is Option D

Explanation:

To solve the inequality, we first find the roots of the quadratic equation 2x^2 + 5x - 3 = 0.

Using the quadratic formula, we get

x = ( -5 ± √(5^2 - 4 * 2 * -3) ) / (2 * 2)

=> x = ( -5 ± √49 ) / 4.

The roots are x = ( -5 + 7 ) / 4 = 1/2 and x = ( -5 - 7 ) / 4 = -3.

We plot these values on a number line and test a value in each interval to check if it satisfies the inequality.

For example, testing x = 0, we find 2(0)^2 + 5(0) - 3 = -3 < 0.

Thus, the solution is x > 1/2 and x < -3.

The correct answer is option D.

Sum of all the roots 3a^{2} +2a-1 = (−x/y) = −2/3

Hence Option A is the correct one.

Product of the roots= (z/x) = −1/3 here, z=-1 and x = 3

p²m²+pqm= q²

pm+(q/2)²= 5q² / 4

pm+ q/2 =  ±√5/4q

If q= 0 then real and equal roots otherwise real and unequal roots

The nature of roots of the given equation can be given by,

D = b²4ac =0,

For equation x² – 5x+3 =0,

D = (-5)²-4(3)(1)

D = 13 > 0

It has two distinct real roots.

2a+3b= 5

4a+pb= Q

For no solution

= 2/4 - 3/q ≠ 5/Q

Therefore p= 6

b² - 13b + f = 0 with roots p and q

Therefore p + q = 13 and p - q= 3

Therefore p= 8 and q= 5

Hence F= pq= 40

Explanation:

To find the value of 'x' that maximizes the volume, we need to find the critical points of the volume function V(x). We take the derivative of V(x) with respect to 'x' and set it to zero to find the critical points.

The derivative of V(x) is dV/dx = 108x^{2} - 60x + 5.

Setting dV/dx = 0 and solving for 'x', we get x = 1/6 and x = 5/6.

To determine whether these are maxima or minima, we can use the second derivative test. Calculating the second derivative, \frac{d^{2}V}{dx^{2}} = 216x - 60, we find that \frac{d^{2}V}{dx^{2}} is positive at x = 1/6 and negative at x = 5/6.

Therefore, x = 1/6 gives the maximum volume. The correct answer is option A.

Given that f(p)= P² – 28p+196= p

Or P² – 28p+196= 0

= (p-14)(p-14)= 0

Therefore p= 14

p + 3 = 0

p = -3

= 3(-3)²+a(-3)+6= 0

= 27-3a+6= 0

Therefore a= 11

a+2 is the factor of equation y(A) so y(-2)= 0

= 16-16-4p+3= 0

= p=3/4

p²-5p+6= 0  have roots 2, 3

When 2 is the common root then f(2)= 0= 2²+2a+3= 0

Therefore a=-7/2

So when 3 is the common root then f(3)= 0= 3²+3a+3=0

= a= -4

a³-a²+2a-17= 0

let the roots be x, y, and z

Therefore xyz= 17

Now since 17 has only 2 factors 1 and 17, and since both these values are not mentioned in the given option. Then, the asnwer is D.

a- (1/a² - 4) = 2- (1/a² - 4)

= a= 2  but if a= 2 then (1/a² - 4)  = infinity

Hence no value for 'a' is possible

Therefore option D is the correct one.

a²+a-6= 0 where x≥0 therefore the root will be 2

Or a²+a-6= 0 if a<0 then root will be -2

Hence the total of roots= 0 (2+(-2))

(x-4)≤6= -2 ≤ x ≤ 10

(x + 4) ≤ 5 = -9 ≤ x ≤ 1

B combining the two equations we get -2 ≤ x< 1

Hence since only 1 element satisfy the condition that is 0

Therefore a+b= -1 and ab= 2

Now = \frac{a^{10} + b^{10}}{a^{-10} + b^{-10}}

=\frac{\frac{a^{10} + b^{10}}{a^{-10} + b^{-10}}} {ab^{10}}

= ab¹⁰= 2¹⁰= 1024

If we follow the given options, then only option A is the perfect square of 44²

And also 1980-44= 1936

m³- m² +m -1 >0

m²(m-1) +(m-1) > 0

(m²+1) +(m-1)>0

As m²+1 >0 for all m therefore m>1

p²+5p+6= 0

Since all the digits on the left side are positive, therefore no root is possible for this equation

p-2= 2^1/3+ 2^2/3= (s-2)²= 2^2/3 + 2.2^1/3+2.2

Now 2^2/3 - 2.2^1/3+2.2 – (2^1/3 + 2.2^2/3- 2= 2

Given a₁, a₂, and a₃ are in the AP

Then 2a₂= a₁+ a₃

Also given that the sum of the roots is a₁ + a₂= -4

From both the equations we can find neither a₁ and a₂

Given p₁, p₂, and p₃ are in the AP

Then 2p₂= p₁+ p₂

Also given that the sum of the roots = p₁ + p₃= - 6

From both the equations, we can find the value of

Therefore option D is the correct one p₂= 3 only

But we cannot determine the root of p₁ nor p₃

Finding the value of p will be possible only if the given quadratic equation is converted to a linear equation. Therefore a= 0

Let the number be a

So a-1/a = 35/6

Therefore a= 6