## If the roots of the equation are -7/4 , 4/7 then the equation will be?

A. 28x^2+43x-28=0
B. 28x^2-43x-28=0
C. 28x^2+49x-16=0
D. 28x^2-33x-28=0

Correct Answer: C. 28x^2 + 49x -16 = 0

### Solution

If the roots of the equation are `-7/4` and `4/7`, then the equation can be written as:

(x + 7/4)(x – 4/7) = 0

Multiplying out the terms gives:

x^2 + (7/4)x – (4/7)x – 7/4 * 4/7 = 0
x^2 + (49x – 16)/28 = 0

Multiplying both sides by `28` to get rid of the denominator, we get:

28x^2 + 49x – 16 = 0

Therefore, the equation with roots `-7/4` and `4/7` is `28x^2 + 49x - 16 = 0`.

## If S=the sum of the roots of the equation and P= Product of the roots of the equation, then their equation can be written as:

A. x^2+Sx+P=0
B. x^2+Sx-P=0
C. x^2-Sx-P=0
D. x^2-Sx+P=0

### Solution

If we have a quadratic equation of the form `ax^2 + bx + c = 0`, the sum of its roots is given by:

S = -b/a

and the product of its roots is given by:

P = c/a

To see why these formulas are true, we can use Vieta’s formulas, which state that for a quadratic equation of the form `ax^2 + bx + c = 0`, the sum of the roots is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, and the product of the roots is equal to the constant term divided by the coefficient of the quadratic term.

So, we can write the quadratic equation as:

a(x – r1)(x – r2) = 0

where `r1` and `r2` are the roots of the equation. Expanding the product and comparing the coefficients with the equation `ax^2 + bx + c = 0`, we get:
a(x^2 – (r1 + r2)x + r1r2) = 0
Equating the coefficients of corresponding powers of `x`, we get:
r1 + r2 = -b/a
r1r2 = c/a

Therefore, we have:

S = -b/a
P = c/a
Thus, if `S` is the sum of the roots of the equation and `P` is the product of the roots of the equation, we can write the equation as:
x^2 – Sx + P = 0

## If the discriminant of ax2+bx+c=0 is not a perfect square then its roots are:

A. Equal
B. Rational
C. Irrational
D. Real

### Solution

If the discriminant of the quadratic equation `ax^2 + bx + c = 0` is not a perfect square, then its roots are `irrational` and `real`.

The discriminant of the quadratic equation is given by the expression `b^2 - 4ac`. If this value is not a perfect square, then the roots of the equation will be of the form:

(-b ± √(b^2 – 4ac)) / 2a

Since the discriminant is not a perfect square, the square root of `b^2 - 4ac` will be irrational, and the roots will be of the form (-b ± irrational number) / 2a, which are both real and irrational. Therefore, the answer is option C: `Irrational`.

## The sum of the roots of the equation 122-7x+4=0 is:

A. 12/7
B. -7/122
C. 1/3
D. 7/122

### Solution

122x^2 – 7x + 4 = 0

We can find the sum of the roots of this equation using the formula:

sum of roots = -b/a

where a and b are the coefficients of x^2 and x, respectively.

Comparing the given equation with the standard form ax^2 + bx + c = 0, we can see that a = 122 and b = -7.

So, the sum of the roots of the equation 122x^2 – 7x + 4 = 0 is:

sum of roots = -b/a = -(-7) / 122 = 7/122

Therefore, the sum of the roots of the equation 122x^2 – 7x + 4 = 0 is 7/122.

## If the discriminant of the quadratic equation is +ve then the roots are:

A. Real & Equal
B. Rational & equal
C. Real & unequal
D. Complex

Correct Answer: C. Real & unequal

### Solution

If the discriminant of a quadratic equation ax^2 + bx + c = 0 is positive (i.e., b^2 – 4ac > 0), then the roots of the quadratic equation are real and unequal.

This is because the discriminant is used to determine the nature of the roots of a quadratic equation. Specifically, if the discriminant is positive, then the roots of the quadratic equation are real and unequal. This is because in this case, the quadratic formula (-b ± sqrt(b^2 – 4ac)) / 2a will yield two distinct real solutions for x.

In contrast, if the discriminant is zero (i.e., b^2 – 4ac = 0), then the roots of the quadratic equation are real and equal. This is because the quadratic formula will yield a single real solution for x.

If the discriminant is negative (i.e., b^2 – 4ac < 0), then the roots of the quadratic equation are complex conjugates. This means that the roots will have the form of a+bi and a-bi, where a and b are real numbers and i is the imaginary unit.

## Sum of the roots of equation ax^2+bx+c=0 is:

A. -b/a
B. -c/a
C. b/a
D. c/a

### Solution

The sum of the roots of a quadratic equation of the form ax^2 + bx + c = 0 is given by -b/a.

To see why this is the case, let the roots of the quadratic be x1 and x2. Then we can write the quadratic equation as:

ax^2 + bx + c = a(x – x1)(x – x2) = 0

Expanding the right-hand side gives:

a(x^2 – (x1 + x2)x + x1x2) = 0

Comparing the coefficients of x^2, x, and the constant term on both sides, we get:

x^2 coefficient: a = a x coefficient: -a(x1 + x2) = b constant term: a(x1)(x2) = c

Solving for x1 + x2, we have:

x1 + x2 = -b/a

Therefore, the sum of the roots of the quadratic equation ax^2 + bx + c = 0 is -b/a.

## The product of the roots of the equation 9x^2-5x-27=0 is:

A. -5/9
B. 5/27
C. -1/3
D. -3

### Solution

To find the product of the roots of the quadratic equation 9x^2-5x-27=0, we can use the fact that the product of the roots of a quadratic equation ax^2+bx+c=0 is equal to c/a.

Therefore, in the given equation 9x^2-5x-27=0, we can see that a=9, b=-5, and c=-27.

So, the product of the roots will be:

c/a = (-27) / (9) = -3

Hence, the product of the roots of the equation 9x^2-5x-27=0 is -3.

## If y^4+y^3-2y^2-8y+2 is divided by (y+1) then the remainder is ______.

A. -8
B. 0
C. 8
D. 11

### Solution

If y^4 + y^3 – 2y^2 – 8y + 2 is divided by y + 1 using synthetic division, we get:

-1 | 1 1 -2 -8 2
| -1 3 -1 9
+————–
| 1 0 -2 -9 11

The remainder is the last number in the last row, which is 11. Therefore, the correct option is D. 11

## Quadratic equation is an equation of degree.

A. 2
B. 3
C. 0
D. 1

### Solution

A quadratic equation is an equation of degree 2. The term “quadratic” comes from the Latin word “quadratus,” which means “square.” A quadratic equation has the form ax^2 + bx + c = 0, where a, b, and c are constants and a is not equal to 0. The highest power of the variable x in a quadratic equation is 2, which makes it a degree 2 polynomial equation. Therefore, the answer is A. 2.

A. Binomial
B. Trinomial
C. Monomial
D. None of these