let the quadratic eq ax^2 +bx + c be such that a,b,c are distinct and each of a,b,c belong to {1,2,3.....n} such that x+1 divides ax^2+ bx +c.If such quad.polynomials are < 99,then max(n)=?

a)14 b)15 c)16 d)18 e)none of these

Since, x+1 satisfies the equation ax^2+ bx+ c=0. Therefore, a(-1)^2+b(-1)+c=0. So, a - b+ c=0.

Now, a + c=b. Minimum value of b is 3. Number of quadratic equations will exceed 99 when b=12. So, the answer is none of these.

How to find the roots of a cubic equation?

There is no formula for solving cubic polynomial you have to factorise them if possible or you have to look for the sum and product of roots here it can be factorised easily

Got it. thanks alot 🙂

The area enclosed by** **f(x) = min {4-x, 4+x} and g(x) = max {-x, x-4} equals

8

12

16

24

A quadratic function attains minimum value of –25 at x = 1. If f(0) = –24, what is the value of f(6)?

0

If a=√(7+4√3), what will be the value of a+1/a?

Hello Nancy!

PFA the solution 🙂

If roots of equation (a^2+b^2)x^2-2b(a+c)x+b^2+c^2=0 are equal then a, b, c are in

A) AP

B) GP

C) HP

D) cannot be determined

PFA the solution 🙂

Solve the inequality

3x^2-7x-6<0

3x^{2} – 7x – 6 < 0.

3x^{2} – 9x - 2x – 6 < 0.

3x ( 3x -2) -2( x-3) < 0.

( 3x – 2) ( x-3) <0

2/3 3

___+ve____|_____-ve _____|____+ve___

2/3 < x < 3

(7/2+a)²+(7/2+a+x)²=7²-----(1)

(7/2+a-x)²+a²=(7/2)²-----(2)

Find values of a and x ?

For a=0.9086, x= 1.0286 and a= -1.6627, x= 4.9173

1. How many integral pairs (x, y), where 0< x, y < 1000, satisfy 141x + 517y = 4158?

A. 0

B. 1

C. 2

D. More than 2

The HCF of 141 and 517 is 47. But 47 does not completely divide 4158. Therefore, there are no integral solutions of the equation. Hence, option A.