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SAT - Math - Quadratic Equations

In the equation above a b and c are constants If the equations is true for all values of x what is the value of b (more)

2x(3x 5) 3(3x 5) = ax2 bx c

In the equation above, a, b, and c are constants. If the equations is true for all values of x, what is the value of b?

2x(3x + 5) + 3(3x + 5) = ax2 + bx + c

expanding LHS
6x2 + 10x + 9x + 15 = ax2 + bx + c

6x2 + 19x + 15 = ax2 + bx + c

comparing coeficients

b = 19

If (ax+2)(bx+7) = 15x2 + cx +14 for all values of x, and a + b = 8, what are the two possible values for c? (more)

If (ax 2)(bx 7) = 15x2 cx 14 for all values of x, and a b = 8, what are the two possible values for c?

(ax+2)(bx+7) = 15x2 + cx +14 

expanding LHS

abx2 + (2b + 7a)x + 14 = 15x2 + cx + 14

ab = 15 ... (1)

2b + 7a = c ... (2)

given: a + b = 8 ... (3)

from (1) and (3)

b = 15/a

a + 15/a = 8

a2 - 8a + 15 = 0

a2 - 5a - 3a + 15 = 0

a(a - 5) - 3(a - 5) = 0

(a - 3) (a - 5) =0

a = 3, a = 5

and

b = 5, b = 3

Case 1:

c = 2b + 7a = 2*3 + 7*5 = 41

Case 2

c = 2b + 7a = 2*5 + 7*3 = 31

 

For which of the following values of a and b does the system of equations have exactly two real solutions? y = 3 y = ax^2 + b (more)

For which of the following values of a and b does the system of equations have exactly two real solutions?
y = 3 
y = ax2 b

A) a= -2, b= 2
B) a= -2, b= 4
C) a= 2, b= 4
D) a= 4, b= 3 

 

Please revise this question. One equation fixes the value of y to be 3. If we follow this, second equation becomes 3=ax^2.b now, this is a quad. eqn. which will always have two real solutions. If the question actually says y=ax^2+b, we have, 3=ax^2+b ax^2+b-3+0 using Shree Dharacharya formula (otherwise known as the quadratic formula), x=[-b+-(b^2-4ac)^1/2]/2a x=[-b+-(b^2+12a)^1/2]/2a for two real values make discriminant , b^2-4ac,greater than zero b^2+12a>0 Easiest way out is to check with the given options, We get, (C) and (D) are both correct.

What is the sum of all values of m that satisfy 2m^2 - 16m + 8 = 0 ? (more)

What is the sum of all values of m that satisfy 2m2 - 16m 8 = 0 ?

For a quadratic equation sum of roots = -b/a = - (-16/2) = 8

(2k2 +17)2 - x = 0 If k > 0 and x = 7 in the equation above, what is the value of k? (more)

(2k 17)2 - x = 0

If k > 0 and x = 7 in the equation above, what is the value of k?

(2k2 +17)2 - x = 0

given x = 7

(2k2 +17)2 = 7

(2k2 +17) = ± 7

case 1

(2k2 +17) =  7

2k2 =  7 - 17 < 0

No values of k

case 2

(2k2 +17) = - 7

2k2 = -  7 - 17 < 0

No values of k