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555-605 (Medium)|   Algebra|      
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Bunuel
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 06 Nov 2014, 09:23
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Tough and Tricky questions: Algebra.



Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

A. 18
B. 15
C. -15
D. -21
E. -24

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D
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 06 Nov 2014, 09:38
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f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

f(6) = 0= 36+6b+c--- Taking 36 to the other side

-> 6b+c= -36

f(-3) = 0= 9-3b+c --- taking -3b+c to the other side

-> 3b-c=9

When we add these 2 equations, we get 9b= -27---> b= -3

And while substituting b= -3 we get c= -18.

b+c= -21--- Answer D
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 10 Nov 2014, 01:37
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\(f(x) = x^2 + bx + c\)

f(6) = 36 + 6b + c = 0

f(-3) = 9 - 3b + c = 0

c = -18; b = -3

b+c = -18-3 = -21

Answer = D
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 17 Aug 2015, 06:45
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6 and -3 are roots of quadratic equation x^2+bx+c=0

c=6*(-3)= -18
b=6+(-3)=3*-1=-3
b+c=-18+(-3)=-21

D
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 19 Jun 2019, 10:46
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Bunuel

Tough and Tricky questions: Algebra.



Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

A. 18
B. 15
C. -15
D. -21
E. -24

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f(6) = 0
—> 6^2 + b(6) + c = 0
—> 36 + 6b + c = 0
—> 6b + c = -36 ...... (1)

and f(-3) = 0
—> (-3)^2 + b(-3) + c = 0
—> 9 - 3b + c = 0
—> -3b + c = -9 ....... (2)

(1) - (2) gives
—> 6b + c - (-3b + c) = -36 - (-9)
—> 9b = -27
—> b = -3

Substitute in (1)
—> 6(-3) + c = -36
—> c = -18

So, b + c = -3 + (-18) = -21


Alternative Method:

f(6) = 0 and f(-3) = 0
—> Roots of the given quadratic equation are 6 & -3

So, the quadratic equation is (x - 6)(x + 3) = 0
—> x^2 - 3x - 18 = 0
Compare the above with given equation x^2 + bx + c = 0
—> b = -3 & c = -18
So, b + c = -3 + (-18) = -21

IMO Option C

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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 23 Jun 2019, 19:35
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Bunuel

Tough and Tricky questions: Algebra.



Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

A. 18
B. 15
C. -15
D. -21
E. -24

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Since f(6) = 0:

36 + 6b + c = 0

c = -6b - 36

Since f(-3) = 0

9 - 3b + c = 0

c = 3b - 9

Setting the two values for c equal:

-6b - 36 =3b - 9

-27 = 9b

-3 = b

Thus, c = -6(-3) - 36 = 18 - 36 = -18

So, b + c = -3 + -18 = -21
Alternate solution:

Since f(6) = 0 and f(-3) = 0, the equation x^2 + bx + c = 0 can be factored as (x - 6)(x + 3) = 0. Expand (x - 6)(x + 3) = 0, we have x^2 - 3x - 18 = 0. We see that b must be -3 and c must be -18. Therefore, b + c = -3 + (-18) = -21.

Answer: D
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 06 Jul 2019, 22:51
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Bunuel

Tough and Tricky questions: Algebra.



Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c =

A. 18
B. 15
C. -15
D. -21
E. -24

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Sol : f(6) = 36 + 6b + c = 0
Or 6b+c = - 36 -- 1 )

F(-3) = 9 - 3B + C = 0
or , 3B - c = 9 -- 2)

Equating 1 & 2
B= -3 & C = -18

B+C = -3 +(-18) = -21
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Let f(x) = x^2 + bx + c. If f(6) = 0 and f(-3) = 0, then b + c = 18 Apr 2026, 01:18
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