The quadratic equation x^2(?+?)x +17 has a minimum value of -8 and

Question

The quadratic equation  x^2−(a+b)x +17  has a minimum value of -8 and a^2− b^2 = 60. If the sum of the roots is positive, what is the value of a^2 + b^2?

A. 64
B. 82
C. 68
D. 58
E. 76

adewale223 · Answer

x^2 - (a + b )x +17

(x^2 - (a + b )x + (a + b)^2/4) - (a + b )^2/4 + 17

Therefore

-(a + b )^2 /4 + 17 = -8

- (a + b )^2 = - 100

(a + b ) ^2 = 100

Since sum of roots is positive

a + b = 10..........................................1

(a + b)(a - b ) = 60

a - b = 6..............................................2

Subtract equation 2 from 1 gives

2b = 4

b = 2

Sub for b in any of equation gives a = 8

a ^2 + b^2 = 68

Answer C

Adewale FAsipe, GMAT Quant instructor from Lagos, Nigeria.

Deepa444 · Answer

chetan2u   Bunuel   Sajjad1994  can anyone help with the solution?

chetan2u · Answer

This is a question related to parabola and the minimum value the parabola takes. Such questions would not be seen on GMAT.

You could differentiate to get the minim or maxima of an equation.

x^2−(a+b)x +17..........\frac{d}{dx}(x^2-(a+b)x+17)=0......2x-(a+b)=0......x=\frac{(a+b)}{2} .....

So the value becomes =>  x^2−(a+b)x +17=(\frac{a+b}{2})^2-(a+b)\frac{(a+b)}{2}+17=-8.....(a+b)^2=100.....a+b=10 , as sum of roots i spositive.

Now  a^2-b^2=60...(a-b)(a+b)=60.......(a-b)*10=60......a-b=6 
a+b=10 and a-b=6. So, (a+b)+(a-b)=10+6......2a=16 or a=8 and b=2
 a^2+b^2=2^2+8^2=64+4=68 
 Deepa444

btsaami · Answer

Method 1: Since minimum value of quadratic eqn x^2−(a+b)x+17 is -8. 
x = -8-17= -25
We have -ve sign we can get this product when one number is 5 and other is -5.
Since x>x-(a+b), x=5 and 5-10=-5 i.e. a+b=10

From a^2− b^2= 60, we get a-b=6. Solve both equations, a=8 , b=2
a^2+b^2= 64+4= 68

Method 2: Use differentiation, since a>0, it will have a minima.
Diff we get,
2x= a+b
substitute the value of x in the equation
we get
(a+b)^2=100
a-b=6
Next steps, same as above

yogeshr2 · Answer

if we assume the original equation to be (x-a)(x-b), then it comes out to x^2 -ax - bx + ab, and the equation in the equation is x^2 -ax -bx + 17. Then, ab = 17. If the minimum value is -8, I also got (a+b)^2 = 100, but my thought process was (a+b)^2 = 100 = a^2 + 2ab + b^2. If ab = 17, 2ab = 34, thus a^2 + b^2 = 100-34 = 66. However, that is not the right answer. I understand how to do it the right way, but where is my lapse in logic?

chetan2u · Answer

You are taking it as  x^2-ax-bx+17=0 , but it is  x^2-ax-bx+17 \geq -8

ZIX · Answer

Here's how I found the answer:

a^2− b^2 = 60    (given)  
  (a+b)(a-b) = 60

I tried 10*6 as one solution so,

(8+2)(8-2) = 60  
 So  a = 8, b =2

a^2 + b^2 = 64 + 4 = 68

EshaFatim · Answer

chetan2u 
is that a right approach to solve this using vertex?
if we do  \frac{-b}{2a} = -8  which is  \frac{(a+b)}{2} = -8  in this case.

I started with this approach. but somehow the answer doesn't match. So I'm guessing there could be a gap in the concept I'm pursuing it with.

Bunuel · Answer

\frac{-b}{2a} = -8 represents the x-coordinate of the vertex. The minimum value is the y-coordinate, given by c - \frac{b^2}{4a} (check here: https://gmatclub.com/forum/math-coordin ... 87652.html). For x^2 − (m + n)x + 17 (I replaced a and b with m and n to avoid confusion with standard quadratic coefficients), c - \frac{b^2}{4a} becomes 17 - \frac{(m + n)^2}{4} . Thus, we have 17 - \frac{(m + n)^2}{4} = -8 , which simplifies to (m + n)^2 = 100 . Since the sum of the roots is positive, we get m + n = 10 . Given m + n = 10 and (m + n)(m - n) = 60 , we find m = 8 and n = 2 . Therefore, m^2 + n^2 = 68 . Answer: C.

The quadratic equation x^2(?+?)x +17 has a minimum value of -8 and

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The quadratic equation x^2(?+?)x +17 has a minimum value of -8 and

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