Problem 17

Algebra Level 3

8 x 2 − 12 x − 39 + 2 x 2 − 3 x − 15 − 2 x 2 − 3 x + 20 = 0 \sqrt { { 8x }^{ 2 }-12x-39 } +\sqrt { { 2x }^{ 2 }-3x-15 } -\sqrt { { 2x }^{ 2 }-3x+20 } =0

Solve the equation, the roots of the above equation can be represented as A ± B 4 , gcd ( A , B ) = 1 \frac { A\pm \sqrt { B } }{ 4 } ,\text{gcd}(A,B)=1 .

Find the value of A + B A+B .


The answer is 140.

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Utkarsh Bansal by Utkarsh Bansal , 23, India

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1 solution

Chew-Seong Cheong
Mar 10, 2015

8 x 2 − 12 x − 39 + 2 x 2 − 3 x − 15 − 2 x 2 − 3 x + 20 = 0 \sqrt{8x^2-12x-39} + \sqrt{2x^2-3x-15} - \sqrt{2x^2-3x+20} = 0

⇒ 8 x 2 − 12 x − 39 = 2 x 2 − 3 x + 20 − 2 x 2 − 3 x − 15 \Rightarrow \sqrt{8x^2-12x-39} = \sqrt{2x^2-3x+20} - \sqrt{2x^2-3x-15}

8 x 2 − 12 x − 39 = 4 x 2 − 6 x + 5 − 2 ( 2 x 2 − 3 x + 20 ) ( 2 x 2 − 3 x − 15 ) \quad 8x^2-12x-39 = 4x^2-6x+5 -2\sqrt{(2x^2-3x+20)(2x^2-3x-15)}

4 x 2 − 6 x − 44 = − 2 ( 2 x 2 − 3 x + 20 ) ( 2 x 2 − 3 x − 15 ) \quad 4x^2-6x-44 = -2\sqrt{(2x^2-3x+20)(2x^2-3x-15)}

2 x 2 − 3 x − 22 = − ( 2 x 2 − 3 x + 20 ) ( 2 x 2 − 3 x − 15 ) \quad 2x^2-3x-22 = -\sqrt{(2x^2-3x+20)(2x^2-3x-15)}

Now, let y = 2 x 2 − 3 x − 22 y = 2x^2-3x-22 then, we have:

y = − ( y + 42 ) ( y + 7 ) ⇒ y 2 = y 2 + 49 y + 294 y = -\sqrt{(y+42)(y+7)} \quad \Rightarrow y^2 = y^2+49y+294

⇒ 0 = 49 y + 294 ⇒ 49 y = − 294 ⇒ y = − 6 \Rightarrow 0 = 49y+294 \quad \Rightarrow 49y = -294 \quad \Rightarrow y = -6

⇒ 2 x 2 − 3 x − 22 = − 6 ⇒ 2 x 2 − 3 x − 16 = 0 \Rightarrow 2x^2-3x-22 = -6 \quad \Rightarrow 2x^2-3x-16 =0

⇒ x = 3 ± 3 2 − 4 ( 2 ) ( − 16 ) 4 = 3 ± 137 4 \Rightarrow x = \dfrac {3\pm \sqrt{3^2-4(2)(-16)}}{4} = \dfrac {3\pm \sqrt{137}}{4}

⇒ A + B = 3 + 137 = 140 \Rightarrow A+B = 3+137 = \boxed{140}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Same approach here and nice solution sir,upvoted

Utkarsh Bansal - 6 years, 3 months ago

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It's a little overrated. 140-150 points would be fine.

Rajdeep Dhingra - 6 years, 2 months ago

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I too agree with you. I first set it to Level 4

Utkarsh Bansal - 6 years, 2 months ago

Yep, same approach kudos to you good sir.

Jesus Ulises Avelar - 6 years, 3 months ago

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