${(2 x^{2} + 3 x + 5)}^{1 / 2} + {(2 x^{2} + 3 x + 20)}^{1 / 2} = 15$ therefore x is

(-8/3)

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(14/5)

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(-11/2)

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Solution

The correct option is C (-11/2)
$(2 x^{2} + 3 x + 5)^{\frac{1}{2}} + (x^{2} + 3 x + 20)^{\frac{1}{2}} = 1$
Let $2 x^{2} + 3 x = a$
Then $(a + 5)^{\frac{1}{2}} + (a + 20)^{\frac{1}{2}} = 1$
Squarer both the side
Then $a + 5 + a + 20 + 2 (a + 5)^{\frac{1}{2}} (a + 20)^{\frac{1}{2}} = 1$
Or $2 a + 25 - 1 = - 2 (a + 5)^{\frac{1}{2}} (a + 20)^{\frac{1}{2}}$
Or $2 a + 24 = - 2 (a + 5)^{\frac{1}{2}} (a + 20)^{\frac{1}{2}}$
Squarer both the side
$a^{2} + 24 a + 144 = (a^{2} + 25 a + 100)$
Or $a^{2} + 24 a + 144 - (a^{2} + 25 a + 100) = 0$
Or $- a + 44 = 0$
Take a= $2 x^{2} + 3 x$
- $(2 x^{2} + 3 x)$ +44=0
Or - $- 2 x^{2} - 11 x + 8 x + 44 = 0 \Rightarrow - 2 x (2 x + 11) + 4 (2 x + 11) = 0 \Rightarrow (2 x + 11) (2 x - 4) = 0$
So $x = 2$ or x= $- \frac{11}{2}$

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(2x2+3x+5)1/2+(2x2+3x+20)1/2=15 therefore x is

${(2 x^{2} + 3 x + 5)}^{1 / 2} + {(2 x^{2} + 3 x + 20)}^{1 / 2} = 15$ therefore x is