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Manipal Entrance Test 2020 Math Paper

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Question : 36 of 70

Marks: +1, -0

The sum of the real solutions of equation

2|x|^{2}+51=|1+20x|

is

5
24
0
None of these

Solution:

Given,

\;\; 2|x|^{2}+51=|1+20x|

Then, following two possible cases arise:

Case I When

1+20x>0

\Rightarrow \;\; x>-\; \frac{1}{20}

\therefore \;\; 2x^{2}+51=(1+20x)

\Rightarrow \;\; 2x^{2}-20x+50=0

\Rightarrow \;\; x^{2}-10x+25=0 \;\; [\because 2\neq 0]

\Rightarrow \;\; (x-5)^{2}=0 \Rightarrow x=5,5

Case II When

1+20x<0

\Rightarrow \;\; x<-\; \frac{1}{20}

\therefore\; 2x^{2}+51=-(1+20x)

\Rightarrow \;\; 2x^{2}-20x+52=0

\Rightarrow \;\; x^{2}-10x+26=0

Here,

\;\; D<0

Thus, roots are imaginary.

Hence, sum of real roots

=5+5=10

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