The sum of twice a whole number and three times the square of the next largest whole number is $83$ . What is the number?

6

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Solution

The correct option is B $4$
Let the number be $x$ .
From the given condition,
$2 x + 3 (x + 1)^{2} = 83$
$3 x^{2} + 8 x + 3 = 83$
$3 x^{2} + 8 x - 80 = 0$
$3 x^{2} - 12 x + 20 x - 80 = 0$
$3 x (x - 4) + 20 (x - 4) + 0$
$(3 x + 20) (x - 4) = 0$
Since it is an integer,
$x = 4$ .

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Match Column I with Column II in the following:

$\begin{matrix} C o l u m n I & C o l u m n I I 1. & T h e d i f f e r e n c e o f 3 a n d a n u m b e r & (a) & 4 - 2 x s q u a r e d 2. & 5 l e s s t h a n t w i c e a n u m b e r s q u a r e d & (b) & n^{2} - 3 3. & F i v e m i n u s t w i c e t h e s q u a r e o f a & (c) & 2 n^{2} - 5 n u m b e r 4. & F o u r m i n u s a n u m b e r m u l t i p l i e d b y 2 & (d) & 5 - 2 n^{2} 5. & S e v e n t i m e s t h e s u m o f a n u m b e r a n d 1 & (e) & 3 - n^{2} 6. & A n u m b e r s q u a r e d p l u s 6 & (f) & 2 (n + 6) 7. & 2 t i m e s t h e s u m o f a n u m b e r a n d 6 & (g) & 7 (n + 1) 8. & T h r e e l e s s t h a n t h e s q u a r e o f a n u m b e r & (h) & n^{2} + 6 \end{matrix}$

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The sum of twice a whole number and three times the square of the next largest whole number is 83. What is the number?

The correct option is B 4Let the number be x.From the given condition,2x+3(x+1)2=833x2+8x+3=833x2+8x−80=03x2−12x+20x−80=03x(x−4)+20(x−4)+0(3x+20)(x−4)=0Since it is an integer,x=4.

The sum of twice a whole number and three times the square of the next largest whole number is $83$ . What is the number?

The correct option is B $4$
Let the number be $x$ .
From the given condition,
$2 x + 3 (x + 1)^{2} = 83$
$3 x^{2} + 8 x + 3 = 83$
$3 x^{2} + 8 x - 80 = 0$
$3 x^{2} - 12 x + 20 x - 80 = 0$
$3 x (x - 4) + 20 (x - 4) + 0$
$(3 x + 20) (x - 4) = 0$
Since it is an integer,
$x = 4$ .