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What is the positive value of $3x^2 - 34x - 98$ ?

I don't know, but whatever it is, it is the answer to the question!

Let $x$ equal the answer to this question.

So,

$3x^2 - 34x - 98 = x \Rightarrow 3x^2 - 35x - 98 = 0 \Rightarrow (3x + 7)(x - 14) = 0 \Rightarrow x = -\frac{3}{7}, 14$

the positive value of

So $x = \boxed{14}$ .

According to the question, the value of the expression is x which is a positive integer. We need to find the positive value of the expression $3x^{2}-34x-98$ . By following the directions in the question, we get the equation, $3x^{2}-34x-98=x$ $\Rightarrow 3x^{2}-35x-98=0$ $\Rightarrow 3x^{2}-42x+7x-98=0$ $\Rightarrow 3x(x-14)+7(x-14)=0$ $\Rightarrow (x-14)(3x+7)=0$ $\Rightarrow x=14$ Since, x is a positive integer. As the value of the expression $3x^{2}-34x-98$ is equal to $x$ , the answer is $\boxed{14}$ .

According to the question, we have,

$3x^{2}-34x-98=x$

$\implies 3x^{2}-35x-98=0$

$\implies 3x^{2}-42x+7x-98=0$

$\implies 3x(x-14)+7(x-14)=0$

$\implies (x-14)(3x+7)=0$

$\implies x-14=0$ [neglecting 3x+7=0 as we require positive value of x]

$\implies x=\boxed{14}$

The question is based on wordplay. If you read carefully, it is telling you that x is the answer, thus we get the equation

$3x^{2} - 34x - 98 = x$

We can then solve it as a normal quadratic equation

$3x^{2} - 35x - 98 = 0$

Using the quadratic formula, the roots will be

$\frac{35 \pm \sqrt{(-35)^2 - (4 \times 3 \times (-98))}}{2 \times 3}$

$\frac{35 \pm \sqrt{1225- ( -1176 )}}{6}$

$\frac{35 \pm \sqrt{1225 + 1176 }}{6}$

$\frac{35 \pm \sqrt{2401 }}{6}$

$\frac{35 \pm 49}{6}$

We want a positive answer

$\frac{35 + 49}{6}$

$\Rightarrow \frac{84}{6}$

$\Rightarrow \boxed{14}$

If $x$ is a positive integer that is the answer to the question, then $3x^{2} - 34x - 98 = x$ because the answer is the positive value of $3x^{2} - 34x - 98$ .Then we subtract $x$ from both sides of the equation and we get $3x^{2} - 35x - 98 = 0$ .Now we can factorize the equation into $(3x+7)(x-14) = 0$ .So either $(3x+7)$ or $(x-14)$ must be 0.Because the question is asking for a positive integer, $x$ is equal to 14.

Since $x$ is a solution to $3x^{2} -34 x - 98$ then
$3x^{2} - 34x - 98= x$
$3x^{2} - 35x - 98 =0$
Factoring this produces:
$(3x+7)(x-14) =0$
$x= -\frac{7}{3} or x= 14$
Since the question requests the positive value, the answer is $\boxed{14}$ .

3x^2 -34x -98= x

=> 3x^2 -35x-98= 0

=> 3x^2 -42x+7x-98 = 0

=> 3x(x-14)+ 7(x-14)=0

=> (x-14)(3x+7)=0

=> x-14=0

=> x=14

The problem amounts to solving for a positive integer $x$ where $3x^2-34x-98=x.$ This yields the quadratic equation $3x^2-35x-98= 0$ which factors as $(3x+7)(x-14) = 0,$ so that $x=-\frac{7}{3}$ or $x= 14$ . Since $x$ is required to be a positive integer, we have $x=\boxed{14}.$

3X^2 - 34X - 98 = 0 <=> X ≈ 13, 72 or X ≈ -2, 38. X is a positive integer <=> X =14 => 3x 14^2 - 34 x14 - 98 = 14.

3x^2 - 34x - 98 = k. 3x^2 - 34x - (98 + k) = 0..(1) [x1 + x2 = 34/3, x1.x2 = - (98 + k)/3.] So, x2 = (34/3 - x1). Thus, x1(34/3 - x1) = -1/3 (k + 98) x1(34 - 3x1)/3 = -1/3 (k + 98) x1(34 - 3x1) = -(k + 98) - 3x1^2 + 34x1 = -(98+k) 3x1^2 - 34x1 - (98+k) = 0...........(2) We know if form (1) = (2).So, 3x^2 - 34x - (98 + k) = 0 [ k = x] 3x^2 - 34x - (98 + x) = 0 3x^2 - 35x - 98 = 0 (3x - 42)(x + 7/3) = 0 So, we have 3x - 42 = 0 , x = 42/3 = 14

3x^2 -34x-98=x 3x^2 -35x-98 =0 factors 14 & -7/3 so answer 14