How do you solve 2x2=x+28?

Question

How do you solve  2x2=x+28?

Kyler Jacobson · Accepted Answer

Step 1
Move all values and variables to the left and either factor, solve with the quadratic equation, or solve by completing the square; you could also leave the expression alone and solve by graphing.
$2 x^{2} = x + 28$
$2 x^{2} - x = 28$
$2 x^{2} - x - 28 = 0$
Next, try to factor this expression
$(2 x + a) (x + b) = 0$
Where $a \times b = - 28$ and $2 b x + a x = - x$ or $2 b + a = - 1$
The factors of -28 are:
$- (a \times b)$ or $- (b \times a)$
$- (1 \times 28)$
$- (2 \times 14)$
$- (4 \times 7)$
Of these factors, only the last pair will allow you to make
$2 b + a = - 1$
by setting
$b = - 4$ and $a = 7$
But this means we can factor the expression above.
$(2 x + a) (x + b) = 0$
$(2 x + 7) (x - 4) = 0$ which is only true when
$x = - \frac{7}{2}$ or $x = 4$
Check your work by plugging these values back into the original expression.

izumrledk · Accepted Answer

Step 1 You could solve this quadratic equation by completing the square. To do that, start by getting your quadratic to the general form x2+bax=−ca by adding -x to both sides of the equation and dividing your terms by 1 2x2−x=⧸x−⧸x+28 ⧸2×x2⧸2−x2=282 x2−x2=14 Now, you need to add a term to both sides of the equation so that the left side of the equation can be written as the square of a binomial. The coefficient of the x-term will tell what term you need to add. More specifically, divide this coefficient by 2 and square the result to get (−12×12)2=116 Add 116 to both sides of the equation x2−x2+116=14+116 The left side of the equation can now be rewritten as x2−2×(14)×x+(14)2=(x−14)2 You now have (x−14)2=22516 Take the square root of both sides of the equation (x−14)2=22516 x−14=±154⇒x1,2=14±154 The two solutions to the equation will thus be x1=14+154=4 and x2=14−154=−72

RizerMix · Accepted Answer

Step 1 1) Solve y=2x2−x−28=0 I use the new Transforming Method. Transformed equation 2) y′=x2−x−56=0 Factor pairs of (−56)→(−4,14)(−7,8) This sum is 1=−b. Two real roots of (2) are: -7 and 8. The 2 real roots of equation (1) are: (−72) and (82=4)

How do you solve 2x^{2}=x+28?

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