Solve for j 2j^2-3j=-1

$2 j^{2} - 3 j = - 1$

Step 1

Add $1$ to both sides of the equation.

$2 j^{2} - 3 j + 1 = 0$

Step 2

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 j$ .

$2 j^{2} - 3 j + 1 = 0$

Rewrite $- 3$ as $- 1$ plus $- 2$

$2 j^{2} + (- 1 - 2) j + 1 = 0$

Apply the distributive property.

$2 j^{2} - 1 j - 2 j + 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 j^{2} - 1 j) - 2 j + 1 = 0$

Factor out the greatest common factor (GCF) from each group.

$j (2 j - 1) - (2 j - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 j - 1$ .

$(2 j - 1) (j - 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 j - 1 = 0$

$j - 1 = 0$

Step 4

Set $2 j - 1$ equal to $0$ and solve for $j$ .

Set $2 j - 1$ equal to $0$ .

$2 j - 1 = 0$

Solve $2 j - 1 = 0$ for $j$ .

Add $1$ to both sides of the equation.

$2 j = 1$

Divide each term in $2 j = 1$ by $2$ and simplify.

Divide each term in $2 j = 1$ by $2$ .

$\frac{2 j}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 j}{2} = \frac{1}{2}$

Divide $j$ by $1$ .

$j = \frac{1}{2}$

Step 5

Set $j - 1$ equal to $0$ and solve for $j$ .

Set $j - 1$ equal to $0$ .

$j - 1 = 0$

Add $1$ to both sides of the equation.

$j = 1$

Step 6

The final solution is all the values that make $(2 j - 1) (j - 1) = 0$ true.

$j = \frac{1}{2}, 1$

Step 7

The result can be shown in multiple forms.

Exact Form:

$j = \frac{1}{2}, 1$

Decimal Form:

$j = 0.5, 1$

Solve for j 2j^2-3j=-1

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