4j2+5j-9=0

For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=4⋅-9=-36 and whose sum is b=5.

Factor 5 out of 5j.

4j2+5(j)-9=0

Rewrite 5 as -4 plus 9

4j2+(-4+9)j-9=0

Apply the distributive property.

4j2-4j+9j-9=0

4j2-4j+9j-9=0

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

(4j2-4j)+9j-9=0

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

4j(j-1)+9(j-1)=0

4j(j-1)+9(j-1)=0

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

(j-1)(4j+9)=0

(j-1)(4j+9)=0

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

j-1=0

4j+9=0

Set the first factor equal to 0.

j-1=0

Add 1 to both sides of the equation.

j=1

j=1

Set the next factor equal to 0.

4j+9=0

Subtract 9 from both sides of the equation.

4j=-9

Divide each term by 4 and simplify.

Divide each term in 4j=-9 by 4.

4j4=-94

Cancel the common factor of 4.

Cancel the common factor.

4j4=-94

Divide j by 1.

j=-94

j=-94

Move the negative in front of the fraction.

j=-94

j=-94

j=-94

The final solution is all the values that make (j-1)(4j+9)=0 true.

j=1,-94

The result can be shown in multiple forms.

Exact Form:

j=1,-94

Decimal Form:

j=1,-2.25

Mixed Number Form:

j=1,-214

Solve for j 4j^2+5j-9=0