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

Math
4j2+5j-9=0
Factor by grouping.
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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.
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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.
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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 and solve.
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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 and solve.
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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.
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Divide each term in 4j=-9 by 4.
4j4=-94
Cancel the common factor of 4.
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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

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