What is 4(m+2)^2+4(m+2)(3m-1)<= 0 ?

The solution to 4(m+2)^2+4(m+2)(3m-1)<= 0 is -2<= m<=-1/4

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4(m+2)^2+4(m+2)(3m-1)<= 0

4 (m + 2)^{2} + 4 (m + 2) (3 m - 1) \leq 0

- 2 \leq m \leq - \frac{1}{4}

Interval Notation

[- 2, - \frac{1}{4}]

Decimal Notation

- 2 \leq m \leq - 0.25

Solution steps

4 (m + 2)^{2} + 4 (m + 2) (3 m - 1) \leq 0

Rewrite in standard form

4 m^{2} + 9 m + 2 \leq 0

Factor

4 m^{2} + 9 m + 2 : (4 m + 1) (m + 2)

(4 m + 1) (m + 2) \leq 0

Identify the intervals

- 2 \leq m \leq - \frac{1}{4}

x^{3} + 3 x^{2} + 5 x + 3 \geq 0

\frac{( 36 x - 7 )}{( 2 + 3 x )} \geq 12

(x^{2} - 9)^{2} \cdot (x + 1) (x^{2} - 2 x - 3) (x - 1) \leq 0

- x^{3} + x \leq 0

\frac{( x ^{2} + 4 ) ( x - 2 )}{( x - 3 )} < 0

What is 4(m+2)^2+4(m+2)(3m-1)<= 0 ?
The solution to 4(m+2)^2+4(m+2)(3m-1)<= 0 is -2<= m<=-1/4