Solve for a:- a3+a2+a-14=0

The given equation is,a3 + a2 + a - 14 = 0Let pa = a3 + a2 + a - 14Put a = 2, we getp2 = 23 + 22 + 2 - 14 = 8 + 4 + 2 - 14 = 14 - 14 = 0So, a - 2 is a factor of pa Now, we will divide pa by a-2 to obtain the other factors So, pa = a-2 a2 + 3a + 7So, a3 + a2 + a - 14 = 0⇒a-2 a2 + 3a + 7 = 0⇒a - 2 = 0 or a2 + 3a + 7 = 0⇒a = 2 or a2 + 3a + 7 = 0Consider, a2 + 3a + 7 = 0We have, A = 1; B = 3; C = 7Now, D = B2 - 4AC = 32 - 4 × 1 × 7 = 9 - 28 = -19 <0So, the given equation has imaginary roots Now, a = -B±D2A = -3±-192 = -3±19i22 = -3±19i2

Solve for a:- a³+a²+a-14=0

The given equation is, a^{3} + a^{2} + a - 14 = 0 Let p (a) = a^{3} + a^{2} + a - 14 Put a = 2, we get p (2) = 2^{3} + 2^{2} + 2 - 14 = 8 + 4 + 2 - 14 = 14 - 14 = 0 So, (a - 2) is a factor of p (a) . Now, we will divide p (a) by (a - 2) to obtain the other factors .

So, p (a) = (a - 2) (a^{2} + 3 a + 7) So, a^{3} + a^{2} + a - 14 = 0 \Rightarrow (a - 2) (a^{2} + 3 a + 7) = 0 \Rightarrow a - 2 = 0 or a^{2} + 3 a + 7 = 0 \Rightarrow a = 2 or a^{2} + 3 a + 7 = 0 Consider, a^{2} + 3 a + 7 = 0 We have, A = 1; B = 3; C = 7 Now, D = B^{2} - 4 AC = {(3)}^{2} - 4 \times 1 \times 7 = 9 - 28 = - 19 < 0 So, the given equation has imaginary roots . Now, a = \frac{- B \pm \sqrt{D}}{2 A} = \frac{- 3 \pm \sqrt{- 19}}{2} = \frac{- 3 \pm \sqrt{19 i^{2}}}{2} = \frac{- 3 \pm \sqrt{19} i}{2}

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