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Byju's Answer

Standard X

Mathematics

Quadratic Formula

Solve for x...

Question

Solve for $x : x^{2} + x + 4 = 0$

\frac{- 1 \pm i \sqrt{15}}{2}

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\frac{- 2 \pm i \sqrt{15}}{2}

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\frac{- 3 \pm i \sqrt{15}}{2}

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None of these

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Solution

The correct option is A $\frac{- 1 \pm i \sqrt{15}}{2}$
$x^{2} + x + 4 = 0$

using quadratic formula

$\Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\Rightarrow x = \frac{- 1 \pm \sqrt{1 - 16}}{2}$

$\Rightarrow x = \frac{- 1 \pm i \sqrt{1} 5}{2}$ .

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Similar questions

Solve the following quadratic equations :

(i) $x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2} i = 0$

(ii) $x^{2} - (5 - i) x + (18 + i) = 0$

(iii) $(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0$

(iv) $x^{2} - (2 + i) x - (1 - 7 i) = 0$

(v) $i x^{2} - 4 x - 4 i = 0$

(vi) $x^{2} + 4 i x - 4 = 0$

(vii) $2 x^{2} + \sqrt{15} i x - i = 0$

(viii) $x^{2} - x + (1 + i) = 0$

(ix) $i x^{2} - x + 12 i = 0$

(x) $x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0$

(xi) $x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0$

(xii) $2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0$

Q. Solve the following quadratic equations:
(i)

x^{2} - (3 \sqrt{2} + 2 i) x + 6 \sqrt{2 i} = 0

(ii)

x^{2} - (5 - i) x + (18 + i) = 0

(iii)

(2 + i) x^{2} - (5 - i) x + 2 (1 - i) = 0

(iv)

x^{2} - (2 + i) x - (1 - 7 i) = 0

(v)

i x^{2} - 4 x - 4 i = 0

(vi)

x^{2} + 4 i x - 4 = 0

(vii)

2 x^{2} + \sqrt{15} i x - i = 0

(viii)

x^{2} - x + (1 + i) = 0

(ix)

i x^{2} - x + 12 i = 0

(x)

x^{2} - (3 \sqrt{2} - 2 i) x - \sqrt{2} i = 0

(xi)

x^{2} - (\sqrt{2} + i) x + \sqrt{2} i = 0

(xii)

2 x^{2} - (3 + 7 i) x + (9 i - 3) = 0

I. $3 x^{2} - 15 x + 18 = 0$

II. $y^{2} - 3 y - 4 = 0$

If $(- 7 - 24 i)^{1 / 2}$ =x-iy, then $x^{2}$ + $y^{2}$ =

If z = \frac{1}{(2 + 3 i)^{2}}, than |z| =

(a)

\frac{1}{13}

(b)

\frac{1}{5}

(c)

\frac{1}{12}

(d) none of these

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Solve for x:x2+x+4=0

The correct option is A −1±i√152x2+x+4=0using quadratic formula⇒x=−b±√b2−4ac2a⇒x=−1±√1−162⇒x=−1±i√152.

Solve for $x : x^{2} + x + 4 = 0$

The correct option is A $\frac{- 1 \pm i \sqrt{15}}{2}$
$x^{2} + x + 4 = 0$

using quadratic formula

$\Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\Rightarrow x = \frac{- 1 \pm \sqrt{1 - 16}}{2}$

$\Rightarrow x = \frac{- 1 \pm i \sqrt{1} 5}{2}$ .