Question

clare solves the quadratic equation $4x^2 + 12x + 58 = 0$. if clare solves it correctly, which of the following solutions will she get?\
\
\\(\bigcirc\\) a) $x = -1.5 \pm \sqrt{28}$\
\\(\bigcirc\\) b) $x = -1.5 \pm i\sqrt{3.5}$\
\\(\bigcirc\\) c) $x = -1.5 \pm 3.5i$\
\\(\bigcirc\\) d) $x = -1.5 \pm 28i$

Explanation:

Step1: Recall quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(4x^{2}+12x + 58=0\), we have \(a = 4\), \(b=12\), \(c = 58\).

Step2: Calculate discriminant \(\Delta=b^{2}-4ac\)

Substitute \(a = 4\), \(b = 12\), \(c=58\) into the discriminant formula:
\(\Delta=(12)^{2}-4\times4\times58=144 - 928=- 784\)

Step3: Apply quadratic formula

\(x=\frac{-12\pm\sqrt{-784}}{2\times4}=\frac{-12\pm28i}{8}\) (since \(\sqrt{-784}=\sqrt{784}\times\sqrt{- 1}=28i\))
Simplify the fraction: \(\frac{-12}{8}\pm\frac{28i}{8}=-1.5\pm3.5i\)

Answer:

c) \(x = -1.5\pm3.5i\)