Question

d) 3.2 + x - 1.2² = 0
a=3.2, b=1, c=-1.2

Explanation:

Step1: Identify the quadratic equation form

The equation is \(3.2 + x - 1.2x^{2}=0\), rewrite it in standard quadratic form \(ax^{2}+bx + c = 0\) as \(- 1.2x^{2}+x + 3.2 = 0\) or multiply both sides by - 1 to get \(1.2x^{2}-x - 3.2 = 0\). Here \(a = 1.2\), \(b=-1\), \(c = - 3.2\) (or if we take the original labeled \(a = 3.2\), \(b = 1\), \(c=-1.2\) there might be a mis - labeling, but we will proceed with the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\))

Step2: Calculate the discriminant \(\Delta\)

The discriminant \(\Delta=b^{2}-4ac\). If we use \(a = 1.2\), \(b=-1\), \(c=-3.2\), then \(\Delta=(-1)^{2}-4\times1.2\times(-3.2)=1 + 15.36=16.36\)

If we use the mis - labeled \(a = 3.2\), \(b = 1\), \(c=-1.2\) (assuming the equation is \(3.2x^{2}+x - 1.2=0\) which might be a mis - writing of the original equation), then \(\Delta=1^{2}-4\times3.2\times(-1.2)=1 + 15.36 = 16.36\)

Step3: Apply the quadratic formula

Using the quadratic formula \(x=\frac{-b\pm\sqrt{\Delta}}{2a}\)

Case 1: If \(a = 1.2\), \(b=-1\), \(c=-3.2\)
\(x=\frac{-(-1)\pm\sqrt{16.36}}{2\times1.2}=\frac{1\pm4.045}{2.4}\)
For the plus sign: \(x=\frac{1 + 4.045}{2.4}=\frac{5.045}{2.4}\approx2.102\)
For the minus sign: \(x=\frac{1-4.045}{2.4}=\frac{-3.045}{2.4}\approx - 1.269\)

Case 2: If \(a = 3.2\), \(b = 1\), \(c=-1.2\)
\(x=\frac{-1\pm\sqrt{16.36}}{2\times3.2}=\frac{-1\pm4.045}{6.4}\)
For the plus sign: \(x=\frac{-1 + 4.045}{6.4}=\frac{3.045}{6.4}\approx0.476\)
For the minus sign: \(x=\frac{-1-4.045}{6.4}=\frac{-5.045}{6.4}\approx - 0.788\)

Since there is a high chance of mis - writing in the original equation (the original equation is written as \(3.2 + x-1.2^{2}=0\) which is a linear equation \(x+3.2 - 1.44=0\) or \(x=-1.76\), but the labels \(a\), \(b\), \(c\) suggest a quadratic equation). If we assume the equation is a quadratic equation \(3.2x^{2}+x - 1.2 = 0\) (correcting the exponent error, maybe the \(x\) term was missed in the square: \(1.2x^{2}\) was written as \(1.2^{2}\) by mistake)

Let's re - express the equation correctly. If the equation is \(3.2x^{2}+x - 1.2=0\) (quadratic equation)

\(a = 3.2\), \(b = 1\), \(c=-1.2\)

\(\Delta=b^{2}-4ac=1^{2}-4\times3.2\times(-1.2)=1 + 15.36 = 16.36\)

\(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-1\pm\sqrt{16.36}}{2\times3.2}=\frac{-1\pm4.045}{6.4}\)

\(x_1=\frac{-1 + 4.045}{6.4}=\frac{3.045}{6.4}\approx0.476\)

\(x_2=\frac{-1-4.045}{6.4}=\frac{-5.045}{6.4}\approx - 0.788\)

If the equation is linear (the original writing \(3.2 + x-1.2^{2}=0\)):

\(3.2+x - 1.44 = 0\)

Step4: Solve the linear equation (if it's a linear equation)

\(x=1.44 - 3.2=-1.76\)

Answer:

If it's a linear equation: \(x=-1.76\); If it's a quadratic equation (\(3.2x^{2}+x - 1.2 = 0\)): \(x\approx0.476\) or \(x\approx - 0.788\) (or with the other quadratic form \(x\approx2.102\) or \(x\approx - 1.269\))