Question

solve for x (round decimals to nearest tenth):
$2x^2 + 6x - 7 = 0$
first x=
second x=
question 6
1 pts
consider the quadratic equation: $x^2 + 4x + 5 = 0$
what is the value of the discriminant?

Explanation:

First Problem: Solve \(2x^2 + 6x - 7 = 0\) for \(x\) (round to nearest tenth)

Step1: Identify coefficients for quadratic formula

For a quadratic equation \(ax^2 + bx + c = 0\), here \(a = 2\), \(b = 6\), \(c = -7\). The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).

Step2: Calculate discriminant (\(D = b^2 - 4ac\))

\(D=(6)^2-4\times2\times(-7)=36 + 56 = 92\)

Step3: Apply quadratic formula for first \(x\) (using \(+\))

\(x_1=\frac{-6+\sqrt{92}}{2\times2}=\frac{-6 + 9.5917}{4}=\frac{3.5917}{4}\approx0.9\)

Step4: Apply quadratic formula for second \(x\) (using \(-\))

\(x_2=\frac{-6-\sqrt{92}}{2\times2}=\frac{-6 - 9.5917}{4}=\frac{-15.5917}{4}\approx - 3.9\)

For a quadratic equation \(ax^2+bx + c = 0\), discriminant \(D=b^2 - 4ac\). Here \(a = 1\), \(b = 4\), \(c = 5\).
So \(D=(4)^2-4\times1\times5=16 - 20=-4\)

Answer:

First \(x\approx0.9\)
Second \(x\approx - 3.9\)

Second Problem: Find discriminant of \(x^2 + 4x + 5 = 0\)