Question

solve each quadratic equation using the quadratic formula.
1.) $2x^2 - 3x - 5 = 0$
2.) $4x^2 + 8x + 7 = 4$
3.) $2x^2 + 9x = -7$
4.) $5x^2 = 80$
5.) $x^2 = 4x - 4$
6.) $x^2 - 31 - 2x = -6 - 3x^2 - 2x$

Explanation:

1. Solve \( 2x^2 - 3x - 5 = 0 \)

Step1: Identify \( a, b, c \)

For \( ax^2 + bx + c = 0 \), here \( a = 2 \), \( b = -3 \), \( c = -5 \).

Step2: Quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)

Substitute values: \( x=\frac{-(-3)\pm\sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} \)

Step3: Simplify discriminant

\( (-3)^2 - 4(2)(-5)=9 + 40 = 49 \), so \( x=\frac{3\pm\sqrt{49}}{4} \)

Step4: Simplify square root and solve

\( \sqrt{49}=7 \), so \( x=\frac{3\pm7}{4} \).
For \( + \): \( x=\frac{3 + 7}{4}=\frac{10}{4}=\frac{5}{2} \)
For \( - \): \( x=\frac{3 - 7}{4}=\frac{-4}{4}=-1 \)

Step1: Rewrite in standard form

Subtract 4: \( 4x^2 + 8x + 3 = 0 \), so \( a = 4 \), \( b = 8 \), \( c = 3 \).

Step2: Quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)

Substitute: \( x=\frac{-8\pm\sqrt{8^2 - 4(4)(3)}}{2(4)} \)

Step3: Simplify discriminant

\( 64 - 48 = 16 \), so \( x=\frac{-8\pm\sqrt{16}}{8} \)

Step4: Simplify square root and solve

\( \sqrt{16}=4 \), so \( x=\frac{-8\pm4}{8} \).
For \( + \): \( x=\frac{-8 + 4}{8}=\frac{-4}{8}=-\frac{1}{2} \)
For \( - \): \( x=\frac{-8 - 4}{8}=\frac{-12}{8}=-\frac{3}{2} \)

Step1: Rewrite in standard form

Add 7: \( 2x^2 + 9x + 7 = 0 \), so \( a = 2 \), \( b = 9 \), \( c = 7 \).

Step2: Quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \)

Substitute: \( x=\frac{-9\pm\sqrt{9^2 - 4(2)(7)}}{2(2)} \)

Step3: Simplify discriminant

\( 81 - 56 = 25 \), so \( x=\frac{-9\pm\sqrt{25}}{4} \)

Step4: Simplify square root and solve

\( \sqrt{25}=5 \), so \( x=\frac{-9\pm5}{4} \).
For \( + \): \( x=\frac{-9 + 5}{4}=\frac{-4}{4}=-1 \)
For \( - \): \( x=\frac{-9 - 5}{4}=\frac{-14}{4}=-\frac{7}{2} \)

Answer:

\( x = \frac{5}{2} \) or \( x = -1 \)

2. Solve \( 4x^2 + 8x + 7 = 4 \)