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algebra 2 ws 3.6 quadratic formula name tyler beethcher hr problems 1 –…

Question

algebra 2 ws 3.6 quadratic formula name tyler beethcher hr problems 1 – 7: solve each equation using the quadratic formula. show work. 1. $x^2 - 21 = 4x$ 2. $4x^2 - 4x + 17 = 0$ 3. $8x - 1 = 4x^2$ 4. $7x^2 - 5x = 0$ 5. $3x^2 = -54$ 6. $x^2 = 4x - 15$

Explanation:

Step 1: Solve \(x^{2}-21 = 4x\)

First, rewrite the equation in standard quadratic form \(ax^{2}+bx + c=0\). Subtract \(4x\) from both sides: \(x^{2}-4x - 21=0\). Here, \(a = 1\), \(b=-4\), \(c=-21\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), substitute the values:
\(x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-21)}}{2(1)}=\frac{4\pm\sqrt{16 + 84}}{2}=\frac{4\pm\sqrt{100}}{2}=\frac{4\pm10}{2}\)
So, \(x=\frac{4 + 10}{2}=\frac{14}{2}=7\) or \(x=\frac{4-10}{2}=\frac{-6}{2}=-3\).

Step 2: Solve \(4x^{2}-4x + 17=0\)

Here, \(a = 4\), \(b=-4\), \(c = 17\).
Calculate the discriminant \(\Delta=b^{2}-4ac=(-4)^{2}-4(4)(17)=16-272=-256\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-(-4)\pm\sqrt{-256}}{2(4)}=\frac{4\pm16i}{8}=\frac{1\pm4i}{2}\).

Step 3: Solve \(8x - 1=4x^{2}\)

Rewrite in standard form: \(-4x^{2}+8x - 1=0\) or \(4x^{2}-8x + 1=0\) (multiplying by - 1). Here, \(a = 4\), \(b=-8\), \(c = 1\).
Discriminant \(\Delta=(-8)^{2}-4(4)(1)=64 - 16 = 48\).
Quadratic formula: \(x=\frac{-(-8)\pm\sqrt{48}}{2(4)}=\frac{8\pm4\sqrt{3}}{8}=\frac{2\pm\sqrt{3}}{2}\).

Step 4: Solve \(7x^{2}-5x=0\)

Factor out \(x\): \(x(7x - 5)=0\). So, either \(x = 0\) or \(7x-5=0\Rightarrow x=\frac{5}{7}\).

Step 5: Solve \(3x^{2}=-54\)

Rewrite as \(3x^{2}+54=0\) or \(x^{2}+18=0\). Here, \(a = 1\), \(b = 0\), \(c = 18\).
Discriminant \(\Delta=0^{2}-4(1)(18)=-72\).
Quadratic formula: \(x=\frac{-0\pm\sqrt{-72}}{2(1)}=\frac{\pm6\sqrt{2}i}{2}=\pm3\sqrt{2}i\).

Step 6: Solve \(x^{2}=4x - 15\)

Rewrite in standard form: \(x^{2}-4x + 15=0\). Here, \(a = 1\), \(b=-4\), \(c = 15\).
Discriminant \(\Delta=(-4)^{2}-4(1)(15)=16 - 60=-44\).
Quadratic formula: \(x=\frac{-(-4)\pm\sqrt{-44}}{2(1)}=\frac{4\pm2\sqrt{11}i}{2}=2\pm\sqrt{11}i\).

Answer:

s:

  1. \(x=-3\) or \(x = 7\)
  2. \(x=\frac{1\pm4i}{2}\)
  3. \(x=\frac{2\pm\sqrt{3}}{2}\)
  4. \(x = 0\) or \(x=\frac{5}{7}\)
  5. \(x=\pm3\sqrt{2}i\)
  6. \(x=2\pm\sqrt{11}i\)