Question

1. for each equation, find the solutions.

a) $4(x - 3)^2+3 = 19$
b) $0 = 49 + 36x^2$
c) $-20 + 36x^2 = 44$
d) $2x^2-3x - 15 = 5$
e) $2x^2+x = 6$
use the quadratic formula:

1. $x^2 + 8x+15 = 2$

b) $2x^2-7x - 13=-10$
c) $x^2-6x + 12=$

Explanation:

Step1: Solve equation a) $4(x - 3)^2+3 = 19$

First, subtract 3 from both sides: $4(x - 3)^2=19 - 3=16$. Then divide both sides by 4: $(x - 3)^2 = 4$. Take the square - root of both sides: $x-3=\pm2$. So $x=3 + 2=5$ or $x=3 - 2 = 1$.

Step2: Solve equation b) $0 = 49+36x^2$

Rearrange to get $36x^2=-49$. Then $x^2=-\frac{49}{36}$. Since the square of a real number cannot be negative, there are no real solutions.

Step3: Solve equation c) $-20 + 36x^2=44$

Add 20 to both sides: $36x^2=44 + 20=64$. Divide both sides by 36: $x^2=\frac{64}{36}=\frac{16}{9}$. Take the square - root: $x=\pm\frac{4}{3}$.

Step4: Solve equation d) $2x^2-3x - 15 = 5$

Rearrange to the standard quadratic form $2x^2-3x-20 = 0$. Factor: $(2x + 5)(x - 4)=0$. Set each factor equal to zero: $2x+5 = 0$ gives $x=-\frac{5}{2}$; $x - 4=0$ gives $x = 4$.

Step5: Solve equation e) $2x^2+x=6$

Rearrange to $2x^2+x - 6=0$. Factor: $(2x - 3)(x + 2)=0$. Set each factor equal to zero: $2x-3 = 0$ gives $x=\frac{3}{2}$; $x + 2=0$ gives $x=-2$.

Step6: Solve equation 5a) $x^2+8x + 15=2$

Rearrange to $x^2+8x + 13=0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = 8$, $c = 13$. $x=\frac{-8\pm\sqrt{8^2-4\times1\times13}}{2\times1}=\frac{-8\pm\sqrt{64 - 52}}{2}=\frac{-8\pm\sqrt{12}}{2}=\frac{-8\pm2\sqrt{3}}{2}=-4\pm\sqrt{3}$.

Step7: Solve equation 5b) $2x^2-7x-13=-10$

Rearrange to $2x^2-7x - 3=0$. Using the quadratic formula with $a = 2$, $b=-7$, $c=-3$, $x=\frac{7\pm\sqrt{(-7)^2-4\times2\times(-3)}}{2\times2}=\frac{7\pm\sqrt{49 + 24}}{4}=\frac{7\pm\sqrt{73}}{4}$.

Step8: Solve equation 5c) $x^2-6x + 12$ (assuming the equation is $x^2-6x + 12 = 0$)

Using the quadratic formula with $a = 1$, $b=-6$, $c = 12$, $x=\frac{6\pm\sqrt{(-6)^2-4\times1\times12}}{2\times1}=\frac{6\pm\sqrt{36 - 48}}{2}=\frac{6\pm\sqrt{-12}}{2}=\frac{6\pm2\sqrt{3}i}{2}=3\pm\sqrt{3}i$.

Answer:

a) $x = 5,x = 1$
b) No real solutions
c) $x=\frac{4}{3},x=-\frac{4}{3}$
d) $x = 4,x=-\frac{5}{2}$
e) $x=-2,x=\frac{3}{2}$
5a) $x=-4+\sqrt{3},x=-4-\sqrt{3}$
5b) $x=\frac{7+\sqrt{73}}{4},x=\frac{7-\sqrt{73}}{4}$
5c) $x=3+\sqrt{3}i,x=3-\sqrt{3}i$