Answer:
[tex]\textsf{1.} \quad x=-6, \quad x=6[/tex]
[tex]\textsf{2.} \quad x=-5, \quad x=2[/tex]
[tex]\textsf{3.} \quad x=-3, \quad x=7[/tex]
[tex]\textsf{4.} \quad x=\dfrac{-5 +\sqrt{57} }{4}, \quad x=\dfrac{-5 -\sqrt{57} }{4}[/tex]
[tex]\textsf{5.} \quad x=\dfrac{7+ \sqrt{309} }{10}, \quad x=\dfrac{7-\sqrt{309} }{10}[/tex]
Step-by-step explanation:
Question 1
Method: Extracting the Square Root
[tex]\begin{aligned}& \textsf{Given}: & x^2 & = 36\\& \textsf{Square root both sides}: & \sqrt{x^2} & = \sqrt{36}\\& \textsf{Simplify}: & x & = \pm 6\\&\textsf{Solution}:&x& = -6, 6\end{aligned}[/tex]
Question 2
Method: Factoring
[tex]\begin{aligned}& \textsf{Given}: & x^2+3x-10 & = 0\\& \textsf{Split the middle term}: & x^2+5x-2x-10 & = 0\\& \textsf{Factor the first two and the last two terms}: & x(x+5)-2(x+5)&=0\\& \textsf{Factor out the common term $(x+5)$}: & (x-2)(x+5)&=0\\& \textsf{Apply the zero-product property}: & (x-2)=0 \implies x&=2\\ &&(x+5)=0 \implies x&=-5\\ & \textsf{Solution}: & x&=-5,2\end{aligned}[/tex]
Question 3
Method: Factoring
[tex]\begin{aligned}& \textsf{Given}: & x^2-4x-21 & = 0\\& \textsf{Split the middle term}: & x^2-7x+3x-21 & = 0\\& \textsf{Factor the first two and the last two terms}: & x(x-7)+3(x-7)&=0\\& \textsf{Factor out the common term $(x-7)$}: & (x+3)(x-7)&=0\\& \textsf{Apply the zero-product property}: & (x+3)=0 \implies x&=-3\\&&(x-7)=0 \implies x&=7\\& \textsf{Solution}: & x & = -3, 7\end{aligned}[/tex]
Question 4
Method: Quadratic Formula
[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}[/tex]
Given:
[tex]5x-4=-2x^2[/tex]
Rearrange into standard form by adding 2x² to both sides:
[tex]\implies 2x^2+5x-4=0[/tex]
Therefore:
[tex]a=2, \quad b=5, \quad c=-4[/tex]
Substitute the values into the quadratic formula and solve for x:
[tex]\implies x=\dfrac{-5 \pm \sqrt{5^2-4(2)(-4)} }{2(2)}[/tex]
[tex]\implies x=\dfrac{-5 \pm \sqrt{25+32} }{4}[/tex]
[tex]\implies x=\dfrac{-5 \pm \sqrt{57} }{4}[/tex]
Therefore, the solutions are:
[tex]x=\dfrac{-5 +\sqrt{57} }{4}, \quad x=\dfrac{-5 -\sqrt{57} }{4}[/tex]
Question 5
Method: Quadratic Formula
[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}[/tex]
Given:
[tex]5x^2-7x-13=0[/tex]
Therefore:
[tex]a=5, \quad b=-7, \quad c=-13[/tex]
Substitute the values into the quadratic formula and solve for x:
[tex]\implies x=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(5)(-13)} }{2(5)}[/tex]
[tex]\implies x=\dfrac{7 \pm \sqrt{49+260} }{10}[/tex]
[tex]\implies x=\dfrac{7 \pm \sqrt{309} }{10}[/tex]
Therefore, the solutions are:
[tex]x=\dfrac{7+ \sqrt{309} }{10}, \quad x=\dfrac{7-\sqrt{309} }{10}[/tex]
