Question

If anyone can help me to solve this!!​

Answer 1

`\textsf{1.} \quad x=-6, \quad x=6`

`\textsf{2.} \quad x=-5, \quad x=2`

`\textsf{3.} \quad x=-3, \quad x=7`

`\textsf{4.} \quad x=(-5 +√(57) )/(4), \quad x=(-5 -√(57) )/(4)`

`\textsf{5.} \quad x=(7+ √(309) )/(10), \quad x=(7-√(309) )/(10)`

Explanation:

Question 1

Method: Extracting the Square Root

`\begin{aligned}& \textsf{Given}: & x^2 & = 36\\& \textsf{Square root both sides}: & √(x^2) & = √(36)\\& \textsf{Simplify}: & x & = \pm 6\\&\textsf{Solution}:&x& = -6, 6\end{aligned}`

Question 2

Method: Factoring

`\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}`

Question 3

Method: Factoring

`\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}`

Question 4

Method: Quadratic Formula

`\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=(-b \pm √(b^2-4ac))/(2a)$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}`

Given:

`5x-4=-2x^2`

Rearrange into standard form by adding 2x² to both sides:

`\implies 2x^2+5x-4=0`

Therefore:

`a=2, \quad b=5, \quad c=-4`

Substitute the values into the quadratic formula and solve for x:

`\implies x=(-5 \pm √(5^2-4(2)(-4)) )/(2(2))`

`\implies x=(-5 \pm √(25+32) )/(4)`

`\implies x=(-5 \pm √(57) )/(4)`

Therefore, the solutions are:

`x=(-5 +√(57) )/(4), \quad x=(-5 -√(57) )/(4)`

Question 5

Method: Quadratic Formula

`\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Quadratic Formula}\end{center}\\\\\begin{center}$x=(-b \pm √(b^2-4ac))/(2a)$\end{center}\\\\\\\begin{center}when $ax^2+bx+c=0$\end{center}\end{minipage}}`

Given:

`5x^2-7x-13=0`

Therefore:

`a=5, \quad b=-7, \quad c=-13`

Substitute the values into the quadratic formula and solve for x:

`\implies x=(-(-7) \pm √((-7)^2-4(5)(-13)) )/(2(5))`

`\implies x=(7 \pm √(49+260) )/(10)`

`\implies x=(7 \pm √(309) )/(10)`

Therefore, the solutions are:

`x=(7+ √(309) )/(10), \quad x=(7-√(309) )/(10)`