Question

Need help ez question

Answer 1

x = 0.25

Explanation:

`9√(x) =5-2x`

`81x = 25 - 20x + 4x^(2)`

`0=4x^(2) -101x+25`

`0 = 4x^(2) -1x-100x+25`

`0=x(4x-1)-25(4x-1)\\0=(x-25)(4x-1)`

`x=25,(1)/(4)`

`2*25+9√(25) =5\\50+9*5=5\\50+45=5\\95=5`

`2*(1)/(4)+9\sqrt{(1)/(4) } =5 \\0.5+9*(1)/(2) =5\\0.5+4.5=5\\5=5`

Answer 2

`x=(1)/(4)`

Explanation:

Given equation:

`2x+9√(x)=5`

Subtract 2x from both sides to isolate the square root:

`\implies 2x+9√(x)-2x=5-2x`

`\implies 9√(x)=5-2x`

Square both sides to eliminate the square root:

`\implies (9√(x))^2=(5-2x)^2`

`\implies 81x=(5-2x)(5-2x)`

`\implies 81x=25-20x+4x^2`

Subtract 81x from both sides:

`\implies 81x-81x=25-20x+4x^2-81x`

`\implies 0=25-101x+4x^2`

`\implies4x^2-101x+25=0`

To factor a quadratic in the form
`ax^2+bx+c`, find two numbers that multiply to
`ac` and sum to
`b`:

`\implies ac=4 \cdot 25=100`

`\implies b=-101`

Therefore, the two numbers are -1 and -100 as:

`\implies -1 \cdot -100=100=a`

`\implies -1+(-100)=-101 =b`

Rewrite
`b` as the sum of these two numbers:

`\implies4x^2-x-100x+25=0`

Factor the first two terms and the last two terms separately:

`\implies x(4x-1)-25(4x-1)=0`

Factor out the common term (4x - 1):

`\implies (x-25)(4x-1)=0`

Apply the zero-product property:

`(x-25)=0 \implies x=25`

`(4x-1)=0 \implies x=(1)/(4)`

Input each solution into the original equation to check its validity:

`\begin{aligned}x=25 \implies 2(25)+9√(25)&=5\\50+9(5)&=5\\50+45&=5\\95 & \\eq 5\end{aligned}`

Therefore, x = 25 is not a valid solution.

`\begin{aligned}x=(1)/(4)\implies 2\left((1)/(4)\right)+9\sqrt{(1)/(4)}&=5\\(1)/(2)+9((1)/(2))&=5\\(1)/(2)+(9)/(2)&=5\\(10)/(2)&=5\\5 & = 5\end{aligned}`

Therefore x = ¹/₄ is a valid solution.

Therefore, the only valid solution to the equation is x = ¹/₄.