Question

Find x: x=8(x)^1/2 + 9 please explain i will make u brainlist

Answer 1

$\large{\boxed{\text{Working Procedure}}}$

$\large{\text{\underline{Step A: Finding the Quadratic Equation}}}$

We will substitute $x^{\frac{1}{2}}$ with $t$.

Let,

$\hookrightarrow t=x^{\frac{1}{2}}$

Then,

$\hookrightarrow t^{2}=x$

Now we can establish an equation for $t$. The newly made equation, $t^{2}=8t+9$ is a quadratic equation.

$\large{\text{\underline{Step B: Solving the Quadratic Equation}}}$

Then there are three methods to solve the quadratic equation,

① factorization method.

② quadratic formula method.

③ completing the square method.

Here we will use factorization.

$\large{\boxed{\text{Solution}}}$

$\large{\text{\underline{Step A: Finding the Quadratic Equation}}}$

The quadratic equation is,

$\hookrightarrow t^{2}=8t+9$

$\hookrightarrow t^{2}-8t-9=0$

$\large{\text{\underline{Step B: Solving the Quadratic Equation}}}$

$\hookrightarrow t^{2}-8t-9=0$

$\hookrightarrow (t+1)(t-9)=0$

$\hookrightarrow t=-1\ \text{or}\ t=9$

Let's substitute the value of $t$.

$\hookrightarrow x^{\frac{1}{2}}=-1\ \text{or}\ x^{\frac{1}{2}}=9$

The first equation doesn't have a solution. Then from the second equation,  we have $x=9^{2}$, or $x=81$ as our solution.

$\hookrightarrow x=81\ \text{\underline{(ANSWER.)}}$

$\large{\boxed{\text{Verification}}}$

We have $x=81$ as a solution.

$\hookrightarrow 81=8\sqrt{81} +9$

$\hookrightarrow 81=8\cdot 9+9$

$\hookrightarrow 81=81$

Thus, the value satisfies the equation. Hence verified!

Answer 2

$\orange{ \boxed{\boxed{\begin{array}{cc} \leadsto \bf \: given \\ \\ \rm \: x = 8 {x}^{ \frac{1}{2} } + 9 \\ \\ \rm \implies\:x - 9= 8 {x}^{ \frac{1}{2} } \\ \\ \rm \implies\: {(x - 9)}^{2} = {(8 {x}^{ \frac{1}{2} } })^{2} \\ \\ \rm \implies\: {x}^{2} - 18x + 81 = 64x \\ \\ \rm \implies\: { {x}^{2} - 18x - 64x + 81 = 0} \\ \\ \rm \implies\: {x}^{2} - 82x + 81 = 0 \\ \\ \rm \implies\: {x}^{2} - 81x - x + 81 = 0 \\ \\ \rm \implies\:x(x - 81) - 1(x - 81) = 0 \\ \\ \rm \implies\:(x - 81)(x - 1) = 0\end{array}}}}$

Now,

$\pink{ \boxed{\boxed{\begin{array}{c | c} \underline{ \rm \: either }& \underline{\rm \: or} \\ \\ \rm \: x - 81 = 0& \rm \: x - 1 = 0 \\ \\ \rm \implies\:x = 81& \rm \implies\:x = 1 \\ \\ \end{array}}}}$

$\therefore \: \rm \: x = 81 \: \: or \: \: 1\:$

But x = 1 does not satisfy the equation.

So,

x = 81