Question

Which equation is related to StartRoot x 10 EndRoot minus 1 = x?.

Answer 1

You can solve this equation by converting to standard form and using the formula for roots of quadratic equation.

The solution to given equation equation is $x = \dfrac{-1 + \sqrt{37}}{2}$

What is standard form of quadratic equation?

The standard form of quadratic equation is given by:

$ax^2 + bx + c =0$

The roots of this equation is obtained by:

$x =\dfrac{-b \pm \sqrt{b^2 -4ac }}{2a}$

How to convert the given equation to standard form?

We can take square on both sides to remove the root.

$\sqrt{x+10} -1 =x\\ \sqrt{x+10} = x + 1\\ \\ \text{Taking square on both sides}\\ \\ x + 10 = (x + 1)^2\\ x + 10 = x^2 + 1 + 2x\\ x^2 + x - 9 = 0$

Know that we took square.

Also know that $(\sqrt{x+10})^2 = (-\sqrt{x+10})^2 = x+10$ This is the reason that both roots we will obtain won't belong to our quadratic equation we obtained. Thus, we will have to check which root is correct solution of the obtained quadratic equation.

Now, comparing with standard form, we get a = 1, b = 1, c = -9, thus:
$x = \dfrac{-1 \pm \sqrt{1 + 36}}{2} = \dfrac{-1\pm\sqrt{37}}{2}$

x = 2.54 or x = -3.54

Putting both values, we get:

For x  =2.54:

$\sqrt{x+10} -1 =x\\ \sqrt{2.54+10}-1 = 2.54\\ 2.5411 \approx 2.54$

For x = -3.54

$\sqrt{x+10} -1 =x\\\sqrt{-3.54+10}-1 = -3.54\\1.5411 \neq -3.54$

Thus, the solution to given equation equation is $x = \dfrac{-1 + \sqrt{37}}{2}$

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