Question: "To calculate: Thesolution set of the equation 2d=1−d+7"

High School
Algebra
Pre-Algebra
Equations, expressions, and inequalitie

shelbs624c

Answered question

2021-11-13

To calculate: Thesolution set of the equation

\sqrt{2 d} = 1 - \sqrt{d + 7}

Answer & Explanation

Mary Darby

Beginner2021-11-14Added 11 answers

Consider the given equation,
$\sqrt{2 d} = 1 - \sqrt{d + 7}$
Now, apply Square on the both sides of the above equation,
$\sqrt{2 d} = 1 - \sqrt{d + 7}$
$(\sqrt{2 d})^{2} = (1 - \sqrt{d + 7})^{2}$
$2 d = 1 - 2 \sqrt{d + 7} + d + 7$
$2 d = 8 + d - 2 \sqrt{d + 7}$
Simplify the above equation,
$2 d = 8 + d - 2 \sqrt{d + 7}$
$2 d - d = 8 - 2 \sqrt{d + 7}$
$2 \sqrt{d + 7} = 8 - d$
Apply square on both sides: of the above equation,
${(2 \sqrt{d + 7})}^{2} = {(8 - d)}^{2}$
$4 (d + 7) = 64 - 16 d + d^{2}$
$4 d + 28 = 64 - 16 d + d^{2}$
Now subtract $4 d + 28$ from both sides of the equation,
$4 d + 28 = 64 - 16 d + d^{2}$
$4 d + 28 - (4 d + 28) = 64 - 16 d + d^{2} - (4 d + 28)$
$0 = 64 - 16 d + d^{2} - 4 d - 28$
$0 = d^{2} - 20 d + 36$
Compare the equation $d^{2} - 20 d + 36 = 0$ with the general form quadratic
equation $a x^{2} + b x + c = 0$ . Then, $a = 1, b = - 20, c = 36$ .
Put the values of a,b,c in the expression, $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ and solve,
$d = \frac{- (- 20) \pm \sqrt{{(- 20)}^{2} - 4 \times 1 \times 36}}{2 \times 1}$
$= \frac{20 \pm \sqrt{400 - 144}}{2}$
$= \frac{20 \pm \sqrt{256}}{2}$
$= \frac{20 \pm 16}{2}$
Therefore, $d = 10 \pm 8$ .
That is, $d = 18 or d = 2$ .
Check:
Put the value $d = 18$ in the provided equation $\sqrt{2 d} = 1 - \sqrt{d + 7}$
$\sqrt{2 \times 18} \overset{?}{=} 1 - \sqrt{18 + 7}$

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