﻿ Solve the equations involving square roots \sqrt{x+5}+\sqrt{20-x} =7

#### Question

Solve the equations involving square roots

\sqrt{x+5}+\sqrt{20-x} =7

Collected in the board: Square Root

Steven Zheng posted 2 years ago

\sqrt{x+5}+\sqrt{20-x} =7

Square both sides of the equation,

x+5+2\sqrt{x+5}\sqrt{20-x}+20-x=49

\sqrt{x+5}\sqrt{20-x}=12

Square both sides of the equation again,

(x+5)(20-x) = 144

x^2-15x+44=0

Factoring the quadratic expression on left hand side

(x-4)(x-11)=0

Now we have found the solution for the equation.

x=4 or x=11

Steven Zheng posted 2 years ago

\sqrt{x+5}+\sqrt{20-x} =7

Let

a = \sqrt{x+5}
(1)
b = \sqrt{20-x}
(2)

Then

a+b = 7
(3)

a^2+b^2 = x+5+20-x=25

then

a^2+b^2 = 25
(4)

Squaring (3) gives

(a+b)^2 = 49
(5)

Subtract (4) from (5)

2ab = 49-25 = 24
(6)

(4)+(6) gives

(a-b)^2 = 1

then

a-b = \pm1
(7)

(3) and (7) gives two system of equaitons

\begin{cases} a+b&=7 \\ a-b &= 1 \end{cases}

and

\begin{cases} a+b&=7 \\ a-b &= -1 \end{cases}

Solving the equations yields

\begin{cases} a&=4 \\ b &= 3 \end{cases}

and

\begin{cases} a&=3 \\ b &= 4 \end{cases}

Substitute back to (1) and (2) and we get the solution to the equation

x = 4 or x = 11

Steven Zheng posted 1 year ago

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