Solve sqrt(x+5)+sqrt(20-x)=7 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for x

$x=11$
$x=4$

Solution Steps

Subtract $\sqrt{20-x}$ from both sides of the equation.

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

Square both sides of the equation.

$$\left(\sqrt{x+5}\right)^{2}=\left(7-\sqrt{20-x}\right)^{2}$$

Calculate $\sqrt{x+5}$ to the power of $2$ and get $x+5$.

$$x+5=\left(7-\sqrt{20-x}\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-\sqrt{20-x}\right)^{2}$.

$$x+5=49-14\sqrt{20-x}+\left(\sqrt{20-x}\right)^{2}$$

Calculate $\sqrt{20-x}$ to the power of $2$ and get $20-x$.

$$x+5=49-14\sqrt{20-x}+20-x$$

Add $49$ and $20$ to get $69$.

$$x+5=69-14\sqrt{20-x}-x$$

Subtract $69-x$ from both sides of the equation.

$$x+5-\left(69-x\right)=-14\sqrt{20-x}$$

To find the opposite of $69-x$, find the opposite of each term.

$$x+5-69+x=-14\sqrt{20-x}$$

Subtract $69$ from $5$ to get $-64$.

$$x-64+x=-14\sqrt{20-x}$$

Combine $x$ and $x$ to get $2x$.

$$2x-64=-14\sqrt{20-x}$$

Square both sides of the equation.

$$\left(2x-64\right)^{2}=\left(-14\sqrt{20-x}\right)^{2}$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x-64\right)^{2}$.

$$4x^{2}-256x+4096=\left(-14\sqrt{20-x}\right)^{2}$$

Expand $\left(-14\sqrt{20-x}\right)^{2}$.

$$4x^{2}-256x+4096=\left(-14\right)^{2}\left(\sqrt{20-x}\right)^{2}$$

Calculate $-14$ to the power of $2$ and get $196$.

$$4x^{2}-256x+4096=196\left(\sqrt{20-x}\right)^{2}$$

Calculate $\sqrt{20-x}$ to the power of $2$ and get $20-x$.

$$4x^{2}-256x+4096=196\left(20-x\right)$$

Use the distributive property to multiply $196$ by $20-x$.

$$4x^{2}-256x+4096=3920-196x$$

Subtract $3920$ from both sides.

$$4x^{2}-256x+4096-3920=-196x$$

Subtract $3920$ from $4096$ to get $176$.

$$4x^{2}-256x+176=-196x$$

Add $196x$ to both sides.

$$4x^{2}-256x+176+196x=0$$

Combine $-256x$ and $196x$ to get $-60x$.

$$4x^{2}-60x+176=0$$

Divide both sides by $4$.

$$x^{2}-15x+44=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+44$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-15$$ $$ab=1\times 44=44$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $44$.

$$-1,-44$$ $$-2,-22$$ $$-4,-11$$

Calculate the sum for each pair.

$$-1-44=-45$$ $$-2-22=-24$$ $$-4-11=-15$$

The solution is the pair that gives sum $-15$.

$$a=-11$$ $$b=-4$$

Rewrite $x^{2}-15x+44$ as $\left(x^{2}-11x\right)+\left(-4x+44\right)$.

$$\left(x^{2}-11x\right)+\left(-4x+44\right)$$

Factor out $x$ in the first and $-4$ in the second group.

$$x\left(x-11\right)-4\left(x-11\right)$$

Factor out common term $x-11$ by using distributive property.

$$\left(x-11\right)\left(x-4\right)$$

To find equation solutions, solve $x-11=0$ and $x-4=0$.

$$x=11$$ $$x=4$$

Substitute $11$ for $x$ in the equation $\sqrt{x+5}+\sqrt{20-x}=7$.

$$\sqrt{11+5}+\sqrt{20-11}=7$$

Simplify. The value $x=11$ satisfies the equation.

$$7=7$$

Substitute $4$ for $x$ in the equation $\sqrt{x+5}+\sqrt{20-x}=7$.

$$\sqrt{4+5}+\sqrt{20-4}=7$$

Simplify. The value $x=4$ satisfies the equation.

$$7=7$$

List all solutions of $\sqrt{x+5}=-\sqrt{20-x}+7$.

$$x=11$$ $$x=4$$