Positive real numbers and satisfy What is the value of ?

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SAT Nonlinear Equations & Systems Question 51 | Free SAT Prep | Aniko

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Question 51·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Positive real numbers $a$ and $b$ satisfy

\begin{cases} a^{2}+b=7,\\[4pt] b^{2}+a=11. \end{cases}

What is the value of $a^{2}+b^{2}$ ?

Your answer

(Express the answer as an integer)

For nonlinear systems with small integer constants on the SAT, first use the fact that variables are positive to get rough upper bounds (for example, from $a^{2}+b=7$ you know $a^{2}<7$ , so $a<3$ ). Then quickly test the few possible small integer values in one equation and check them in the other; this is usually much faster and less error-prone than doing full algebraic substitution that leads to high-degree polynomials. Once you have the solution pair, plug directly into the expression the question asks for, rather than solving for any extra quantities you do not need.

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Relate the two equations

Try adding the two equations together and see what expression you get. How does that relate to $a^{2}+b^{2}$ ?

Use positivity to narrow down possibilities

Because $a$ and $b$ are positive, from $a^{2}+b=7$ and $b^{2}+a=11$ you can find upper bounds on $a$ and $b$ . Think about which small positive integers they could be close to.

Try simple integer values

Once you know $a$ must be less than $3$ and $b$ less than $4$ , try plugging in the few possible integer values for $a$ into $a^{2}+b=7$ to get $b$ , then check the second equation.

Finish with the target expression

After you find a pair $(a,b)$ that satisfies both equations, substitute into $a^{2}+b^{2}$ and simplify.

Desmos Guide

Graph the system

In Desmos, let $x$ represent $a$ and $y$ represent $b$ . Enter the two equations as

x^2 + y = 7
y^2 + x = 11 Desmos will graph these as implicit curves.

Find the positive intersection point

Look for the intersection point where both $x$ and $y$ are positive. Click on that intersection to see its coordinates; these coordinates are the values of $a$ and $b$ that satisfy the system.

Compute the desired value

Use the positive intersection coordinates from Desmos (the $x$ - and $y$ -values you just found) and type a new expression like (x_value)^2 + (y_value)^2. The output of this expression is the value of $a^{2}+b^{2}$ .

Step-by-step Explanation

Use the equations to bound possible values of a and b

From $a^{2}+b=7$ and $b>0$ , we get $a^{2}<7$ , so $a<\sqrt{7}$ , which is a little less than $3$ .

From $b^{2}+a=11$ and $a>0$ , we get $b^{2}<11$ , so $b<\sqrt{11}$ , which is a little more than $3$ but less than $4$ .

So possible integer values (if they exist) must satisfy:

$a$ is a positive integer less than $3$ : so $a=1$ or $a=2$ .
$b$ is a positive integer less than $4$ : so $b=1,2$ , or $3$ .

Test small integer values for a and find b

Use the first equation $a^{2}+b=7$ to express $b$ in terms of $a$ :

b = 7 - a^{2}.

Now test the possible integer values of $a$ :

If $a=1$ , then $b = 7 - 1^{2} = 6$ . Check in the second equation: $b^{2}+a = 6^{2}+1 = 36+1 = 37 \neq 11$ , so $a=1$ does not work.
If $a=2$ , then $b = 7 - 2^{2} = 7 - 4 = 3$ .

Now check $a=2$ , $b=3$ in the second equation:

$b^{2}+a = 3^{2}+2 = 9+2 = 11$ , which matches the given equation.

So the positive pair that satisfies both equations is $a=2$ and $b=3$ .

Compute the required expression

We are asked to find $a^{2}+b^{2}$ .

Using $a=2$ and $b=3$ :

$a^{2} = 2^{2} = 4$
$b^{2} = 3^{2} = 9$

a^{2} + b^{2} = 4 + 9 = 13.

Therefore, the value of $a^{2}+b^{2}$ is $13$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation