I’m confused with how to work out “a” in this equation..

a + root a =132

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• (I'm assuming your use of "root" is the principal square root denoted by the √ symbol.)

I solved it using a substitution:

a + √a = 132

√a(√a + 1) = 132

Let x = √a

x(x + 1) = 132

At this point, we can solve by inspection knowing that 11 * 12 = 132 or -12 * -11 = 132

That means x = 11 or x = -12

Since x = √a → a = x²

a = 11² = 121

or

a = (-12)² = 144

However in testing the answers, we see that 144 is an extraneous result and reject it.

121 + √121 =? 132

121 + 11 = 132 ✓ (accept a = 121)

144 + √144 =? 132

144 + 12 =? 132

156 ≠ 132 (reject a = 144)

a = 121

If you didn't notice the way to solve for x by inspection, you could solve it by factoring:

x(x + 1) = 132

x² + x - 132 = 0

(x + 12)(x - 11) = 0

Hence:

x = -12 or x = 11

a = 144 or a = 121

Again reject 144 by testing.

a = 121

Update: The radical sign (√) is defined to return the principal (non-negative) square root.

• ==> sqrt(a) = 132 - a ... now square both sides to get rid of sqrt(a)

==> a = (132-a)^2

==> a = 132^2 - 264a + a^2

==> a^2 - 265a + 132^2 = 0

solve for a

==> a = (265 +/- sqrt(265^2 - 4* 132^2))/ 2

==> a = (265 +/- sqrt(265^2 - 264^2))/2

==> a = (265 +/- sqrt((265 + 264)(265 -264)))/2

==> a = (265 +/- sqrt(529))/2

==> a = (265 +/- 23)/2

==> a = 242/2 or 188/2

==> a = 121 or 144

• this picture should help: • a = 121 root a = 11

It is just by looking at numbers bigger than 10.The first one,11,gives the answer.

• a + root a = 132

sqrt (a) = 132 - a

a = (132 - a)^2

a = (a - 132)^2

Solutions:

a = 121

a = 144

• Let a = x^2, (as suggested by 121 + 11 = 132)

x^2 + x – 132 = 0 = (x – 11)(x + 12)

a was 121

• a + root a = 132

sqrt (a) = 11

a = 121

• If that is:

a + √a = 132

Let's first subtract "a" from both sides to get the radical by itself:

√a = 132 - a

Now square both sides:

a = (132 - a)²

Expand the right side and simplify into a quadratic:

a = 17424 - 264a + a²

a = a² - 264a + 17424

0 = a² - 265a + 17424

Now use the quadratic equation to find possible solutions for a:

a = [ -b ± √(b² - 4ac)] / (2a)

a = [ -(-265) ± √((-265)² - 4(1)(17424))] / (2 * 1)

a = [ 265 ± √(70225 - 69696)] / 2

a = (265 ± √529) / 2

a = (265 ± 23) / 2

a = 242/2 and 288/2

a = 121 and 144

Now we test both possible solutions to see if any do not make sense:

a + √a = 132

121 + √121 = 132 and 144 + √144 = 132

121 + 11 = 132 and 144 + 12 = 132

132 = 132 and 156 = 132

TRUE and FALSE

Since the second is FALSE, we can throw that out as an extraneous solution.  There is one solution to your equation:

a = 121

• Let sqrt(a) = x.  Now you have:

x^2 + x = 132

x^2 + x - 132 = 0

x = (-1 +/- sqrt(1 + 4 * 132)) / 2

x = (-1 +/- sqrt(1 + 528)) / 2

x = (-1 +/- sqrt(529)) / 2

x = (-1 +/- 23) / 2

x = -24/2 , 22/2

x = -12 , 11

sqrt(a) = -12 , 11

a = (-12)^2 , 11^2

a = 144 , 121

Some will try to say that a = 144 is extraneous, because if you graph the function, non-principal roots aren't defined.  Of course, they fail to point out when or where this problem ever said that this was a function.

a = 144 , 121

Try them out.

• I assume it is square root.

root a = 132 - a

a = (132-a) ^2