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Hello nick,

Can you please mention why you squared both the roots ? I did not understand the logic behind that. Kindly explain.

Thanks.


nick1816
a+b = 4; a*b=1

\((\sqrt{a}+\sqrt{b})^2\)

\(= a+b+2 \sqrt{a*b}\)

\(= 4+2*\sqrt{1}\)

\(=6\)

\(\sqrt{a}+\sqrt{b}\) = \(\sqrt{6}\)



MathRevolution
[GMAT math practice question]

If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?

A. \(2\sqrt{2} \)

B. \(±2\sqrt{2} \)

C. \(\sqrt{6}\)

D. \(±\sqrt{6} \)

E. \(±8\)
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We know the value of a+b and a*b from the quadratic equation.

So i transformed the term(\(\sqrt{a}+\sqrt{b}\)), whose value is asked in the question, in terms of a+b and a*b.




subramanya1991
Hello nick,

Can you please mention why you squared both the roots ? I did not understand the logic behind that. Kindly explain.

Thanks.


nick1816
a+b = 4; a*b=1

\((\sqrt{a}+\sqrt{b})^2\)

\(= a+b+2 \sqrt{a*b}\)

\(= 4+2*\sqrt{1}\)

\(=6\)

\(\sqrt{a}+\sqrt{b}\) = \(\sqrt{6}\)



MathRevolution
[GMAT math practice question]

If \(a\) and \(b\) are roots of the quadratic equation \(x^2 - 4x + 1 = 0\), what is \(\sqrt{a}+\sqrt{b}\)?

A. \(2\sqrt{2} \)

B. \(±2\sqrt{2} \)

C. \(\sqrt{6}\)

D. \(±\sqrt{6} \)

E. \(±8\)
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Standard quadratic equation is \({ax}^2\) + bx + c = 0.

On comapring the given equation with standard we get: a = 1, b = -4 and c = 1

Since 'a' and 'b' are the roots of the quadratic equation \(x^2\)- 4x + 1=0

=> Sum of roots a + b = - \(\frac{b}{a}\) = - \(\frac{(-4) }{ 1}\) = 4

=> Product of roots a * b = \(\frac{c}{a}\) = \(\frac{1 }{ 1}\) = 1

Squaring (a + b) => \((\sqrt{a}+\sqrt{b})^2\) = \((\sqrt{a})^2 + (\sqrt{b})^2 + 2 \sqrt{a}\sqrt{b}\)

=> \((\sqrt{a}+\sqrt{b})^2\) = a + b + 2 \(\sqrt{a*b}\)

=> \((\sqrt{a}+\sqrt{b})^2\) = 4 + 2 \(\sqrt{1}\)

=> \((\sqrt{a}+\sqrt{b})^2\) = 6

=> (a+b) = \(\sqrt{6}\)

Answer C
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why not +/- root 6 and just + root 6?
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namya2.0
why not +/- root 6 and just + root 6?

The square root of a number, or more generally, the even root of a number, is always non-negative. Therefore, the sum of the square roots of two numbers ( \(\sqrt{a}+\sqrt{b}\)) cannot be negative.
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I think its because we only consider positive root when taking square roots. In GMAT \sqrt{x} results in positive root only whereas for x^2 both positive and negative roots are considered
namya2.0
why not +/- root 6 and just + root 6?
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