Surds (Leaving Cert Mathematics): Revision Notes

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Solving Surd Equations

Up until now we have dealt with polynomial equations. Now we explore the other types of algebraic equations found in algebra, starting with surd equations.

Surd equations are those involving an algebraic root :

Example

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Solve for xx in the equation 3x+7−x=1\sqrt{3x+7}-x=1.

If there is one surd, it is best to isolate it.

3x+7=1+x\sqrt{3x+7}=1+x

Once isolated, we can get rid of the surd by squaring both sides.

(3x+7)2=(1+x)2\left(\sqrt{3x+7}\right)^2=(1+x)^23x+7=x2+2x+13x+7=x^2+2x+1

Solve the quadratic :

0=x2−x−60=x^2-x-6x=−2,x=3x=-2,x=3

With the roots found, we need to check our answers :

For x=−2x=-2 :

3(−2)+7−(−2)=1\sqrt{3(-2)+7}-(-2)=13≠1!3 \ne 1 !

For x=3x=3 :

3(3)+7−(3)=1\sqrt{3(3)+7}-(3)=11=11 = 1

With both answered tested, we conclude that x=3 is the only solution to the original surd equation

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Marking both answers as correct without testing can result in lost marks. Clearly indicate to the examiner that one answer does not satisfy the equation.

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