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Class 10 Optional Mathematics Surds: 21 Solved Questions (SEE)
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1. Class 10 Optional Mathematics Surds: Unit Overview and Key Formulas

Unit 5: Surds (Optional Mathematics, Grade 10, SEE, Nepal)

Class 10 Optional Mathematics Surds is the fifth unit of the Grade 10 Optional Mathematics course prescribed by the Curriculum Development Centre (CDC), Nepal. These class 10 optional math surds notes solve every question of Exercise 5.1.

Every question is written as in the textbook: the main question first (Q.1, Q.2, Q.3 …) with its parts (a, b, c …) listed under it and the Nepali question in brackets. Each solution follows the textbook example format, with the working under Solution and the check under Verifying. All solutions are in English because the steps and symbols are the same in both languages.

  • Simple surd equations: isolate the surd, then square (or cube) both sides.
  • Two surds: isolate one surd, square, isolate the other surd and square again.
  • Check every root: squaring can create extraneous roots that must be rejected.

Key formulas to remember

\(\left(\sqrt{a}\right)^2=a\ (a\ge 0)\), \(\left(\sqrt[3]{a}\right)^3=a\), and \(\sqrt{a^2}=|a|\). Also \((a\pm b)^2=a^2\pm 2ab+b^2\) and \(a^2-b^2=(a-b)(a+b)\).
Componendo and dividendo: if \(\dfrac{a}{b}=\dfrac{c}{d}\), then \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\).
Step 1Isolatethe surdStep 2Squareboth sidesStep 3Solvethe equationStep 4VerifyrootsAlways reject extraneous roots after squaring.

Figure: Four steps for solving an equation with surds.

For more chapters, visit the complete SEE Class 10 Optional Mathematics notes. You can also read about roots of numbers for extra understanding.

2. Exercise 5.1 Surds: Class 10 Optional Mathematics Surds

Exercise 5.1 Surds: Written Questions (लेख्ने प्रश्न)

Q.1 Solve the given equations and verify.

  • a. \(\sqrt{2x}=6\)
  • b. \(\sqrt{3x-5}=1\)
  • c. \(6-\sqrt{3x+4}=1\)
  • d. \(\sqrt[3]{5x-3}=3\)
  • e. \(\sqrt{3x+6}-3\sqrt{x-4}=0\)
  • f. \(\sqrt{11x^2+45}=4x\)

(दिइएका समीकरणहरूको हल गर्नुहोस् र जाँचेर हेर्नुहोस् : )

a. \(\sqrt{2x}=6\)

Solution

Here, \(\sqrt{2x}=6\)

Squaring on both sides

or, \((\sqrt{2x})^2=6^2\)

or, \(2x=36\)

or, \(x=\dfrac{36}{2}\)

\(\therefore\ x=18\)

Verifying

Here, \(\sqrt{2x}=6\)

Now, substituting \(x=18\)

or, \(\sqrt{2\times 18}=6\)

or, \(\sqrt{36}=6\)

or, \(6=6\), which is true.

Thus, \(x=18\)

\(x=18\)
b. \(\sqrt{3x-5}=1\)

Solution

Here, \(\sqrt{3x-5}=1\)

Squaring on both sides

or, \((\sqrt{3x-5})^2=1^2\)

or, \(3x-5=1\)

or, \(3x=1+5\)

or, \(3x=6\)

or, \(x=\dfrac{6}{3}\)

\(\therefore\ x=2\)

Verifying

Here, \(\sqrt{3x-5}=1\)

Now, substituting \(x=2\)

or, \(\sqrt{3\times 2-5}=1\)

or, \(\sqrt{6-5}=1\)

or, \(\sqrt{1}=1\)

or, \(1=1\), which is true.

Thus, \(x=2\)

\(x=2\)
c. \(6-\sqrt{3x+4}=1\)

Solution

Here, \(6-\sqrt{3x+4}=1\)

or, \(-\sqrt{3x+4}=1-6\)

or, \(-\sqrt{3x+4}=-5\)

or, \(\sqrt{3x+4}=5\)

Squaring on both sides

or, \((\sqrt{3x+4})^2=5^2\)

or, \(3x+4=25\)

or, \(3x=25-4\)

or, \(3x=21\)

or, \(x=\dfrac{21}{3}\)

\(\therefore\ x=7\)

Verifying

Here, \(6-\sqrt{3x+4}=1\)

Now, substituting \(x=7\)

or, \(6-\sqrt{3\times 7+4}=1\)

or, \(6-\sqrt{25}=1\)

or, \(6-5=1\)

or, \(1=1\), which is true.

Thus, \(x=7\)

\(x=7\)
d. \(\sqrt[3]{5x-3}=3\)

Solution

Here, \(\sqrt[3]{5x-3}=3\)

Cubing on both sides

or, \((\sqrt[3]{5x-3})^3=3^3\)

or, \(5x-3=27\)

or, \(5x=27+3\)

or, \(5x=30\)

or, \(x=\dfrac{30}{5}\)

\(\therefore\ x=6\)

Verifying

Here, \(\sqrt[3]{5x-3}=3\)

Now, substituting \(x=6\)

or, \(\sqrt[3]{5\times 6-3}=3\)

or, \(\sqrt[3]{30-3}=3\)

or, \(\sqrt[3]{27}=3\)

or, \(3=3\), which is true.

Thus, \(x=6\)

\(x=6\)
e. \(\sqrt{3x+6}-3\sqrt{x-4}=0\)

Solution

Here, \(\sqrt{3x+6}-3\sqrt{x-4}=0\)

or, \(\sqrt{3x+6}=3\sqrt{x-4}\)

Squaring on both sides

or, \((\sqrt{3x+6})^2=(3\sqrt{x-4})^2\)

or, \(3x+6=9(x-4)\)

or, \(3x+6=9x-36\)

or, \(6+36=9x-3x\)

or, \(42=6x\)

or, \(x=\dfrac{42}{6}\)

\(\therefore\ x=7\)

Verifying

Here, \(\sqrt{3x+6}-3\sqrt{x-4}=0\)

Now, substituting \(x=7\)

or, \(\sqrt{3\times 7+6}-3\sqrt{7-4}=0\)

or, \(\sqrt{27}-3\sqrt{3}=0\)

or, \(3\sqrt{3}-3\sqrt{3}=0\)

or, \(0=0\), which is true.

Thus, \(x=7\)

\(x=7\)
f. \(\sqrt{11x^2+45}=4x\)

Solution

Here, \(\sqrt{11x^2+45}=4x\)

Squaring on both sides

or, \((\sqrt{11x^2+45})^2=(4x)^2\)

or, \(11x^2+45=16x^2\)

or, \(45=16x^2-11x^2\)

or, \(5x^2=45\)

or, \(x^2=9\)

\(\therefore\ x=\pm 3\)

Verifying

Here, \(\sqrt{11x^2+45}=4x\)

Now, substituting \(x=3\)

or, \(\sqrt{11\times 3^2+45}=4\times 3\)

or, \(\sqrt{144}=12\)

or, \(12=12\), which is true.

Again, substituting \(x=-3\)

or, \(\sqrt{11\times (-3)^2+45}=4\times(-3)\)

or, \(\sqrt{144}=-12\)

or, \(12=-12\), which is false. Thus, \(x\neq -3\)

Thus, \(x=3\)

\(x=3\) (reject \(x=-3\))

Q.2 Solve the given equations and verify.

  • a. \(\sqrt{x+7}+\sqrt{x}=7\)
  • b. \(\sqrt{x+5}=5-\sqrt{x}\)
  • c. \(\sqrt{3x-2}-\sqrt{x-2}=2\)
  • d. \(\sqrt{3x+10}-x=0\)
  • e. \(\sqrt{3x-5}=x-1\)
  • f. \(\sqrt{2x^2-7}=x+3\)

(दिइएका समीकरणहरूको हल गर्नुहोस् र जाँचेर हेर्नुहोस् : )

a. \(\sqrt{x+7}+\sqrt{x}=7\)

Solution

Here, \(\sqrt{x+7}+\sqrt{x}=7\)

or, \(\sqrt{x+7}=7-\sqrt{x}\)

Squaring on both sides

or, \((\sqrt{x+7})^2=(7-\sqrt{x})^2\)

or, \(x+7=49-14\sqrt{x}+x\)

or, \(14\sqrt{x}=49-7\)

or, \(14\sqrt{x}=42\)

or, \(\sqrt{x}=\dfrac{42}{14}\)

or, \(\sqrt{x}=3\)

Squaring on both sides again

or, \((\sqrt{x})^2=3^2\)

\(\therefore\ x=9\)

Verifying

Here, \(\sqrt{x+7}+\sqrt{x}=7\)

Now, substituting \(x=9\)

or, \(\sqrt{9+7}+\sqrt{9}=7\)

or, \(\sqrt{16}+3=7\)

or, \(4+3=7\)

or, \(7=7\), which is true.

Thus, \(x=9\)

\(x=9\)
b. \(\sqrt{x+5}=5-\sqrt{x}\)

Solution

Here, \(\sqrt{x+5}=5-\sqrt{x}\)

Squaring on both sides

or, \((\sqrt{x+5})^2=(5-\sqrt{x})^2\)

or, \(x+5=25-10\sqrt{x}+x\)

or, \(10\sqrt{x}=25-5\)

or, \(10\sqrt{x}=20\)

or, \(\sqrt{x}=2\)

Squaring on both sides again

or, \((\sqrt{x})^2=2^2\)

\(\therefore\ x=4\)

Verifying

Here, \(\sqrt{x+5}=5-\sqrt{x}\)

Now, substituting \(x=4\)

or, \(\sqrt{4+5}=5-\sqrt{4}\)

or, \(\sqrt{9}=5-2\)

or, \(3=3\), which is true.

Thus, \(x=4\)

\(x=4\)
c. \(\sqrt{3x-2}-\sqrt{x-2}=2\)

Solution

Here, \(\sqrt{3x-2}-\sqrt{x-2}=2\)

or, \(\sqrt{3x-2}=2+\sqrt{x-2}\)

Squaring on both sides

or, \((\sqrt{3x-2})^2=(2+\sqrt{x-2})^2\)

or, \(3x-2=4+4\sqrt{x-2}+x-2\)

or, \(3x-x-2-2=4\sqrt{x-2}\)

or, \(2x-4=4\sqrt{x-2}\)

or, \(2(x-2)=4\sqrt{x-2}\)

or, \(x-2=2\sqrt{x-2}\)

Squaring on both sides again

or, \((x-2)^2=4(x-2)\)

or, \(x^2-4x+4=4x-8\)

or, \(x^2-8x+12=0\)

or, \(x^2-6x-2x+12=0\)

or, \(x(x-6)-2(x-6)=0\)

or, \((x-2)(x-6)=0\)

or, \(x-2=0\), \(\therefore x=2\)

or, \(x-6=0\), \(\therefore x=6\)

\(\therefore\ x=2\text{ and }6\)

Verifying

Here, \(\sqrt{3x-2}-\sqrt{x-2}=2\)

Now, substituting \(x=2\)

or, \(\sqrt{3\times 2-2}-\sqrt{2-2}=2\)

or, \(\sqrt{4}-\sqrt{0}=2\)

or, \(2-0=2\)

or, \(2=2\), which is true.

Again, substituting \(x=6\)

or, \(\sqrt{3\times 6-2}-\sqrt{6-2}=2\)

or, \(\sqrt{16}-\sqrt{4}=2\)

or, \(4-2=2\)

or, \(2=2\), which is true.

Thus, \(x=2\text{ and }6\)

\(x=2 and 6\)
d. \(\sqrt{3x+10}-x=0\)

Solution

Here, \(\sqrt{3x+10}-x=0\)

or, \(\sqrt{3x+10}=x\)

Squaring on both sides

or, \((\sqrt{3x+10})^2=x^2\)

or, \(3x+10=x^2\)

or, \(x^2-3x-10=0\)

or, \(x^2-x(5-2)-10=0\)

or, \(x^2-5x+2x-10=0\)

or, \(x(x-5)+2(x-5)=0\)

or, \((x+2)(x-5)=0\)

or, \(x+2=0\), \(\therefore x=-2\)

or, \(x-5=0\), \(\therefore x=5\)

\(\therefore\ x=-2\text{ and }5\)

Verifying

Here, \(\sqrt{3x+10}-x=0\)

Now, substituting \(x=-2\)

or, \(\sqrt{3(-2)+10}-(-2)=0\)

or, \(\sqrt{-6+10}+2=0\)

or, \(\sqrt{4}+2=0\)

or, \(2+2=0\)

or, \(4\neq 0\), which is false. Thus, \(x\neq -2\)

Again, substituting \(x=5\)

or, \(\sqrt{3(5)+10}-5=0\)

or, \(\sqrt{25}-5=0\)

or, \(5-5=0\)

or, \(0=0\), which is true.

Thus, \(x=5\)

\(x=5\) (reject \(x=-2\))
e. \(\sqrt{3x-5}=x-1\)

Solution

Here, \(\sqrt{3x-5}=x-1\)

Squaring on both sides

or, \((\sqrt{3x-5})^2=(x-1)^2\)

or, \(3x-5=x^2-2x+1\)

or, \(x^2-2x-3x+1+5=0\)

or, \(x^2-5x+6=0\)

or, \(x^2-3x-2x+6=0\)

or, \(x(x-3)-2(x-3)=0\)

or, \((x-3)(x-2)=0\)

or, \(x-3=0\), \(\therefore x=3\)

or, \(x-2=0\), \(\therefore x=2\)

\(\therefore\ x=2\text{ and }3\)

Verifying

Here, \(\sqrt{3x-5}=x-1\)

Now, substituting \(x=2\)

or, \(\sqrt{3\times 2-5}=2-1\)

or, \(\sqrt{1}=1\)

or, \(1=1\), which is true.

Again, substituting \(x=3\)

or, \(\sqrt{3\times 3-5}=3-1\)

or, \(\sqrt{4}=2\)

or, \(2=2\), which is true.

Thus, \(x=2\text{ and }3\)

\(x=2 and 3\)
f. \(\sqrt{2x^2-7}=x+3\)

Solution

Here, \(\sqrt{2x^2-7}=x+3\)

Squaring on both sides

or, \((\sqrt{2x^2-7})^2=(x+3)^2\)

or, \(2x^2-7=x^2+6x+9\)

or, \(2x^2-x^2-6x-7-9=0\)

or, \(x^2-6x-16=0\)

or, \(x^2-8x+2x-16=0\)

or, \(x(x-8)+2(x-8)=0\)

or, \((x-8)(x+2)=0\)

or, \(x-8=0\), \(\therefore x=8\)

or, \(x+2=0\), \(\therefore x=-2\)

\(\therefore\ x=-2\text{ and }8\)

Verifying

Here, \(\sqrt{2x^2-7}=x+3\)

Now, substituting \(x=-2\)

or, \(\sqrt{2(-2)^2-7}=-2+3\)

or, \(\sqrt{8-7}=1\)

or, \(1=1\), which is true.

Again, substituting \(x=8\)

or, \(\sqrt{2(8)^2-7}=8+3\)

or, \(\sqrt{128-7}=11\)

or, \(\sqrt{121}=11\)

or, \(11=11\), which is true.

Thus, \(x=-2\text{ and }8\)

\(x=-2 and 8\)

Q.3 Solve the given equations and verify.

  • a. \(\sqrt{x-1}+\sqrt{x}=\dfrac{2}{\sqrt{x}}\)
  • b. \(\dfrac{x-1}{\sqrt{x}+1}=3+\dfrac{\sqrt{x}+1}{2}\)
  • c. \(\sqrt{x^2-3x+5}-\sqrt{x^2-4x+4}=1\)
  • d. \(\dfrac{\sqrt{x+4}+\sqrt{x-4}}{\sqrt{x+4}-\sqrt{x-4}}=2\)
  • e. \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{\sqrt{4x+8}}=2\)
  • f. \(\sqrt{x^2-3x+3}+\sqrt{x^2-x+1}=2\)
  • g. \(\sqrt{x^2-2x-4}-\sqrt{x^2-3x-3}=1\)

(दिइएका समीकरणहरूको हल गर्नुहोस् र जाँचेर हेर्नुहोस् : )

a. \(\sqrt{x-1}+\sqrt{x}=\dfrac{2}{\sqrt{x}}\)

Solution

Here, \(\sqrt{x-1}+\sqrt{x}=\dfrac{2}{\sqrt{x}}\)

Multiplying both sides by \(\sqrt{x}\)

or, \(\sqrt{x}\left(\sqrt{x-1}+\sqrt{x}\right)=2\)

or, \(\sqrt{x^2-x}+x=2\)

or, \(\sqrt{x^2-x}=2-x\)

Squaring on both sides

or, \(x^2-x=(2-x)^2\)

or, \(x^2-x=4-4x+x^2\)

or, \(-x+4x=4\)

or, \(3x=4\)

or, \(x=\dfrac{4}{3}\)

\(\therefore\ x=\dfrac{4}{3}\)

Verifying

Here, \(\sqrt{x-1}+\sqrt{x}=\dfrac{2}{\sqrt{x}}\)

Now, substituting \(x=\dfrac{4}{3}\)

or, \(\sqrt{\dfrac{4}{3}-1}+\sqrt{\dfrac{4}{3}}=\dfrac{2}{\sqrt{4/3}}\)

or, \(\dfrac{1}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}=\dfrac{2}{2/\sqrt{3}}\)

or, \(\dfrac{3}{\sqrt{3}}=\sqrt{3}\)

or, \(\sqrt{3}=\sqrt{3}\), which is true.

Thus, \(x=\dfrac{4}{3}\)

\(x=\dfrac{4}{3}\)
b. \(\dfrac{x-1}{\sqrt{x}+1}=3+\dfrac{\sqrt{x}+1}{2}\)

Solution

Here, \(\dfrac{x-1}{\sqrt{x}+1}=3+\dfrac{\sqrt{x}+1}{2}\)

or, \(\dfrac{(\sqrt{x})^2-1^2}{\sqrt{x}+1}=\dfrac{6+\sqrt{x}+1}{2}\)

or, \(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}+1}=\dfrac{\sqrt{x}+7}{2}\)

or, \(\sqrt{x}-1=\dfrac{\sqrt{x}+7}{2}\)

or, \(2\sqrt{x}-2=\sqrt{x}+7\)

or, \(2\sqrt{x}-\sqrt{x}=7+2\)

or, \(\sqrt{x}=9\)

Squaring on both sides

or, \((\sqrt{x})^2=9^2\)

\(\therefore\ x=81\)

Verifying

Here, \(\dfrac{x-1}{\sqrt{x}+1}=3+\dfrac{\sqrt{x}+1}{2}\)

Now, substituting \(x=81\)

or, \(\dfrac{81-1}{\sqrt{81}+1}=3+\dfrac{\sqrt{81}+1}{2}\)

or, \(\dfrac{80}{9+1}=3+\dfrac{9+1}{2}\)

or, \(\dfrac{80}{10}=3+5\)

or, \(8=8\), which is true.

Thus, \(x=81\)

Note: The textbook answer key prints \(25\) for this question. That is the answer of Example 2, whose denominator is \(\sqrt{x}-1\). For the question as printed (denominator \(\sqrt{x}+1\)) the verified solution is \(x=81\).

\(x=81\)
c. \(\sqrt{x^2-3x+5}-\sqrt{x^2-4x+4}=1\)

Solution

Here, \(\sqrt{x^2-3x+5}-\sqrt{x^2-4x+4}=1\)

or, \(\sqrt{x^2-3x+5}=1+\sqrt{x^2-4x+4}\)

or, \(\sqrt{x^2-3x+5}=1+\sqrt{(x-2)^2}\)

or, \(\sqrt{x^2-3x+5}=1+x-2\)

or, \(\sqrt{x^2-3x+5}=x-1\)

Squaring on both sides

or, \((\sqrt{x^2-3x+5})^2=(x-1)^2\)

or, \(x^2-3x+5=x^2-2x+1\)

or, \(x^2-x^2-3x+2x=1-5\)

or, \(x^2-x^2-3x+2x=-4\)

or, \(-x=-4\)

\(\therefore\ x=4\)

Verifying

Here, \(\sqrt{x^2-3x+5}-\sqrt{x^2-4x+4}=1\)

Now, substituting \(x=4\)

or, \(\sqrt{4^2-3\times 4+5}-\sqrt{4^2-4\times 4+4}=1\)

or, \(\sqrt{16-12+5}-\sqrt{16-16+4}=1\)

or, \(\sqrt{9}-\sqrt{4}=1\)

or, \(3-2=1\)

or, \(1=1\), which is true.

Thus, \(x=4\)

Note: Here \(\sqrt{(x-2)^2}=x-2\) is taken for \(x\ge 2\), as in the textbook. If \(x<2\), then \(\sqrt{(x-2)^2}=2-x\) and the same working gives \(x=\frac{4}{3}\), which also satisfies the equation: \(\sqrt{\frac{25}{9}}-\sqrt{\frac{4}{9}}=\frac{5}{3}-\frac{2}{3}=1\). The textbook answer key lists \(x=4\) only.

\(x=4\)
d. \(\dfrac{\sqrt{x+4}+\sqrt{x-4}}{\sqrt{x+4}-\sqrt{x-4}}=2\)

Solution

Here, \(\dfrac{\sqrt{x+4}+\sqrt{x-4}}{\sqrt{x+4}-\sqrt{x-4}}=2\)

Applying componendo and dividendo

or, \(\dfrac{(\sqrt{x+4}+\sqrt{x-4})+(\sqrt{x+4}-\sqrt{x-4})}{(\sqrt{x+4}+\sqrt{x-4})-(\sqrt{x+4}-\sqrt{x-4})}=\dfrac{2+1}{2-1}\)

or, \(\dfrac{2\sqrt{x+4}}{2\sqrt{x-4}}=3\)

or, \(\dfrac{\sqrt{x+4}}{\sqrt{x-4}}=3\)

Squaring on both sides

or, \(\dfrac{x+4}{x-4}=9\)

or, \(x+4=9x-36\)

or, \(4+36=9x-x\)

or, \(40=8x\)

or, \(x=\dfrac{40}{8}\)

\(\therefore\ x=5\)

Verifying

Here, \(\dfrac{\sqrt{x+4}+\sqrt{x-4}}{\sqrt{x+4}-\sqrt{x-4}}=2\)

Now, substituting \(x=5\)

or, \(\dfrac{\sqrt{5+4}+\sqrt{5-4}}{\sqrt{5+4}-\sqrt{5-4}}=2\)

or, \(\dfrac{\sqrt{9}+\sqrt{1}}{\sqrt{9}-\sqrt{1}}=2\)

or, \(\dfrac{3+1}{3-1}=2\)

or, \(\dfrac{4}{2}=2\)

or, \(2=2\), which is true.

Thus, \(x=5\)

Note: The textbook answer key prints \(3\) for this question, but \(x=3\) makes \(\sqrt{x-4}=\sqrt{-1}\), which is not a real number. The verified solution is \(x=5\).

\(x=5\)
e. \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{\sqrt{4x+8}}=2\)

Solution

Here, \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{\sqrt{4x+8}}=2\)

or, \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{\sqrt{4(x+2)}}=2\)

or, \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{2\sqrt{x+2}}=2\)

or, \(\dfrac{2-1}{2\sqrt{x+2}}=2\)

or, \(\dfrac{1}{2\sqrt{x+2}}=2\)

or, \(1=4\sqrt{x+2}\)

or, \(\sqrt{x+2}=\dfrac{1}{4}\)

Squaring on both sides

or, \(x+2=\dfrac{1}{16}\)

or, \(x=\dfrac{1}{16}-2\)

or, \(x=\dfrac{1-32}{16}\)

\(\therefore\ x=-\dfrac{31}{16}\)

Verifying

Here, \(\dfrac{1}{\sqrt{x+2}}-\dfrac{1}{\sqrt{4x+8}}=2\)

Now, substituting \(x=-\dfrac{31}{16}\)

or, \(\dfrac{1}{\sqrt{-\frac{31}{16}+2}}-\dfrac{1}{\sqrt{-\frac{31}{4}+8}}=2\)

or, \(\dfrac{1}{\sqrt{1/16}}-\dfrac{1}{\sqrt{1/4}}=2\)

or, \(\dfrac{1}{1/4}-\dfrac{1}{1/2}=2\)

or, \(4-2=2\)

or, \(2=2\), which is true.

Thus, \(x=-\dfrac{31}{16}\)

\(x=-\dfrac{31}{16}\)
f. \(\sqrt{x^2-3x+3}+\sqrt{x^2-x+1}=2\)

Solution

Here, \(\sqrt{x^2-3x+3}+\sqrt{x^2-x+1}=2\)

or, \(\sqrt{x^2-3x+3}=2-\sqrt{x^2-x+1}\)

Squaring on both sides

or, \(x^2-3x+3=4-4\sqrt{x^2-x+1}+x^2-x+1\)

or, \(x^2-3x+3-x^2+x-5=-4\sqrt{x^2-x+1}\)

or, \(-2x-2=-4\sqrt{x^2-x+1}\)

or, \(x+1=2\sqrt{x^2-x+1}\)

Squaring on both sides again

or, \((x+1)^2=4(x^2-x+1)\)

or, \(x^2+2x+1=4x^2-4x+4\)

or, \(3x^2-6x+3=0\)

or, \(x^2-2x+1=0\)

or, \((x-1)^2=0\)

\(\therefore\ x=1\)

Verifying

Here, \(\sqrt{x^2-3x+3}+\sqrt{x^2-x+1}=2\)

Now, substituting \(x=1\)

or, \(\sqrt{1^2-3\times 1+3}+\sqrt{1^2-1+1}=2\)

or, \(\sqrt{1-3+3}+\sqrt{1}=2\)

or, \(\sqrt{1}+\sqrt{1}=2\)

or, \(1+1=2\)

or, \(2=2\), which is true.

Thus, \(x=1\)

\(x=1\)
g. \(\sqrt{x^2-2x-4}-\sqrt{x^2-3x-3}=1\)

Solution

Here, \(\sqrt{x^2-2x-4}-\sqrt{x^2-3x-3}=1\)

or, \(\sqrt{x^2-2x-4}=1+\sqrt{x^2-3x-3}\)

Squaring on both sides

or, \(x^2-2x-4=1+2\sqrt{x^2-3x-3}+x^2-3x-3\)

or, \(x^2-2x-4-x^2+3x+2=2\sqrt{x^2-3x-3}\)

or, \(x-2=2\sqrt{x^2-3x-3}\)

Squaring on both sides again

or, \((x-2)^2=4(x^2-3x-3)\)

or, \(x^2-4x+4=4x^2-12x-12\)

or, \(3x^2-8x-16=0\)

or, \(3x^2-12x+4x-16=0\)

or, \(3x(x-4)+4(x-4)=0\)

or, \((x-4)(3x+4)=0\)

or, \(x-4=0\), \(\therefore x=4\)

or, \(3x+4=0\), \(\therefore x=-\dfrac{4}{3}\)

\(\therefore\ x=4\text{ and }-\dfrac{4}{3}\)

Verifying

Here, \(\sqrt{x^2-2x-4}-\sqrt{x^2-3x-3}=1\)

Now, substituting \(x=4\)

or, \(\sqrt{4^2-2\times 4-4}-\sqrt{4^2-3\times 4-3}=1\)

or, \(\sqrt{16-8-4}-\sqrt{16-12-3}=1\)

or, \(\sqrt{4}-\sqrt{1}=1\)

or, \(2-1=1\)

or, \(1=1\), which is true.

Again, substituting \(x=-\dfrac{4}{3}\)

or, \(\sqrt{\dfrac{16}{9}+\dfrac{8}{3}-4}-\sqrt{\dfrac{16}{9}+4-3}=1\)

or, \(\sqrt{\dfrac{4}{9}}-\sqrt{\dfrac{25}{9}}=1\)

or, \(\dfrac{2}{3}-\dfrac{5}{3}=1\)

or, \(-1=1\), which is false. Thus, \(x\neq -\dfrac{4}{3}\)

Thus, \(x=4\)

\(x=4\) (reject \(x=-\frac{4}{3}\))
Q.4 Prove that the value of \(x\) is \(81\) while solving the equation involving surd \(\dfrac{x-1}{\sqrt{x}+1}=4+\dfrac{\sqrt{x}-1}{2}\). (सर्डयुक्त समीकरण \(\dfrac{x-1}{\sqrt{x}+1}=4+\dfrac{\sqrt{x}-1}{2}\) लाई हल गर्दा \(x\) को मान \(81\) आउँछ भनी प्रमाणित गर्नुहोस् । उक्त समीकरण कुन अवस्थामा अपरिभाषित हुन्छ, कारण लेख्नुहोस् ।)

Solution

Here, \(\dfrac{x-1}{\sqrt{x}+1}=4+\dfrac{\sqrt{x}-1}{2}\)

or, \(\dfrac{(\sqrt{x})^2-1^2}{\sqrt{x}+1}=\dfrac{8+\sqrt{x}-1}{2}\)

or, \(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}+1}=\dfrac{\sqrt{x}+7}{2}\)

or, \(\sqrt{x}-1=\dfrac{\sqrt{x}+7}{2}\)

or, \(2\sqrt{x}-2=\sqrt{x}+7\)

or, \(2\sqrt{x}-\sqrt{x}=7+2\)

or, \(\sqrt{x}=9\)

Squaring on both sides

or, \((\sqrt{x})^2=9^2\)

\(\therefore\ x=81\)

Verifying

Here, \(\dfrac{x-1}{\sqrt{x}+1}=4+\dfrac{\sqrt{x}-1}{2}\)

Now, substituting \(x=81\)

or, \(\dfrac{81-1}{\sqrt{81}+1}=4+\dfrac{\sqrt{81}-1}{2}\)

or, \(\dfrac{80}{9+1}=4+\dfrac{9-1}{2}\)

or, \(\dfrac{80}{10}=4+4\)

or, \(8=8\), which is true.

Thus, \(x=81\)

Hence proved that \(x=81\).

Reason for the undefined condition (asked in the Nepali version)

The fraction \(\dfrac{x-1}{\sqrt{x}+1}\) would be undefined if its denominator were zero, that is, if \(\sqrt{x}+1=0\), so \(\sqrt{x}=-1\). A square root cannot be negative, so for real \(x\ge 0\) we always have \(\sqrt{x}+1\ge 1\). The denominator never becomes zero, so the cancellation of \((\sqrt{x}+1)\) is valid.

The equation becomes undefined only when \(\sqrt{x}+1=0\), which is impossible for real \(x\).
Q.5 Construct any five questions as given in the exercise and solve, then present them in the classroom. (अभ्यासमा दिइएजस्तै अन्य कुनै पाँचओटा प्रश्नहरू निर्माण गरी तिनीहरूको समाधानसहित कक्षाकोठामा प्रस्तुत गर्नुहोस् ।)

Five sample questions with solutions are given below.

Sample question 1: \(\sqrt{4x+1}=5\)

Solution

Here, \(\sqrt{4x+1}=5\)

Squaring on both sides

or, \((\sqrt{4x+1})^2=5^2\)

or, \(4x+1=25\)

or, \(4x=24\)

or, \(x=\dfrac{24}{4}\)

\(\therefore\ x=6\)

Verifying

Here, \(\sqrt{4x+1}=5\)

Now, substituting \(x=6\)

or, \(\sqrt{4\times 6+1}=5\)

or, \(\sqrt{25}=5\)

or, \(5=5\), which is true.

Thus, \(x=6\)

Sample question 2: \(\sqrt{x+3}+\sqrt{x}=3\)

Solution

Here, \(\sqrt{x+3}+\sqrt{x}=3\)

or, \(\sqrt{x+3}=3-\sqrt{x}\)

Squaring on both sides

or, \(x+3=9-6\sqrt{x}+x\)

or, \(6\sqrt{x}=9-3\)

or, \(\sqrt{x}=1\)

Squaring on both sides again

\(\therefore\ x=1\)

Verifying

Here, \(\sqrt{x+3}+\sqrt{x}=3\)

Now, substituting \(x=1\)

or, \(\sqrt{1+3}+\sqrt{1}=3\)

or, \(2+1=3\)

or, \(3=3\), which is true.

Thus, \(x=1\)

Sample question 3: \(\sqrt{2x+3}=x\)

Solution

Here, \(\sqrt{2x+3}=x\)

Squaring on both sides

or, \(2x+3=x^2\)

or, \(x^2-2x-3=0\)

or, \(x^2-3x+x-3=0\)

or, \(x(x-3)+1(x-3)=0\)

or, \((x+1)(x-3)=0\)

\(\therefore\ x=-1\text{ and }3\)

Verifying

Here, \(\sqrt{2x+3}=x\)

Now, substituting \(x=-1\)

or, \(\sqrt{2(-1)+3}=-1\)

or, \(\sqrt{1}=-1\)

or, \(1=-1\), which is false. Thus, \(x\neq -1\)

Again, substituting \(x=3\)

or, \(\sqrt{2(3)+3}=3\)

or, \(\sqrt{9}=3\)

or, \(3=3\), which is true.

Thus, \(x=3\)

Sample question 4: \(\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=3\)

Solution

Here, \(\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=3\)

Applying componendo and dividendo

or, \(\dfrac{2\sqrt{x+2}}{2\sqrt{x-2}}=\dfrac{3+1}{3-1}\)

or, \(\dfrac{\sqrt{x+2}}{\sqrt{x-2}}=2\)

Squaring on both sides

or, \(\dfrac{x+2}{x-2}=4\)

or, \(x+2=4x-8\)

or, \(10=3x\)

\(\therefore\ x=\dfrac{10}{3}\)

Verifying

Here, \(\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=3\)

Now, substituting \(x=\dfrac{10}{3}\)

or, \(\dfrac{\sqrt{16/3}+\sqrt{4/3}}{\sqrt{16/3}-\sqrt{4/3}}=3\)

or, \(\dfrac{4/\sqrt{3}+2/\sqrt{3}}{4/\sqrt{3}-2/\sqrt{3}}=3\)

or, \(\dfrac{6}{2}=3\)

or, \(3=3\), which is true.

Thus, \(x=\dfrac{10}{3}\)

Sample question 5: \(\sqrt[3]{2x+4}=2\)

Solution

Here, \(\sqrt[3]{2x+4}=2\)

Cubing on both sides

or, \((\sqrt[3]{2x+4})^3=2^3\)

or, \(2x+4=8\)

or, \(2x=4\)

\(\therefore\ x=2\)

Verifying

Here, \(\sqrt[3]{2x+4}=2\)

Now, substituting \(x=2\)

or, \(\sqrt[3]{2\times 2+4}=2\)

or, \(\sqrt[3]{8}=2\)

or, \(2=2\), which is true.

Thus, \(x=2\)

Five constructed surd equations solved as shown above.

3. Frequently Asked Questions on Class 10 Optional Mathematics Surds

What is covered in Class 10 Optional Mathematics Surds (Unit 5)? (कक्षा १० ऐच्छिक गणित सर्ड (एकाइ ५) मा के के समावेश छ ?)

Class 10 Optional Mathematics Surds is Unit 5 of the Grade 10 Optional Mathematics book of Nepal. Exercise 5.1 covers solving equations that contain surds (square roots and cube roots), including equations with fractions, componendo and dividendo, and rejecting extraneous roots. This page solves all 21 questions.

What is an equation involving surds? (सर्डयुक्त समीकरण भनेको के हो ?)

It is an equation in which the unknown appears under a root sign, such as \(\sqrt{2x}=6\) or \(\sqrt[3]{5x-3}=3\).

How do I solve a surd equation? (सर्डयुक्त समीकरण कसरी हल गर्ने ?)
  1. Isolate the surd on one side.
  2. Square both sides (cube both sides for a cube root).
  3. Solve the resulting equation.
  4. Verify every answer in the original equation.

Example: \(\sqrt{3x-5}=1\) gives \(3x-5=1\), so \(x=2\) (Exercise 5.1, Question 1b).

Why must I verify the answers? (किन उत्तर जाँच गर्नुपर्छ ?)

Squaring can create extra roots that do not satisfy the original equation. Such a root is called an extraneous root. For example, in \(\sqrt{11x^2+45}=4x\) the value \(x=-3\) is rejected because the right side becomes negative (Exercise 5.1, Question 1f).

What do I do when there are two surds? (दुईओटा सर्ड भएमा के गर्ने ?)

Isolate one surd, square both sides, then isolate the remaining surd and square again. Example: \(\sqrt{x+7}+\sqrt{x}=7\) gives \(x=9\) (Exercise 5.1, Question 2a).

How do I solve a cube root equation? (घनमूलयुक्त समीकरण कसरी हल गर्ने ?)

Cube both sides to remove the cube root. For \(\sqrt[3]{5x-3}=3\), we get \(5x-3=27\), so \(x=6\) (Exercise 5.1, Question 1d).

When can I use componendo and dividendo? (कम्पोनेन्डो र डिभिडेन्डो कहिले प्रयोग गर्ने ?)

Use it when an equation is a ratio equal to a number. If \(\dfrac{a}{b}=\dfrac{c}{d}\), then \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\). It simplifies equations such as Exercise 5.1, Question 3d, which gives \(x=5\).

Why is \(\sqrt{x^2-4x+4}\) equal to \(|x-2|\)? (\(\sqrt{x^2-4x+4}\) किन \(|x-2|\) बराबर हुन्छ ?)

Because \(x^2-4x+4=(x-2)^2\), and the square root of a square is the absolute value: \(\sqrt{a^2}=|a|\). This is why Exercise 5.1, Question 3c has two cases, \(x\ge 2\) and \(x<2\), and two solutions.

When is an equation with a surd undefined? (सर्डयुक्त समीकरण कहिले अपरिभाषित हुन्छ ?)

It is undefined when a denominator is zero or when the expression under a square root is negative. In \(\dfrac{x-1}{\sqrt{x}+1}\), the denominator \(\sqrt{x}+1\) is at least \(1\) for real \(x\ge 0\), so it is never zero (Exercise 5.1, Question 4).

Are these questions useful for SEE preparation in Nepal? (के यी प्रश्नहरू नेपालमा SEE को तयारीका लागि उपयोगी छन् ?)

Yes, for students who study Optional Mathematics. This set covers every question type of Exercise 5.1: simple surd equations, equations with two surds, equations with fractions and verification of roots. Practising them builds the skills needed for SEE-style questions.

Are the solutions available in Nepali? (के समाधान नेपाली भाषामा पनि उपलब्ध छन् ?)

Every question is shown in English with the Nepali question in brackets. All solutions are in English only, because the steps and symbols of mathematics are the same in both languages.

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