Exercise 1.4 Class 10 – Expert Solutions You Need
In this exercise, students will learn how to solve radical equations—equations that include square roots of algebraic expressions. The questions in Exercise 1.4 Class 10 Math strengthen your understanding of isolating radicals, squaring both sides carefully, and checking for extraneous roots that can appear when removing radicals. These concepts are essential for mastering higher topics in algebra and building confidence in logical problem-solving.
Question 1, Question 2, Question 3, Question 4, Question 5, Question 6, Question 7, Question 8, Question 9, Question 10, Question 11.
Solve the following questions
Question 1 of Exercise 1.4 Class 10. \(\boldsymbol{2x + 5 = \sqrt{7x + 16}}\)
\[
2x + 5 = \sqrt{7x + 16}
\]
\(\textbf{Squaring both sides:}\)
\[
(2x + 5)^2 = (\sqrt{7x + 16})^2
\]
\[
4x^2 + 20x + 25 = 7x + 16
\]
\[
4x^2 + 13x + 9 = 0
\]
\(\textbf{Factoring:}\)
\[
(4x + 9)(x + 1) = 0
\]
\[
x = -1 \quad \text{or} \quad x = \frac{-9}{4}
\]
\(\textbf{Checking:}\)
For \(x = -1\):
\[
2(-1) + 5 = \sqrt{7(-1) + 16} \quad \Rightarrow \quad 3 = \sqrt{9} \quad \Rightarrow \quad 3 = 3 \, \text{(True)}
\]
For \(x = \frac{-9}{4}\):
\[
2\left(\frac{-9}{4}\right) + 5 = \sqrt{7\left(\frac{-9}{4}\right) + 16} \quad \Rightarrow \quad \frac{1}{2} = \frac{1}{2} \, \text{(True)}
\]
\(\textbf{Solution set:}\left\{-1, \frac{-9}{4}\right\}\)
Question 2 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{x + 3} = 3x – 1}\)
To solve the equation \(\sqrt{x + 3} = 3x – 1\):
Square both sides:
\[
x + 3 = (3x – 1)^2
\]
Expand:
\[
x + 3 = 9x^2 – 6x + 1
\]
Rearrange:
\[
0 = 9x^2 – 7x – 2
\]
Now, factor the quadratic:
\[
0 = (9x + 2)(x – 1)
\]
Setting each factor to zero gives:
\[
9x + 2 = 0 \quad \Rightarrow \quad x = -\frac{2}{9}
\]
\[
x – 1 = 0 \quad \Rightarrow \quad x = 1
\]
Check for extraneous solutions:
For \(x = 1\):
\[
\sqrt{1 + 3} = 3(1) – 1 \quad \Rightarrow \quad 2 = 2 \quad \text{(True)}
\]
For \(x = -\frac{2}{9}\):
\[
\sqrt{\frac{-2}{9} + 3} = 3\left(\frac{-2}{9}\right) – 1
\]
Left-hand side:
\[
\sqrt{\frac{25}{9}} = \frac{5}{3}
\]
Right-hand side:
\[
\frac{-6}{9} – 1 = \frac{-5}{3}
\]
Since \(\frac{5}{3} \neq \frac{-5}{3}\), \(x = -\frac{2}{9}\) is extraneous.
Final solution:
\[
x = 1
\]
Question 3 of Exercise 1.4 Class 10. \(\boldsymbol{4x = \sqrt{13x + 14} – 3}\)
To solve the equation \(4x = \sqrt{13x + 14} – 3\):
First, isolate the square root:
\[
4x + 3 = \sqrt{13x + 14}
\]
Taking square both sides:
\[
(4x + 3)^2 = 13x + 14
\]
Expanding square of left side:
\[
16x^2 + 24x + 9 = 13x + 14
\]
Rearranging gives: \(\quad 16x^2 + 24x + 9 – 13x – 14 = 0\)
\[
16x^2 + 11x – 5 = 0
\]
\[
(16x – 5)(x + 1) = 0
\]
Setting each factor to zero gives:
\[
16x – 5 = 0 \quad \Rightarrow \quad x = \frac{5}{16}
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
Check for extraneous solutions:
For \(x = \frac{5}{16}\):
\[
4\left(\frac{5}{16}\right) = \sqrt{13\left(\frac{5}{16}\right) + 14} – 3
\]
Calculating:
\[
\frac{20}{16} = \sqrt{\frac{65}{16} + 14} – 3 \rightarrow \frac{5}{4} = \sqrt{\frac{65 + 224}{16}} – 3
\]
\[
\frac{5}{4} = \sqrt{\frac{289}{16}} – 3 \Rightarrow \frac{5}{4} = \frac{17}{4} – 3
\]
\[
\frac{5}{4} = \frac{17}{4} – \frac{12}{4} \quad \Rightarrow \quad \frac{5}{4} = \frac{5}{4} \quad \text{(True)}
\]
For \(x = -1\):
\[
4(-1) = \sqrt{13(-1) + 14} – 3
\]
Calculating:
\[
-4 = \sqrt{-13 + 14} – 3
\]
\[
-4 = \sqrt{1} – 3
\]
\[
-4 = 1 – 3 \quad \Rightarrow \quad -4 = -2 \quad \text{(False)}
\]
\(\textbf{Final solution: }\boldsymbol x = \frac{5}{16}\)
Question 4 of Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{3x + 100} – x = 4}\)
\[
\begin{align*}
\quad & \sqrt{3x + 100} – x = 4 \quad \text{(i)} \\
\text{Solution:} \\
& \sqrt{3x + 100} = x + 4 \quad \text{(ii)} \\
& \text{Squaring both sides of (ii)} \\
& (\sqrt{3x + 100})^2 = (x + 4)^2 \\
& 3x + 100 = x^2 + 8x + 16 \\
& x^2 + 5x – 84 = 0 \\
& x^2 + 12x – 7x – 84 = 0 \\
& x(x + 12) – 7(x + 12) = 0 \\
& (x – 7)(x + 12) = 0 \\
& x = 7, \; x = -12 \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{Putting } x = 7 \text{ in the equation (i) we have} \\
& \sqrt{3(7) + 100} – 7 = 4 \\
& \sqrt{121} – 7 = 4 \\
& 11 – 7 = 4 \\
& 4 = 4 \\
& \text{which is true} \\
& \text{So, } -12 \text{ is an erroneous root} \\
& \text{Putting } x = -12 \text{ in the equation (i) we have} \\
& \sqrt{3(-12) + 100} – (-12) = 4 \\
& \sqrt{64} + 12 = 4 \\
& 8 + 12 \neq 4 \\
& \text{So, the solution set is } \{7\}
\end{align*}
\]
Question 5 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{x + 5} + \sqrt{x + 21} = \sqrt{x + 60}}\)
\[
\begin{align*}
\quad & \sqrt{x + 5} + \sqrt{x + 21} = \sqrt{x + 60} \\
\text{Solution:} \\
& \sqrt{x + 5} + \sqrt{x + 21} = \sqrt{x + 60} \quad \text{(i)} \\
& \text{Squaring both sides:} \\
& (\sqrt{x + 5} + \sqrt{x + 21})^2 = (\sqrt{x + 60})^2 \\
& x + 5 + x + 21 + 2\sqrt{(x + 5)(x + 21)} = x + 60 \\
& 2x + 26 + 2\sqrt{(x + 5)(x + 21)} = x + 60 \\
& 2\sqrt{(x + 5)(x + 21)} = x + 34 – 2x – 26 \\
& 2\sqrt{(x + 5)(x + 21)} = -x + 34 \\
& \text{Squaring both sides again:} \\
& 4(x + 5)(x + 21) = (34 – x)^2 \\
& 4(x^2 + 26x + 105) = 1156 + x^2 – 68x \\
& 4x^2 + 104x + 420 = 1156 + x^2 – 68x \\
& 3x^2 + 172x – 736 = 0 \\
& \text{Solving the quadratic equation:} \\
& a = 3, \; b = 172, \; c = -736 \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
& x = \frac{-172 \pm \sqrt{29584 + 8832}}{6} \\
& x = \frac{-172 \pm \sqrt{38416}}{6} \\
& x = \frac{-172 \pm 196}{6} \\
& x = \frac{24}{6} \quad \text{or} \quad x = \frac{-368}{6} \\
& x = 4 \quad \text{or} \quad x = -\frac{184}{3} \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{Putting } x = 4 \text{ in equation (i)} \\
& \sqrt{4 + 5} + \sqrt{4 + 21} = \sqrt{4 + 60} \\
& \sqrt{9} + \sqrt{25} = \sqrt{64} \\
& 3 + 5 = 8 \\
& 8 = 8 \quad \text{which is true} \\
& \text{Putting } x = -\frac{184}{3} \text{ in equation (ii)} \\
& \sqrt{-\frac{184}{3} + 5} + \sqrt{-\frac{184}{3} + 21} = \sqrt{-\frac{184}{3} + 60} \\
& \sqrt{-\frac{169}{3}} + \sqrt{-\frac{121}{3}} \neq \sqrt{-\frac{4}{3}} \\
& \text{which is not true} \\
& \text{As } x = -\frac{184}{3} \text{ is an extraneous root, the solution set is } \{4\}
\end{align*}
\]
Question 6 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{x + 1} + \sqrt{x – 2} = \sqrt{x + 6}}\)
\[
\begin{align*}
\text{Solution:} \\
& \sqrt{x + 1} + \sqrt{x – 2} = \sqrt{x + 6} \quad \text{(i)} \\
& \text{Squaring both sides:} \\
& (\sqrt{x + 1} + \sqrt{x – 2})^2 = (\sqrt{x + 6})^2 \\
& (\sqrt{x + 1})^2 + (\sqrt{x – 2})^2 + 2\sqrt{(x + 1)(x – 2)} = x + 6 \\
& x + 1 + x – 2 + 2\sqrt{(x + 1)(x – 2)} = x + 6 \\
& 2x – 1 + 2\sqrt{(x + 1)(x – 2)} = x + 6 \\
& 2\sqrt{(x + 1)(x – 2)} = 7 – x \\
& \text{Squaring both sides again:} \\
& 4(x + 1)(x – 2) = (7 – x)^2 \\
& 4(x^2 – x – 2) = 49 – 14x + x^2 \\
& 4x^2 – 4x – 8 = 49 – 14x + x^2 \\
& 3x^2 + 10x – 57 = 0 \\
& \text{Solving the quadratic equation:} \\
& a = 3, \; b = 10, \; c = -57 \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
& x = \frac{-10 \pm \sqrt{100 + 684}}{6} \\
& x = \frac{-10 \pm \sqrt{784}}{6} \\
& x = \frac{-10 \pm 28}{6} \\
& x = 3 \quad \text{or} \quad x = -\frac{19}{3} \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{For } x = 3: \\
& \sqrt{3 + 1} + \sqrt{3 – 2} = \sqrt{3 + 6} \\
& \sqrt{4} + \sqrt{1} = \sqrt{9} \\
& 2 + 1 = 3 \\
& 3 = 3 \quad \text{which is true} \\
& \text{For } x = -\frac{19}{3}: \\
& \sqrt{-\frac{19}{3} + 1} + \sqrt{-\frac{19}{3} – 2} = \sqrt{-\frac{19}{3} + 6} \\
& \sqrt{-\frac{16}{3}} + \sqrt{-\frac{25}{3}} \neq \sqrt{-\frac{1}{3}} \\
& \text{which is not true} \\
& \text{So, the solution set is } \{3\}
\end{align*}
\]
Question 7 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{11 – x} + \sqrt{6 – x} = \sqrt{27 – x}}\)
\[
\begin{align*}
\text{Solution:} \\
& \sqrt{11 – x} + \sqrt{6 – x} = \sqrt{27 – x} \quad \text{(i)} \\
& \text{Squaring both sides:} \\
& (\sqrt{11 – x} + \sqrt{6 – x})^2 = (\sqrt{27 – x})^2 \\
& (11 – x) + (6 – x) + 2\sqrt{(11 – x)(6 – x)} = 27 – x \\
& 17 – 2x + 2\sqrt{(11 – x)(6 – x)} = 27 – x \\
& 2\sqrt{(11 – x)(6 – x)} = x + 10 \\
& \text{Squaring both sides again:} \\
& (2\sqrt{(11 – x)(6 – x)})^2 = (x + 10)^2 \\
& 4(11 – x)(6 – x) = (x + 10)^2 \\
& 4(66 – 17x + x^2) = x^2 + 20x + 100 \\
& 264 – 68x + 4x^2 = x^2 + 20x + 100 \\
& 3x^2 – 88x + 164 = 0 \\
& \text{Solving the quadratic equation:} \\
& a = 3, \; b = -88, \; c = 164 \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
& x = \frac{88 \pm \sqrt{7744 – 1968}}{6} \\
& x = \frac{88 \pm \sqrt{5776}}{6} \\
& x = \frac{88 \pm 76}{6} \\
& x = \frac{164}{6} \quad \text{or} \quad x = \frac{12}{6} \\
& x = \frac{82}{3} \quad \text{or} \quad x = 2 \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{For } x = 2: \\
& \sqrt{11 – 2} + \sqrt{6 – 2} = \sqrt{27 – 2} \\
& \sqrt{9} + \sqrt{4} = \sqrt{25} \\
& 3 + 2 = 5 \\
& 5 = 5 \quad \text{which is true} \\
& \text{For } x = \frac{82}{3}: \\
& \sqrt{11 – \frac{82}{3}} + \sqrt{6 – \frac{82}{3}} = \sqrt{27 – \frac{82}{3}} \\
& \sqrt{\frac{33 – 82}{3}} + \sqrt{\frac{18 – 82}{3}} = \sqrt{\frac{81 – 82}{3}} \\
& \sqrt{\frac{-49}{3}} + \sqrt{\frac{-64}{3}} \neq \sqrt{\frac{-1}{3}} \\
& \text{which is not true} \\
& \text{So, the valid solution is } \{2\}
\end{align*}
\]
Question 8 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{4a + x} – \sqrt{a – x} = \sqrt{a}}\)
\[
\begin{align*}
\text{Solution:} \\
& \sqrt{4a + x} – \sqrt{a – x} = \sqrt{a} \quad \text{(i)} \\
& \text{Squaring both sides:} \\
& (\sqrt{4a + x} – \sqrt{a – x})^2 = (\sqrt{a})^2 \\
& (4a + x) + (a – x) – 2\sqrt{(4a + x)(a – x)} = a \\
& 4a + x + a – x – 2\sqrt{(4a + x)(a – x)} = a \\
& 5a – 2\sqrt{(4a + x)(a – x)} = a \\
& 5a – a = 2\sqrt{(4a + x)(a – x)} \\
& 4a = 2\sqrt{(4a + x)(a – x)} \\
& 2a = \sqrt{(4a + x)(a – x)} \\
& \text{Squaring both sides again:} \\
& (2a)^2 = (4a + x)(a – x) \\
& 4a^2 = 4a^2 – ax + xa – x^2 \\
& 4a^2 = 4a^2 – x^2 \\
& x^2 = 0 \\
& x = 0 \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{For } x = 0: \\
& \sqrt{4a + 0} – \sqrt{a – 0} = \sqrt{a} \\
& \sqrt{4a} – \sqrt{a} = \sqrt{a} \\
& 2\sqrt{a} – \sqrt{a} = \sqrt{a} \\
& \sqrt{a} = \sqrt{a} \quad \text{which is true} \\
& \text{So, the valid solution is } \{0\}
\end{align*}
\]
Question 9 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{x^2 + x + 1} – \sqrt{x^2 + x – 1} = 1}\)
\[
\text{Solution:} \quad \sqrt{x^2 + x + 1} – \sqrt{x^2 + x – 1} = 1 \quad \text{(i)}
\]
\[
\begin{align*}
& \text{Let} \quad x^2 + x = y \quad \text{(ii)} \\
& \text{Substitute it into equation (i):} \\
& \sqrt{y + 1} – \sqrt{y – 1} = 1 \\
& \text{Squaring both sides:} \\
& (\sqrt{y + 1} – \sqrt{y – 1})^2 = (1)^2 \\
& (\sqrt{y + 1})^2 + (\sqrt{y – 1})^2 – 2(\sqrt{y + 1})(\sqrt{y – 1}) = 1 \\
& y + 1 + y – 1 – 2\sqrt{(y + 1)(y – 1)} = 1 \\
& 2y – 2\sqrt{y^2 – 1} = 1 \\
& 2y – 1 = 2\sqrt{y^2 – 1} \\
& \text{Squaring both sides again:} \\
& (2y – 1)^2 = (2\sqrt{y^2 – 1})^2 \\
& (2y)^2 + (1)^2 – 2(2y)(1) = 4(y^2 – 1) \\
& 4y^2 + 1 – 4y = 4y^2 – 4 \\
& 4y^2 + 1 – 4y = 4y^2 – 4 \\
& 1 – 4y = -4 \\
& 1 + 4 = 4y \\
& 5 = 4y \\
& y = \frac{5}{4} \\
& \text{Substitute back} \ y = x^2 + x: \\
& x^2 + x = \frac{5}{4} \\
& 4x^2 + 4x = 5 \\
& 4x^2 + 4x – 5 = 0 \\
& \text{Solving by the quadratic formula:} \\
& a = 4, \quad b = 4, \quad c = -5 \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
& x = \frac{-4 \pm \sqrt{(4)^2 – 4(4)(-5)}}{2(4)} \\
& x = \frac{-4 \pm \sqrt{16 + 80}}{8} \\
& x = \frac{-4 \pm \sqrt{96}}{8} \\
& x = \frac{-1 \pm \sqrt{6}}{2} \\
& \text{Thus, the solutions are:} \quad x = \frac{-1 + \sqrt{6}}{2} \quad \text{or} \quad x = \frac{-1 – \sqrt{6}}{2}
\end{align*}
\]
Question 10 Exercise 1.4 Class 10. \(\boldsymbol{\quad \sqrt{x^2 + 3x + 8} + \sqrt{x^2 + 3x + 2} = 3}\)
\(\text{Solution:} \quad \sqrt{x^2 + 3x + 8} + \sqrt{x^2 + 3x + 2} = 3 \quad \text{(i)}\)
\[
\begin{align*}
& \text{Let} \quad x^2 + 3x = y \quad \text{(ii)} \\
& \text{Substitute it into equation (i):} \\
& \sqrt{y + 8} + \sqrt{y + 2} = 3 \\
& \text{Squaring both sides:} \\
& (\sqrt{y + 8} + \sqrt{y + 2})^2 = (3)^2 \\
& (\sqrt{y + 8})^2 + (\sqrt{y + 2})^2 + 2\sqrt{(y + 8)(y + 2)} = 9 \\
& y + 8 + y + 2 + 2\sqrt{(y + 8)(y + 2)} = 9 \\
& 2y + 10 + 2\sqrt{y^2 + 10y + 16} = 9 \\
& 2\sqrt{y^2 + 10y + 16} = -2y – 1 \\
& \text{Squaring both sides again:} \\
& (2\sqrt{y^2 + 10y + 16})^2 = (-2y – 1)^2 \\
& 4(y^2 + 10y + 16) = (2y)^2 + (1)^2 + 2(2y)(1) \\
& 4y^2 + 40y + 64 = 4y^2 + 1 + 4y \\
& 4y^2 + 40y + 64 = 4y^2 + 4y + 1 \\
& 40y – 4y = 1 – 64 \\
& 36y = -63 \\
& y = \frac{-63}{36} = -\frac{7}{4} \\
& \text{Substitute back} \ y = x^2 + 3x: \\
& x^2 + 3x = -\frac{7}{4} \\
& 4x^2 + 12x + 7 = 0 \\
\text{Using the quadratic formula:} \\
& x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
\text{where } a = 4, \; b = 12, \; c = 7 \\
& x = \frac{-12 \pm \sqrt{12^2 – 4 \cdot 4 \cdot 7}}{2 \cdot 4} \\
& x = \frac{-12 \pm \sqrt{144 – 112}}{8} \\
& x = \frac{-12 \pm \sqrt{32}}{8} \\
& x = \frac{-12 \pm 4\sqrt{2}}{8} \\
& x = \frac{-12}{8} \pm \frac{4\sqrt{2}}{8} \\
& x = -\frac{3}{2} \pm \frac{\sqrt{2}}{2} \\
& x = -\frac{3}{2} + \frac{\sqrt{2}}{2} \quad \text{or} \quad x = -\frac{3}{2} – \frac{\sqrt{2}}{2}
\end{align*}
\]
Question 11 Exercise 1.4 Class 10. \(\boldsymbol{\sqrt{x^2 + 3x + 9} + \sqrt{x^2 + 3x + 4} = 5}\)
\[
\begin{align*}
\text{Solution:} \\
& \sqrt{x^2 + 3x + 9} + \sqrt{x^2 + 3x + 4} = 5 \\
& \text{Let } y = \sqrt{x^2 + 3x + 4} \\
& \text{Then } \sqrt{x^2 + 3x + 9} = y + 1 \\
& \text{So, the equation becomes:} \\
& y + 1 + y = 5 \\
& 2y + 1 = 5 \\
& 2y = 4 \\
& y = 2 \\
& \text{Since } y = \sqrt{x^2 + 3x + 4}, \text{ we have:} \\
& \sqrt{x^2 + 3x + 4} = 2 \\
& x^2 + 3x + 4 = 4 \\
& x^2 + 3x = 0 \\
& x(x + 3) = 0 \\
& x = 0 \quad \text{or} \quad x = -3 \\
\end{align*}
\]
\[
\begin{align*}
\text{Checking:} \\
& \text{For } x = 0: \\
& \sqrt{0^2 + 3(0) + 9} + \sqrt{0^2 + 3(0) + 4} = 5 \\
& \sqrt{9} + \sqrt{4} = 5 \\
& 3 + 2 = 5 \\
& 5 = 5 \quad \text{which is true} \\
& \text{For } x = -3: \\
& \sqrt{(-3)^2 + 3(-3) + 9} + \sqrt{(-3)^2 + 3(-3) + 4} = 5 \\
& \sqrt{9 – 9 + 9} + \sqrt{9 – 9 + 4} = 5 \\
& \sqrt{9} + \sqrt{4} = 5 \\
& 3 + 2 = 5 \\
& 5 = 5 \quad \text{which is true} \\
& \text{So, the solution set is } \{0, -3\}
\end{align*}
\]
That takes us to the end of Unit 1 by completing exercise 1.4 class 10. Looking to review earlier concepts? Go back to the solutions for Exercise 1.1 to strengthen your foundation in quadratic equations. For a comprehensive understanding of radical equations and effective solving strategies, check out this Purplemath Radical Equations Tutorial.
⚠️ Disclaimer Paragraph
Disclaimer: The textbook used for these solutions is published by the Punjab Curriculum and Textbook Board (PCTB). The notes and solutions provided here are created only for educational support and to help students understand each step clearly. As the new Class 10 Math book has been recently introduced, some questions or numerical values may contain minor errors or updates. For official curriculum details and authentic book versions, please visit pctb.punjab.gov.pk.
Take a look at this video lecture of Exercise 1.4 class 10.
