Exercise 1.3 Class 10 Easy Solutions
In Exercise 1.3 Class 10 math, we will deal with the equations that are reducible to quadratic form and are typically of five types understanding these equations will help in solving exercise 1.3 questions.
Question 1, Question 2, Question 3, Question 4, Question 5, Question 6, Question 7, Question 8, Question 9, Question 10, Question 11, Question 12, Question 13, Question 14, Question 15, Question 16,
Solve the following equations of Exercise 1.3 Class 10
Question 1 of Exercise 1.3 Class 10. \( 2x^4 – 11x^2 + 5 = 0 \)
To solve 2x⁴ – 11x² + 5 = 0:
Substitute y = x²:
2y² – 11y + 5 = 0
Factor:
(2y – 1)(y – 5) = 0
Solve for y:
2y – 1 = 0 and y – 5 = 0
y = 1/2 and y = 5
Substitute x² back:
For y = 1/2: x = ±√(1/2) = ±√2/2
For y = 5: x = ±√5
The solutions are:
x = ±√2/2, x = ±√5
Question 2 of Exercise 1.3 Class 10. \( 2x^4 = 9x^2 – 4 \)
To solve 2x⁴ = 9x² – 4:
Let y = x². Substitute y into the equation:
2y² = 9y – 4
Rearrange the equation: 2y² – 9y + 4 = 0
(2y – 1)(y – 4) = 0
Solve for y: 2y – 1 = 0 → y = 1/2
For y – 4 = 0 → y = 4
Substitute back x² for y:
For y = 1/2 → x² = 1/2
x = ±√(1/2)
x = ±1/√2
For y = 4:
x² = 4 → x = ±2
So the solutions are x = ±1/√2 and x = ±2.
Question 3 of Exercise 1.3 Class 10. \(5x^{1/2} = 7x^{1/4} – 2\)
Question 4 of Exercise 1.3 Class 10. \(x^{2/3} + 54 = 15x^{1/3}\)
To solve the equation \(x^{2/3} + 54 = 15x^{1/3}\)
Start by substituting \(x^{1/3} = y\). This gives \(x^{2/3} = y^2\),
so the equation becomes: \(y^2 + 54 = 15y\)
Rearrange the equation: \(y^2 – 15y + 54 = 0\)
Now, factorize the quadratic equation: \((y – 9)(y – 6) = 0\)
Solve for \(y\):
\[y = 9 \quad \text{or} \quad y = 6\]
Substitute back \(y = x^{1/3}\) to find \(x\):
- For \(y = 9\):
\[x^{1/3} = 9 \implies x = 9^3 = 729\] - For \(y = 6\):
\[x^{1/3} = 6 \implies x = 6^3 = 216\]
Thus, the solutions are \(x = 729\) and \(x = 216\).
Question 5 of Exercise 1.3 Class 10. \(3x^{-2} + 5= 8x^{-1}\)
Substitute \( y = x^{-1} \), so the equation becomes:
\[3y^2 + 5 = 8y\]
Rearrange: \(3y^2 – 8y + 5 = 0\)
Now, to factorize \( 3y^2 – 8y + 5 = 0 \), we look for two numbers that multiply to \( 3 \times 5 = 15 \) and add up to \( -8 \). These numbers are \( -3 \) and \( -5 \).
Rewrite the middle term: \(3y^2 – 3y – 5y + 5 = 0\)
Group the terms: \((3y^2 – 3y) – (5y – 5) = 0\)
Factor each group: \(3y(y – 1) – 5(y – 1) = 0\)
Factor out the common factor \( (y – 1) \):
\[
(y – 1)(3y – 5) = 0
\]
Solve for \( y \):
\[
y = 1 \quad \text{or} \quad y = \frac{5}{3}
\]
Since \( y = x^{-1} \), substitute back:
\[
x = 1 \quad \text{or} \quad x = \frac{3}{5}
\]
Question 6 of Exercise 1.3 Class 10. \((2x^2 + 1) + \frac{3}{2x^2 + 1} = 4\)
Given the equation: \((2x^2 + 1) + \frac{3}{2x^2 + 1} = 4\)
Let \( y = 2x^2 + 1 \), so the equation becomes: \(y + \frac{3}{y} = 4\)
Multiply through by \( y \) to eliminate the fraction: \(y^2 + 3 = 4y\)
Rearrange the equation: \(y^2 – 4y + 3 = 0\)
Factorize: \((y – 1)(y – 3) = 0\)
Solve for \( y \):
\[
y = 1 \quad \text{or} \quad y = 3
\]
Substitute \( y = 2x^2 + 1 \) back into the equation:
For \( y = 1 \):
\[2x^2 + 1 = 1\rightarrow 2x^2 = 0\]
\[x = 0\]
\[\text{For } y = 3 \]:
\[2x^2 + 1 = 3\]
\[2x^2 = 2 \rightarrow x^2 = 1\]
\[x = \pm 1\]
The solutions are: \(x = 0, \quad x = 1, \quad x = -1\)
Question 7 of Exercise 1.3 Class 10. \(\frac{x}{x-3} + 4\left(\frac{x-3}{x}\right) = 4\)
To solve the equation \(\frac{x}{x-3} + 4\left(\frac{x-3}{x}\right) = 4\),
First, let \( y = \frac{x}{x-3} \), which implies \(\frac{x-3}{x} = \frac{1}{y}\).
Substituting this into the equation gives:
\[y + 4 \cdot \frac{1}{y} = 4.\]
Multiply both sides by \( y \) to eliminate the fraction:
\[y^2 + 4 = 4y.\]
Rearranging the equation: \(y^2 – 4y + 4 = 0.\)
Now, factor the quadratic equation: \((y – 2)^2 = 0,\)
which gives \( y = 2 \).
Since \( y = \frac{x}{x-3} \), substitute back to get:
\[\frac{x}{x-3} = 2.\]
Solve for \( x \) by cross-multiplying: \(x = 2(x – 3),\)
which simplifies to: \(x = 2x – 6.\)
Finally, solving for \( x \):
\[x = 6.\]
The solution is \( x = 6 \).
Question 8 of Exercise 1.3 Class 10. \(\frac{4x+1}{4x-1} + \frac{4x-1}{4x+1} = 2 \frac{1}{6}\)
To solve the equation
\[\frac{4x+1}{4x-1} + \frac{4x-1}{4x+1} = 2\frac{1}{6},\]
first, recognize that \(2 \frac{1}{6}\) is a mixed fraction, which equals \( \frac{13}{6} \).
The equation now becomes:
\[\frac{4x+1}{4x-1} + \frac{4x-1}{4x+1} = \frac{13}{6}.\]
We can combine the fractions on the left-hand side by finding a common denominator. The common denominator is \((4x-1)(4x+1)\).
Expanding the numerators, we get:
\[(4x+1)^2 = 16x^2 + 8x + 1,\]
\[(4x-1)^2 = 16x^2 – 8x + 1.\]
Adding these expressions:
\[(16x^2 + 8x + 1) + (16x^2 – 8x + 1) = 32x^2 + 2.\]
So the equation becomes:
\[\frac{32x^2 + 2}{(4x-1)(4x+1)} = \frac{13}{6}.\]
The denominator \((4x-1)(4x+1)\) simplifies to \(16x^2 – 1\), so the equation is now:
\[\frac{32x^2 + 2}{16x^2 – 1} = \frac{13}{6}.\]
Cross-multiply:
\[6(32x^2 + 2) = 13(16x^2 – 1).\]
Expanding both sides gives:
\[192x^2 + 12 = 208x^2 – 13.\]
Rearranging the terms:
\[192x^2 – 208x^2 + 12 + 13 = 0,\]
\[-16x^2 + 25 = 0.\]
Solve for \(x\):
\[16x^2 = 25,\]
\[x^2 = \frac{25}{16},\]
\[x = \pm \frac{5}{4}.\]
Thus, the solution is:
\[x = \pm \frac{5}{4}.\]
Question 9 of Exercise 1.3 Class 10. \(\frac{x-a}{x+a} – \frac{x+a}{x-a} = \frac{7}{12},\)
\[
\frac{x-a}{x+a} – \frac{x+a}{x-a} = \frac{7}{12},
\]
let’s substitute \( y = \frac{x-a}{x+a} \). This means that:
\[
\frac{x+a}{x-a} = \frac{1}{y}.
\]
The equation now becomes:
\[
y – \frac{1}{y} = \frac{7}{12}.
\]
Next, multiply through by \( y \) to eliminate the fraction:
\[
y^2 – 1 = \frac{7}{12}y.
\]
Rearranging this gives:
\[
12y^2 – 7y – 12 = 0.
\]
Now we can factor this quadratic equation. We are looking for two numbers that multiply to \( 12 \times -12 = -144 \) and add to \( -7 \). The numbers are \( -16 \) and \( 9 \).
Rewriting the equation:
\[
12y^2 – 16y + 9y – 12 = 0.
\]
Grouping the terms:
\[
(12y^2 – 16y) + (9y – 12) = 0.
\]
Factoring by grouping:
\[
4y(3y – 4) + 3(3y – 4) = 0.
\]
Factoring out the common factor (\(3y – 4)\):
\[
(3y – 4)(4y + 3) = 0.
\]
Setting each factor to zero gives:
\[
3y – 4 = 0 \quad \text{or} \quad 4y + 3 = 0.
\]
Solving for \( y \):
\[
y = \frac{4}{3} \quad \text{or} \quad y = -\frac{3}{4}.
\]
Recalling that \( y = \frac{x-a}{x+a} \), we can set up two equations:
- \(\frac{x-a}{x+a} = \frac{4}{3}\)
- \(\frac{x-a}{x+a} = -\frac{3}{4}\)
For the first equation, cross-multiplying gives:
\[
3(x-a) = 4(x+a).
\]
Expanding and rearranging gives:
\[
3x – 3a = 4x + 4a \quad \Rightarrow \quad -x = 7a \quad \Rightarrow \quad x = -7a.
\]
For the second equation, cross-multiplying gives:
\[
4(x-a) = -3(x+a).
\]
Expanding and rearranging gives:
\[
4x – 4a = -3x – 3a \quad \Rightarrow \quad 7x = a \quad \Rightarrow \quad x = \frac{a}{7}.
\]
Thus, the solutions are:
\[
x = -7a \quad \text{and} \quad x = \frac{a}{7}.
\]
Question 10 of Exercise 1.3 Class 10. \( x^4 – 2x^3 – 2x^2 + 2x + 1 = 0 \)
Let’s divide the equation \( x^4 – 2x^3 – 2x^2 + 2x + 1 = 0 \) by \( x^2 \):
\[
x^2 – 2x – 2 + \frac{2}{x} + \frac{1}{x^2} = 0
\]
Now, rearranging gives:
\[
x^2 + \frac{1}{x^2} – 2\left(x – \frac{1}{x}\right) – 2 = 0
\]
Next, we can substitute \( y = x – \frac{1}{x} \). Then, we also know that:
\[
x^2 + \frac{1}{x^2} = (x – \frac{1}{x})^2 + 2 = y^2 + 2
\]
Substituting this into the equation gives us:
\[
(y^2 + 2) – 2y – 2 = 0
\]
Simplifying this results in:
\[
y^2 – 2y = 0
\]
Factoring out \( y \):
\[
y(y – 2) = 0
\]
Thus, \( y = 0 \) or \( y = 2 \).
- If \( y = 0 \):
\[
x – \frac{1}{x} = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1
\] - If \( y = 2 \):
\[
x – \frac{1}{x} = 2 \implies x^2 – 2x – 1 = 0
\]
Using the quadratic formula:
\[
x = \frac{2 \pm \sqrt{(-2)^2 – 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}
\]
So, the complete set of solutions is:
\[
{ x = 1, \, x = -1, \, x = 1 + \sqrt{2}, \, x = 1 – \sqrt{2} }
\]
Question 11 of Exercise 1.3 Class 10. \( 2x^4 + x^3 – 6x^2 + x + 2 = 0 \)
Given:
\[ 2x^4 + x^3 – 6x^2 + x + 2 = 0 \]
Dividing each term by \( x^2 \):
\[ \frac{2x^4 + x^3 – 6x^2 + x + 2}{x^2} = 0 \]
\[ 2x^2 + x – 6 + \frac{1}{x} + \frac{2}{x^2} = 0 \]
Let: \( y = x + \frac{1}{x} \)
Then:\( y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \)
Substituting \( y \) into the equation:
\[ 2 \left( y^2 – 2 \right) + y – 6 = 0 \]
\[ 2y^2 – 4 + y – 6 = 0 \]
\[ 2y^2 + y – 10 = 0 \]
Now, solve for \( y \) by factoring:
\[ 2y^2 + 5y – 4y – 10 = 0 \]
\[ y(2y + 5) – 2(2y + 5) = 0 \]
\[ (y – 2)(2y + 5) = 0 \]
So, \( y = 2 \) or \( y = -\frac{5}{2} \).
For \( y = 2 \):
\[ 2 = x + \frac{1}{x} \]
\[ 2x = x^2 + 1 \]
\[ x^2 – 2x + 1 = 0 \]
\[ (x – 1)^2 = 0 \]
\[ x = 1 \]
For \( y = -\frac{5}{2} \):
\[ -\frac{5}{2} = x + \frac{1}{x} \]
\[ -\frac{5}{2}x = x^2 + 1 \]
\[ 2x^2 + 5x + 2 = 0 \]
\[ (2x + 1)(x + 2) = 0 \]
\[ x = -\frac{1}{2} \text{ or } x = -2 \]
So, the solutions are:
\[ x = 1, x = -\frac{1}{2}, \text{ and } x = -2 \]
Question 12 of Exercise 1.3 Class 10. \( 4 \cdot 2^{x+1} = 4 \cdot (2^x) \cdot 2 = 8 \cdot 2^x \)
\[ 4 \cdot 2^{x+1} = 4 \cdot (2^x) \cdot 2 = 8 \cdot 2^x \]
\[ 9 \cdot 2^x = 9 \cdot 2^x \]
\[\text{Let } y = 2^x \text{ then equation } (i) \text{ becomes:}\]
\[ 8y – 9y + 1 = 0 \quad \quad \quad (ii) \]
\[ y – y + 1 = 0 \]
\[ (y – 1)(8y – 1) = 0 \]
\[ y – 1 = 0 \text{ or } 8y – 1 = 0 \]
\[ y = 1 \text{ or } y = \frac{1}{8} \]
\[\text{From equation } (ii) \text{: Put } y = 2^x\]
\[ \Rightarrow 2^x = 1 \text{ or } 2^x = \frac{1}{8} \]
\[ \Rightarrow x = 0 \quad \text{or} \quad x = -3 \]
\[ \therefore x = \{0, -3\} \]
Question 13 of Exercise 1.3 Class 10. \(3^{2x+2} = 12 \cdot 3^x – 3 \)
\(Solution:\)
\[ 3^{2x+2} = 12 \cdot 3^x – 3 \]
First, rewrite the equation:
\[ 3^{2x+2} = 3^{2x} \cdot 3^2 = 9 \cdot 3^{2x} \]
So the equation becomes:
\[ 9 \cdot 3^{2x} = 12 \cdot 3^x – 3 \]
Let \( y = 3^x \). Then \( 3^{2x} = (3^x)^2 = y^2 \). Substituting \( y \) into the equation, we get:
\[ 9y^2 = 12y – 3 \]
Rearrange the equation to form a standard quadratic equation:
\[ 9y^2 – 12y + 3 = 0 \]
Now, factorize the quadratic equation:
\[ 9y^2 – 12y + 3 = 0 \]
\[ 9y^2 – 9y – 3y + 3 = 0 \]
Group the terms:
\[ 9y(y – 1) – 3(y – 1) = 0 \]
Factor out the common factor:
\[ (9y – 3)(y – 1) = 0 \]
Set each factor to zero:
\[ 9y – 3 = 0 \quad \text{or} \quad y – 1 = 0 \]
Solve for \( y \):
\[ y = \frac{3}{9} = \frac{1}{3} \quad \text{or} \quad y = 1 \]
Since \( y = 3^x \), we have:
\[ 3^x = 1 \text { or } 3^x = \frac{1}{3} \]
For \( 3^x = 1 \rightarrow 3^x=3^0\) so \( x = 0 \)
For \( 3^x = \frac{1}{3}\rightarrow 3^x=3^{-1} \) so \( x = -1 \)
Therefore, the solutions are:
\[ x = 0 \text{ and } x = -1 \]
Question 14 of Exercise 1.3 Class 10. \( 2^x + 64 \cdot 2^{-x} = 20 \)
$$
2^x + 64 \cdot 2^{-x} = 20
$$
Let \( 2^x = y \), then the equation becomes:
$$
y + \frac{64}{y} = 20
$$
Multiply both sides by \( y \):
$$
y^2 + 64 = 20y
$$
$$
y^2 – 20y + 64 = 0
$$
Solve the quadratic equation:
$$
(y – 16)(y – 4) = 0
$$
Thus, \( y = 16 \) or \( y = 4 \).
Since \( y = 2^x \):
$$
2^x = 16 \quad \text{or} \quad 2^x = 4
$$
$$ 2^x= 2^4 \quad \text{|} \quad 2^x=2^2$$
Therefore:
$$
x = 4 \quad \text{or} \quad x = 2
$$
Question 15 of Exercise 1.3 Class 10. \(\quad (x+1)(x+3)(x-5)(x-7) = 192\)
\[ \quad (x+1)(x+3)(x-5)(x-7) = 192\]
\[
\therefore \quad \boxed{a+b=c+d}
\]
\[
1 – 5 = 3 – 7
\]
\[
-4 = -4
\]
\[
(x+1)(x-5)(x+3)(x-7) = 192
\]
\[
(x^2 – 5x + 1x – 5)(x^2 – 7x + 3x – 21) = 192
\]
\[
(x^2 – 4x – 5)(x^2 – 4x – 21) = 192 \quad \text{….(i)}
\]
\(\text{Let} \quad x^2 – 4x = y\)
\(\text{The equation (i) becomes}\)
\[
(y – 5)(y – 21) = 192
\]
\[
y^2 – 21y – 5y + 105 = 192
\]
\[
y^2 – 26y + 105 – 192 = 0
\]
\[
y^2 – 26y – 87 = 0
\]
\[
y^2 – 29y + 3y – 87 = 0
\]
\[
y(y – 29) + 3(y – 29) = 0
\]
\[
(y – 29)(y + 3) = 0
\]
\[
y – 29 = 0 \quad \text{or} \quad y + 3 = 0
\]
\[
y = 29 \quad \text{or} \quad y = -3
\]
\(\text{From equation (ii) Put} \quad y = x^2 – 4x\)
\[
x^2 – 4x = 29 \quad \text{or} \quad x^2 – 4x = -3
\]
\[
x^2 – 4x – 29 = 0 \quad \text{or} \quad x^2 – 4x + 3 = 0
\]
\[
\boxed{x^2 – 4x – 29 = 0} \quad \text{or} \quad \boxed{x^2 – 4x + 3 = 0}
\]
\(\text{First we solve} \quad x^2 – 4x – 29 = 0 \quad \text{by quadratic formula}\)
\[
a = 1, \quad b = -4, \quad c = -29
\]
\[
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
\]
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-29)}}{2(1)}
\]
\[
x = \frac{4 \pm \sqrt{16 + 116}}{2}
\]
\[
x = \frac{4 \pm \sqrt{132}}{2}
\]
\[
x = \frac{4 \pm \sqrt{4 \times 33}}{2}
\]
\[
x = \frac{4 \pm 2\sqrt{33}}{2}
\]
\[
x = \frac{4}{2} \pm \sqrt{33}
\]
\[
\boxed{x = 2 \pm \sqrt{33}}
\]
\(\text{Now we solve} \quad x^2 – 4x + 3 = 0 \quad \text{by factorization}\)
\[
x^2 – 4x + 3 = 0
\]
\[
x^2 – 3x – x + 3 = 0
\]
\[
x(x – 3) – 1(x – 3) = 0
\]
\[
(x – 3)(x – 1) = 0
\]
\[
x – 3 = 0 \quad \text{or} \quad x – 1 = 0
\]
\[
x = 3 \quad \text{or} \quad x = 1
\]
\(\text{Solution set is} \quad \{1, 3, 2 \pm \sqrt{33}\}\)
Question 16 of Exercise 1.3 Class 10. \( (x-1)(x-2)(x-8)(x+5) + 360 = 0 \)
\[
[(x-1)(x-2)][(x-8)(x+5)] + 360 = 0
\]
\[
[x^2 – 2x – 1x + 2][x^2 + 5x – 8x – 40] + 360 = 0
\]
\[
(x^2 – 3x + 2)(x^2 – 3x – 40) + 360 = 0 \quad \text{….(i)}
\]
\(\text{Let} \quad x^2 – 3x = y\)
\(\text{Put it in equation (i)}\)
\[
(y+2)(y-40) + 360 = 0
\]
\[
y^2 – 40y + 2y – 80 + 360 = 0
\]
\[
y^2 – 38y + 280 = 0
\]
\[
y^2 – 28y – 10y + 280 = 0
\]
\[
y(y – 28) – 10(y – 28) = 0
\]
\[
(y – 28)(y – 10) = 0
\]
\[
y – 28 = 0 \quad \text{or} \quad y – 10 = 0
\]
\[
y = 28 \quad \text{or} \quad y = 10
\]
\(\text{From equation (ii) Put} \quad y = x^2 – 3x\)
\[
x^2 – 3x = 28 \quad \text{ and } \quad x^2 – 3x = 10
\]
\[
x^2 – 3x – 28 = 0
\]
\[
x^2 – 7x + 4x – 28 = 0
\]
\[
x(x – 7) + 4(x – 7) = 0
\]
\[
(x – 7)(x + 4) = 0
\]
\[
x – 7 = 0 \quad \text{or} \quad x + 4 = 0
\]
\[
x = 7 \quad \text{or} \quad x = -4
\]
\(\text {and now for } x^2 – 3x = 10\)
\[
x^2 – 3x – 10 = 0
\]
\[
x^2 – 5x + 2x – 10 = 0
\]
\[
x(x – 5) + 2(x – 5) = 0
\]
\[
(x – 5)(x + 2) = 0
\]
\[
x – 5 = 0 \quad \text{or} \quad x + 2 = 0
\]
\[
x = 5 \quad \text{or} \quad x = -2
\]
\(\text{The solution set is} \quad \{5, -2, 7, -4\}\)
That’s all for the Exercise 1.3 Class 10 math now lets move on to the Exercise 1.4
Here’s video lecture of complete exercise 1.3 if you are any having problems understanding these solutions. Check this online quadratic equation solver tool to make math interesting.
