Quadratic EquationsClass 10 Mathematics NCERT Solutions

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Solution: Expanding the terms: LHS: $(x+1)^2 = x^2 + 2x + 1$ RHS: $2(x-3) = 2x - 6$

Equating LHS and RHS: $x^2 + 2x + 1 = 2x - 6$ $x^2 + 2x - 2x + 1 + 6 = 0$ $x^2 + 7 = 0$ This can be written as $x^2 + 0x + 7 = 0$. It is in the form $ax^2 + bx + c = 0$ where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: Expanding the RHS: RHS: $(-2)(3-x) = -6 + 2x$

Equating LHS and RHS: $x^2 - 2x = -6 + 2x$ $x^2 - 2x - 2x + 6 = 0$ $x^2 - 4x + 6 = 0$ This is in the form $ax^2 + bx + c = 0$ where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: Expanding both sides: LHS: $(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$ RHS: $(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$

Equating LHS and RHS: $x^2 - x - 2 = x^2 + 2x - 3$ $x^2 - x^2 - x - 2x - 2 + 3 = 0$ $-3x + 1 = 0$ This is a linear equation, not a quadratic equation, as the coefficient of $x^2$ is 0.

Final Answer: No, the given equation is not a quadratic equation.

Solution: Expanding both sides: LHS: $(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$ RHS: $x(x+5) = x^2 + 5x$

Equating LHS and RHS: $2x^2 - 5x - 3 = x^2 + 5x$ $2x^2 - x^2 - 5x - 5x - 3 = 0$ $x^2 - 10x - 3 = 0$ This is in the form $ax^2 + bx + c = 0$ where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: Expanding both sides: LHS: $(2x-1)(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$ RHS: $(x+5)(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5$

Equating LHS and RHS: $2x^2 - 7x + 3 = x^2 + 4x - 5$ $2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$ $x^2 - 11x + 8 = 0$ This is in the form $ax^2 + bx + c = 0$ where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: Expanding the RHS: RHS: $(x-2)^2 = x^2 - 4x + 4$

Equating LHS and RHS: $x^2 + 3x + 1 = x^2 - 4x + 4$ $x^2 - x^2 + 3x + 4x + 1 - 4 = 0$ $7x - 3 = 0$ This is a linear equation, not a quadratic equation, as the coefficient of $x^2$ is 0.

Final Answer: No, the given equation is not a quadratic equation.

Solution: Expanding both sides using the formula $(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$: LHS: $(x+2)^3 = x^3 + 2^3 + 3(x^2)(2) + 3(x)(2^2) = x^3 + 8 + 6x^2 + 12x$ RHS: $2x(x^2 - 1) = 2x^3 - 2x$

Equating LHS and RHS: $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$ $2x^3 - x^3 - 6x^2 - 12x - 2x - 8 = 0$ $x^3 - 6x^2 - 14x - 8 = 0$ This is a cubic equation, not a quadratic equation, as the highest power of $x$ is 3.

Final Answer: No, the given equation is not a quadratic equation.

Solution: Expanding the RHS using the formula $(a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2$: RHS: $(x-2)^3 = x^3 - 2^3 - 3(x^2)(2) + 3(x)(2^2) = x^3 - 8 - 6x^2 + 12x$

Equating LHS and RHS: $x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$ $x^3 - x^3 - 4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$ $2x^2 - 13x + 9 = 0$ This is in the form $ax^2 + bx + c = 0$ where $a=2 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Solution: We simplify the given equation: $LHS = (x+1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1$ $RHS = 2(x-3) = 2x - 6$ Equating LHS and RHS: $x^2 + 2x + 1 = 2x - 6$ $x^2 + 2x - 2x + 1 + 6 = 0$ $x^2 + 7 = 0$ This equation can be written as $1x^2 + 0x + 7 = 0$. It is of the form $ax^2 + bx + c = 0$, where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: We simplify the given equation: $LHS = x^2 - 2x$ $RHS = (-2)(3-x) = -6 + 2x$ Equating LHS and RHS: $x^2 - 2x = -6 + 2x$ $x^2 - 2x - 2x + 6 = 0$ $x^2 - 4x + 6 = 0$ This equation is of the form $ax^2 + bx + c = 0$, where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: We simplify the given equation: $LHS = (x-2)(x+1) = x(x+1) - 2(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$ $RHS = (x-1)(x+3) = x(x+3) - 1(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$ Equating LHS and RHS: $x^2 - x - 2 = x^2 + 2x - 3$ $x^2 - x^2 - x - 2x - 2 + 3 = 0$ $-3x + 1 = 0$ This equation is not of the form $ax^2 + bx + c = 0$ because the coefficient of $x^2$ is 0.

Final Answer: No, the given equation is not a quadratic equation.

Solution: We simplify the given equation: $LHS = (x-3)(2x+1) = x(2x+1) - 3(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$ $RHS = x(x+5) = x^2 + 5x$ Equating LHS and RHS: $2x^2 - 5x - 3 = x^2 + 5x$ $2x^2 - x^2 - 5x - 5x - 3 = 0$ $x^2 - 10x - 3 = 0$ This equation is of the form $ax^2 + bx + c = 0$, where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: We simplify the given equation: $LHS = (2x-1)(x-3) = 2x(x-3) - 1(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$ $RHS = (x+5)(x-1) = x(x-1) + 5(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5$ Equating LHS and RHS: $2x^2 - 7x + 3 = x^2 + 4x - 5$ $2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$ $x^2 - 11x + 8 = 0$ This equation is of the form $ax^2 + bx + c = 0$, where $a=1 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: We simplify the given equation: $LHS = x^2 + 3x + 1$ $RHS = (x-2)^2 = x^2 - 2(x)(2) + 2^2 = x^2 - 4x + 4$ Equating LHS and RHS: $x^2 + 3x + 1 = x^2 - 4x + 4$ $x^2 - x^2 + 3x + 4x + 1 - 4 = 0$ $7x - 3 = 0$ This equation is not of the form $ax^2 + bx + c = 0$ because the coefficient of $x^2$ is 0.

Final Answer: No, the given equation is not a quadratic equation.

Solution: We simplify the given equation using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. $LHS = (x+2)^3 = x^3 + 2^3 + 3(x)(2)(x+2) = x^3 + 8 + 6x(x+2) = x^3 + 8 + 6x^2 + 12x$ $RHS = 2x(x^2-1) = 2x^3 - 2x$ Equating LHS and RHS: $x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$ $2x^3 - x^3 - 6x^2 - 12x - 2x - 8 = 0$ $x^3 - 6x^2 - 14x - 8 = 0$ This is a cubic equation, not a quadratic equation, as the highest power of $x$ is 3.

Final Answer: No, the given equation is not a quadratic equation.

Solution: We simplify the given equation using the identity $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$. $LHS = x^3 - 4x^2 - x + 1$ $RHS = (x-2)^3 = x^3 - 2^3 - 3(x)(2)(x-2) = x^3 - 8 - 6x(x-2) = x^3 - 8 - 6x^2 + 12x$ Equating LHS and RHS: $x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$ $x^3 - x^3 - 4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$ $2x^2 - 13x + 9 = 0$ This equation is of the form $ax^2 + bx + c = 0$, where $a=2 \neq 0$.

Final Answer: Yes, the given equation is a quadratic equation.

Solution: Let the breadth of the rectangular plot be $x$ metres. According to the problem, the length of the plot is one more than twice its breadth. So, length $= (2x + 1)$ metres.

Area of a rectangle = length $\times$ breadth Given, Area = $528 \text{ m}^2$

Therefore, we can write the equation: $(2x + 1) \times x = 528$ $2x^2 + x = 528$ $2x^2 + x - 528 = 0$

This is the required quadratic equation to find the length and breadth of the plot.

Final Answer: The required quadratic equation is $2x^2 + x - 528 = 0$.

Solution: Let the two consecutive positive integers be $x$ and $(x+1)$. According to the problem, their product is 306.

Therefore, we can write the equation: $x(x+1) = 306$ $x^2 + x = 306$ $x^2 + x - 306 = 0$

This is the required quadratic equation to find the integers.

Final Answer: The required quadratic equation is $x^2 + x - 306 = 0$.

Solution: Let Rohan's present age be $x$ years. Since his mother is 26 years older, her present age is $(x + 26)$ years.

After 3 years: Rohan's age will be $(x + 3)$ years. His mother's age will be $(x + 26 + 3) = (x + 29)$ years.

According to the problem, the product of their ages after 3 years will be 360. Therefore, we can write the equation: $(x + 3)(x + 29) = 360$ $x(x+29) + 3(x+29) = 360$ $x^2 + 29x + 3x + 87 = 360$ $x^2 + 32x + 87 - 360 = 0$ $x^2 + 32x - 273 = 0$

This is the required quadratic equation to find Rohan's present age.

Final Answer: The required quadratic equation is $x^2 + 32x - 273 = 0$.

Solution: Let the uniform speed of the train be $x$ km/h. Distance to be covered = 480 km.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Time taken to cover 480 km at speed $x$ is $T_1 = \frac{480}{x}$ hours.

If the speed had been 8 km/h less, the new speed would be $(x - 8)$ km/h. Time taken to cover 480 km at the new speed is $T_2 = \frac{480}{x-8}$ hours.

According to the problem, $T_2$ is 3 hours more than $T_1$. So, $T_2 = T_1 + 3$. $\frac{480}{x-8} = \frac{480}{x} + 3$ $\frac{480}{x-8} - \frac{480}{x} = 3$ $480 \left( \frac{1}{x-8} - \frac{1}{x} \right) = 3$ $480 \left( \frac{x - (x-8)}{x(x-8)} \right) = 3$ $480 \left( \frac{x - x + 8}{x^2 - 8x} \right) = 3$ $480 \left( \frac{8}{x^2 - 8x} \right) = 3$ $\frac{3840}{x^2 - 8x} = 3$ $3840 = 3(x^2 - 8x)$ $1280 = x^2 - 8x$ $x^2 - 8x - 1280 = 0$

This is the required quadratic equation to find the speed of the train.

Final Answer: The required quadratic equation is $x^2 - 8x - 1280 = 0$.

Solution: Let the breadth of the rectangular plot be $x$ metres. According to the problem, the length of the plot is one more than twice its breadth. So, length of the plot $= (2x + 1)$ metres.

We are given that the area of the rectangular plot is $528 \text{ m}^2$. Area of a rectangle = length $\times$ breadth $528 = (2x + 1) \times x$ $528 = 2x^2 + x$ Rearranging the terms to form a standard quadratic equation: $2x^2 + x - 528 = 0$

Final Answer: The required quadratic equation is $2x^2 + x - 528 = 0$.

Solution: Let the two consecutive positive integers be $x$ and $(x+1)$. According to the problem, their product is 306. $x(x+1) = 306$ $x^2 + x = 306$ Rearranging the terms to form a standard quadratic equation: $x^2 + x - 306 = 0$

Final Answer: The required quadratic equation is $x^2 + x - 306 = 0$.

Solution: Let Rohan's present age be $x$ years. Since his mother is 26 years older than him, his mother's present age is $(x + 26)$ years.

3 years from now: Rohan's age will be $(x + 3)$ years. His mother's age will be $(x + 26 + 3) = (x + 29)$ years.

According to the problem, the product of their ages 3 years from now will be 360. $(x+3)(x+29) = 360$ $x(x+29) + 3(x+29) = 360$ $x^2 + 29x + 3x + 87 = 360$ $x^2 + 32x + 87 - 360 = 0$ $x^2 + 32x - 273 = 0$

Final Answer: The required quadratic equation is $x^2 + 32x - 273 = 0$.

Solution: Let the uniform speed of the train be $x$ km/h. Distance to be covered = 480 km.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$. Time taken to cover 480 km at speed $x$ km/h is $T_1 = \frac{480}{x}$ hours.

If the speed had been 8 km/h less, the new speed would be $(x - 8)$ km/h. Time taken to cover 480 km at this new speed is $T_2 = \frac{480}{x-8}$ hours.

According to the problem, the new time taken ($T_2$) is 3 hours more than the original time ($T_1$). $T_2 = T_1 + 3$ $\frac{480}{x-8} = \frac{480}{x} + 3$ $\frac{480}{x-8} - \frac{480}{x} = 3$ $480 \left( \frac{1}{x-8} - \frac{1}{x} \right) = 3$ $480 \left( \frac{x - (x-8)}{x(x-8)} \right) = 3$ $480 \left( \frac{x - x + 8}{x^2 - 8x} \right) = 3$ $\frac{3840}{x^2 - 8x} = 3$ $3840 = 3(x^2 - 8x)$ $1280 = x^2 - 8x$ Rearranging the terms to form a standard quadratic equation: $x^2 - 8x - 1280 = 0$

Final Answer: The required quadratic equation is $x^2 - 8x - 1280 = 0$.

Solution: We need to find two numbers whose product is -10 and sum is -3. These numbers are -5 and 2. Splitting the middle term: $x^2 - 5x + 2x - 10 = 0$ $x(x - 5) + 2(x - 5) = 0$ $(x - 5)(x + 2) = 0$ So, either $x - 5 = 0$ or $x + 2 = 0$. If $x - 5 = 0$, then $x = 5$. If $x + 2 = 0$, then $x = -2$.

Final Answer: The roots are 5 and -2.

Solution: We need to find two numbers whose product is $2 \times (-6) = -12$ and sum is 1. These numbers are 4 and -3. Splitting the middle term: $2x^2 + 4x - 3x - 6 = 0$ $2x(x + 2) - 3(x + 2) = 0$ $(2x - 3)(x + 2) = 0$ So, either $2x - 3 = 0$ or $x + 2 = 0$. If $2x - 3 = 0$, then $x = \frac{3}{2}$. If $x + 2 = 0$, then $x = -2$.

Final Answer: The roots are $\frac{3}{2}$ and -2.

Solution: We need to find two numbers whose product is $\sqrt{2} \times 5\sqrt{2} = 10$ and sum is 7. These numbers are 5 and 2. Splitting the middle term: $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$ $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$ $(x + \sqrt{2})(\sqrt{2}x + 5) = 0$ So, either $x + \sqrt{2} = 0$ or $\sqrt{2}x + 5 = 0$. If $x + \sqrt{2} = 0$, then $x = -\sqrt{2}$. If $\sqrt{2}x + 5 = 0$, then $x = -\frac{5}{\sqrt{2}}$.

Final Answer: The roots are $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$.

Solution: To eliminate the fraction, multiply the entire equation by 8: $8(2x^2 - x + \frac{1}{8}) = 8(0)$ $16x^2 - 8x + 1 = 0$ We can write this as a perfect square: $(4x)^2 - 2(4x)(1) + (1)^2 = 0$ $(4x - 1)^2 = 0$ So, $4x - 1 = 0$. This gives $x = \frac{1}{4}$. Since the factor is repeated, the equation has two equal roots.

Final Answer: The roots are $\frac{1}{4}$ and $\frac{1}{4}$.

Solution: We can write this as a perfect square: $(10x)^2 - 2(10x)(1) + (1)^2 = 0$ $(10x - 1)^2 = 0$ So, $10x - 1 = 0$. This gives $x = \frac{1}{10}$. Since the factor is repeated, the equation has two equal roots.

Final Answer: The roots are $\frac{1}{10}$ and $\frac{1}{10}$.

Solution: To factorise the equation, we need to find two numbers whose product is $1 \times (-10) = -10$ and whose sum is $-3$. These numbers are $-5$ and $2$. Splitting the middle term: $x^2 - 5x + 2x - 10 = 0$ Factoring by grouping: $x(x - 5) + 2(x - 5) = 0$ $(x - 5)(x + 2) = 0$ For the product of two factors to be zero, at least one of them must be zero. So, either $x - 5 = 0$ or $x + 2 = 0$. If $x - 5 = 0$, then $x = 5$. If $x + 2 = 0$, then $x = -2$.

Final Answer: The roots of the equation are $5$ and $-2$.

Solution: To factorise the equation, we need to find two numbers whose product is $2 \times (-6) = -12$ and whose sum is $1$. These numbers are $4$ and $-3$. Splitting the middle term: $2x^2 + 4x - 3x - 6 = 0$ Factoring by grouping: $2x(x + 2) - 3(x + 2) = 0$ $(x + 2)(2x - 3) = 0$ So, either $x + 2 = 0$ or $2x - 3 = 0$. If $x + 2 = 0$, then $x = -2$. If $2x - 3 = 0$, then $2x = 3$, so $x = \frac{3}{2}$.

Final Answer: The roots of the equation are $-2$ and $\frac{3}{2}$.

Solution: To factorise the equation, we need to find two numbers whose product is $(\sqrt{2}) \times (5\sqrt{2}) = 5 \times 2 = 10$ and whose sum is $7$. These numbers are $5$ and $2$. Splitting the middle term: $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$ Factoring by grouping: $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$ $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$ So, either $\sqrt{2}x + 5 = 0$ or $x + \sqrt{2} = 0$. If $\sqrt{2}x + 5 = 0$, then $\sqrt{2}x = -5$, so $x = -\frac{5}{\sqrt{2}}$. If $x + \sqrt{2} = 0$, then $x = -\sqrt{2}$.

Final Answer: The roots of the equation are $-\frac{5}{\sqrt{2}}$ and $-\sqrt{2}$.

Solution: To eliminate the fraction, we can multiply the entire equation by 8: $8(2x^2 - x + \frac{1}{8}) = 8(0)$ $16x^2 - 8x + 1 = 0$ To factorise this equation, we need two numbers whose product is $16 \times 1 = 16$ and whose sum is $-8$. These numbers are $-4$ and $-4$. Splitting the middle term: $16x^2 - 4x - 4x + 1 = 0$ Factoring by grouping: $4x(4x - 1) - 1(4x - 1) = 0$ $(4x - 1)(4x - 1) = 0$ $(4x - 1)^2 = 0$ So, $4x - 1 = 0$, which gives $4x = 1$, so $x = \frac{1}{4}$. Since the factor is repeated, the equation has two equal roots.

Final Answer: The roots of the equation are $\frac{1}{4}$ and $\frac{1}{4}$.

Solution: This equation is in the form of a perfect square trinomial, $(a-b)^2 = a^2 - 2ab + b^2$. We can write the equation as: $(10x)^2 - 2(10x)(1) + (1)^2 = 0$ $(10x - 1)^2 = 0$ So, $10x - 1 = 0$, which gives $10x = 1$, so $x = \frac{1}{10}$. Since the factor is repeated, the equation has two equal roots.

Alternatively, by splitting the middle term: We need two numbers whose product is $100 \times 1 = 100$ and sum is $-20$. These numbers are $-10$ and $-10$. $100x^2 - 10x - 10x + 1 = 0$ $10x(10x - 1) - 1(10x - 1) = 0$ $(10x - 1)(10x - 1) = 0$ This gives $x = \frac{1}{10}$ (repeated root).

Final Answer: The roots of the equation are $\frac{1}{10}$ and $\frac{1}{10}$.

Given: From Example 1(i), the quadratic equation representing the situation is: $x^{2}-45 x+324=0$ where $x$ is the number of marbles John had.

To Find: The number of marbles they had to start with.

Solution: We need to solve the equation $x^2 - 45x + 324 = 0$ by factorisation. We need two numbers whose product is 324 and sum is -45. These numbers are -36 and -9. Splitting the middle term: $x^2 - 36x - 9x + 324 = 0$ $x(x - 36) - 9(x - 36) = 0$ $(x - 36)(x - 9) = 0$ So, either $x - 36 = 0$ or $x - 9 = 0$. This gives $x = 36$ or $x = 9$.

Case 1: If John had $x = 36$ marbles, then Jivanti had $45 - x = 45 - 36 = 9$ marbles. Case 2: If John had $x = 9$ marbles, then Jivanti had $45 - x = 45 - 9 = 36$ marbles.

Final Answer: The number of marbles they had to start with were 9 and 36.

Given: From Example 1(ii), the quadratic equation representing the situation is: $x^{2}-55 x+750=0$ where $x$ is the number of toys produced on that day.

To Find: The number of toys produced on that day.

Solution: We need to solve the equation $x^2 - 55x + 750 = 0$ by factorisation. We need two numbers whose product is 750 and sum is -55. These numbers are -30 and -25. Splitting the middle term: $x^2 - 30x - 25x + 750 = 0$ $x(x - 30) - 25(x - 30) = 0$ $(x - 30)(x - 25) = 0$ So, either $x - 30 = 0$ or $x - 25 = 0$. This gives $x = 30$ or $x = 25$.

Final Answer: The number of toys produced on that day was either 25 or 30.

The problems in Example 1 require finding the roots of the quadratic equations derived.

(i) From Example 1(i), the equation is $x^{2}-45 x+324=0$. This problem was about John and Jivanti's marbles, where $x$ is the number of marbles John had.

Solution: We need to solve the equation $x^2 - 45x + 324 = 0$ by factorisation. We need to find two numbers whose product is 324 and whose sum is $-45$. Let's find the factors of 324: $1, 2, 3, 4, 6, 9, 12, 27, 36, ...$ We can see that $-36 + (-9) = -45$ and $(-36) \times (-9) = 324$. So, the numbers are $-36$ and $-9$. Splitting the middle term: $x^2 - 36x - 9x + 324 = 0$ $x(x - 36) - 9(x - 36) = 0$ $(x - 36)(x - 9) = 0$ So, either $x - 36 = 0$ or $x - 9 = 0$. This gives $x = 36$ or $x = 9$.