Exercise 2.1 Class 10 for Effortless Understanding

Question 1(part i, part ii, part iii, part iv), Question 2(part i, part ii, part iii, part iv ),Question 3, Question 4(part i, part ii, part iii ), Question 5, Question 6, Question 7, Question 8(part i, part ii), Question 9, Question 10.

Exercise 2.1 class 10 explores the concept of determining the discriminant of quadratic equations and using it to analyze the nature of the roots. By calculating the discriminant, you’ll learn how to classify the roots as real and distinct, real and equal, or complex. This concept is essential for understanding the behavior of quadratic equations and forms a key part of solving them efficiently.

Question 1 of Exercise 2.1 Class 10. find the discriminant of the following quadratic equations.

1(i) 𝟮𝘅² + 𝟯𝘅 − 𝟭 = 𝟬

For the given equation 𝟮𝘅² + 𝟯𝘅 − 𝟭 = 𝟬, the coefficients are:
𝘢 = 𝟮, 𝘣 = 𝟯, 𝘤 = −𝟭

Δ = 𝘣² − 𝟰𝘢𝘤
Δ = 𝟯² − 𝟰(𝟮)(−𝟭)
Δ = 𝟵 + 𝟴
Δ = 𝟭𝟽

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1(ii) 𝟲𝘹² − 𝟴𝘹 + 𝟯 = 𝟬
Given:
𝘢 = 𝟲, 𝘣 = −𝟴, 𝘤 = 𝟯
Δ = 𝘣² − 𝟰𝘢𝘤
Δ = (−𝟴)² − 𝟰(𝟲)(𝟯)
Δ = 𝟲𝟰 − 𝟳𝟮
Δ = −𝟴

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1(iii) 𝟵𝘹² − 𝟯𝟬𝘹 + 𝟮𝟱 = 𝟬
In the equation 𝟵𝘹² − 𝟯𝟬𝘹 + 𝟮𝟱 = 𝟬
𝘢 = 𝟵, 𝘣 = −𝟯𝟬, 𝘤 = 𝟮𝟱

Δ = 𝘣² − 𝟰𝘢𝘤
Δ = (−𝟯𝟬)² − 𝟰(𝟵)(𝟮𝟱)
Δ = 𝟵𝟬𝟬 − 𝟵𝟬𝟬
Δ = 𝟬

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1(iv) 𝟰𝘹² − 𝟳𝘹 − 𝟮 = 𝟬
In the equation 𝟰𝘹² − 𝟳𝘹 − 𝟮 = 𝟬
𝘢 = 𝟰, 𝘣 = −𝟳, 𝘤 = −𝟮

Δ = 𝘣² − 𝟰𝘢𝘤
Δ = (−𝟳)² − 𝟰(𝟰)(−𝟮)
Δ = 𝟰𝟵 + 𝟯𝟮
Δ = 𝟴𝟭

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Question 2 of Exercise 2.1 Class 10.
Find the nature of the roots of the following given quadratic equations and verify the result by solving the equations.

2(i) 𝘹² − 𝟮𝟯𝘹 + 𝟭𝟮𝟬 = 𝟬
For the quadratic equation 𝘢𝘹² + 𝘣𝘹 + 𝘤 = 𝟬, the discriminant Δ is given by:

𝘢 = 𝟭, 𝘣 = −𝟮𝟯, 𝘤 = 𝟭𝟮𝟬

Δ = 𝘣² − 𝟰𝘢𝘤
Δ = (−𝟮𝟯)² − 𝟰(𝟭)(𝟭𝟮𝟬)
Δ = 𝟱𝟮𝟵 − 𝟰𝟴𝟬
Δ = 𝟰𝟵

Since Δ = 𝟰𝟵, which is positive and a perfect square,
the equation has two distinct real roots.

𝘹 = (−𝘣 ± √Δ) / 𝟮𝘢
𝘹 = (−(−𝟮𝟯) ± √𝟰𝟵) / 𝟮(𝟭)
𝘹 = (𝟮𝟯 ± 𝟳) / 𝟮

𝘹 = (𝟮𝟯 + 𝟳) / 𝟮 = 𝟭𝟱
𝘹 = (𝟮𝟯 − 𝟳) / 𝟮 = 𝟴

𝐒𝐨, 𝐭𝐡𝐞 𝐫𝐨𝐨𝐭𝐬 𝐚𝐫𝐞 𝐫𝐞𝐚𝐥, 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐚𝐧𝐝 𝐮𝐧𝐞𝐪𝐮𝐚𝐥.

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2(ii) 𝟮𝘹² + 𝟯𝘹 + 𝟳 = 𝟬

Here, $a=2$,
$b=3$, and
$c=7$.

$∆=b^2–4ac$

$∆=\left(3\right)2^{}–4\left(2\right)\left(7\right)$

$∆=9–56$

$∆=–47$

Since $∆=–47$, the equation has two complex conjugate roots.

$x=\frac{–b\pm \sqrt{∆}}{2a}$

$x=\frac{–3\pm \sqrt{–47}}{2\left(2\right)}$

$x=\frac{–3\pm i\sqrt{47}}{4}$

So, the roots are imaginary:

$x=\frac{–3\pm i\sqrt{47}}{4}$

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2(iii) 𝟭𝟲𝘹² − 𝟮𝟰𝘹 + 𝟵 = 𝟬

Here, $a=16$,
$b=–24$, and
$c=9$.

$∆=b^2–4ac$

$∆=(–24)2^{}–4\left(16\right)\left(9\right)$

$∆=576–576$

$∆=0$

Since $∆=0$, the equation has one real repeated root.

$x=\frac{–b\pm \sqrt{∆}}{2a}$

$x=\frac{–(–24)\pm \sqrt{0}}{2\left(16\right)}$

$x=\frac{24}{32}$

$x=\frac{3}{4}$

So, the roots are rational (real) and equal:

$x=\frac{3}{4}$

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2(iv) 𝟯𝘹² + 𝟳𝘹 − 𝟭𝟯 = 𝟬

Given:
$a=3$,
$b=7$,
$c=–13$

$$∆=b^2–4ac$$

$$∆=\left(7\right)2^{}–4\left(3\right)(–13)$$

$$∆=49+156$$

$$∆=205$$

Since
$∆=205$, which is positive but not a perfect square, the equation has two distinct real irrational roots.

$$x=\frac{–b\pm \sqrt{∆}}{2a}$$

$$x=\frac{–7\pm \sqrt{205}}{2\left(3\right)}$$

$$x=\frac{–7\pm \sqrt{205}}{6}$$

$$x_1=\frac{–7+\sqrt{205}}{6}$$

$$x_2=\frac{–7–\sqrt{205}}{6}$$

Hence, the equation has two distinct real irrational roots.

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Question 3 Exercise 2.1 Class 10.
𝐅𝐨𝐫 𝐰𝐡𝐚𝐭 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐤, 𝐭𝐡𝐞 𝐞𝐱𝐩𝐫𝐞𝐬𝐬𝐢𝐨𝐧 𝐤² + 𝟐(𝐤 + 𝟏)𝐱 + 𝐱² 𝐢𝐬 𝐚 𝐩𝐞𝐫𝐟𝐞𝐜𝐭 𝐬𝐪𝐮𝐚𝐫𝐞?

Solution:

Given:

$$k^2+2(k+1)x+x^2$$

Comparing with:

$$a{x}^2+bx+c=0$$

We calculate the discriminant:

$$b^2–4ac=\left[2\right(k+1\left)2^{}\right]–4\left(k^2\right)\left(1\right)$$

$$=4(k+1)2^{}–4k^2$$

$$=4[k^2+2k+1]–4k^2$$

$$=4k^2+8k+4–4k^2$$

As given that this expression is a perfect square, the discriminant should be zero:

$$b^2–4ac=0$$

Dividing by:

$–4$

$$3k^2–2k–1=0$$

Simplifying:

$$3k^2–2k–1=0$$

For what value of:

$k$

will make it zero?

$$(k–1)(3k+1)=0$$

or:

$$k=1$$
$$k=–\frac{1}{3}$$

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Question 4 of Exercise 2.1 Class 10.
Find the value of k if the roots of the following equations are equal.

4(i) (𝟐𝐤−𝟏)𝐱² + 𝟑𝐤𝐱 + 𝟑 = 𝟎

Given equation:

$$(2k–1)x^2+3kx+3=0$$

Coefficients:

$a=(2k–1)$,
$b=3k$,
$c=3$

Discriminant:

$$∆=b^2–4ac$$

$$∆={\left(}^3–4(2k–1)\left(3\right)$$

$$∆=9k^2–12(2k–1)$$

$$∆=9k^2–24k+12$$

Set the discriminant to zero:

$$9k^2–24k+12=0$$

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4(ii) \(\boldsymbol{x^2 + 2(k+2)x +(3k+4)=0}\)

\begin{align*}
\text{Given equation:} & \quad x^2 + 2(k+2)x + (3k+4) = 0 \\
\text{Coefficients:} & \quad a = 1, \quad b = 2(k+2), \quad c = 3k+4 \\
\text{Discriminant:} & \quad \Delta = b^2 – 4ac \\
\Delta &= [2(k+2)]^2 – 4(3k+4) \\
&= 4(k+2)^2 – 4(3k+4) \\
&= 4(k^2 + 4k + 4) – 4(3k + 4) \\
&= 4k^2 + 16k + 16 – 12k – 16 \\
&= 4k^2 + 4k \\
\text{Set the discriminant to zero:} & \quad 4k^2 + 4k = 0 \\
\text{Factorize:} & \quad 4k(k + 1) = 0 \\
\text{Solve for } k: & \\
4k &= 0 \quad \Rightarrow \quad k = 0 \\
k + 1 &= 0 \quad \Rightarrow \quad k = -1 \\
\text{Values of } k: & \quad k = 0 \quad \text{or} \quad k = -1
\end{align*}

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4(iii) \(\boldsymbol{(3k+2)x^2 – 5(k+1)x + (2k+3) = 0}\)

Find the value of \( k \) if the roots of the following equation are equal:

\[ (3k+2)x^2 – 5(k+1)x + (2k+3) = 0 \]

Given the equation \( (3k+2)x^2 – 5(k+1)x + (2k+3) = 0 \), we identify the coefficients as:
\[ a = 3k + 2, b = -5(k + 1), c = 2k + 3 \]

Now, we calculate the discriminant:

\[ \Delta = b^2 – 4ac \]

Substitute the values of \( a \), \( b \), and \( c \):

\begin{align*}
\Delta &= [-5(k + 1)]^2 – 4(3k + 2)(2k + 3) \\
&= 25(k + 1)^2 – 4(3k + 2)(2k + 3) \\
&= 25(k^2 + 2k + 1) – 4(6k^2 + 13k + 6) \\
&= 25k^2 + 50k + 25 – 24k^2 – 52k – 24 \\
&= k^2 – 2k + 1
\end{align*}

Set the discriminant to zero: \( k^2 – 2k + 1 = 0 \)

Factorize the quadratic equation:\( (k – 1)^2 = 0 \)

Solve for \( k \):

\[ k – 1 = 0 \quad \Rightarrow \quad k = 1 \]

Therefore, the value of \( k \) for which the roots of the equation are equal is:

\[ k = 1 \]

Question 5 of Exercise 2.1 Class 10.
Show that the equation \(\boldsymbol{x^2 + (mx+c)^2 =a^2}\) has equal roots if \(\boldsymbol{c^2 = a^2 (1 + m^2)}\).

\(\text{Expand } (mx + c)^2:\)
\begin{align*}
x^2 + m^2x^2 + 2mxc + c^2 &= a^2
\end{align*}

\(\text{Combine like terms:}\)
\begin{align*}
(1 + m^2)x^2 + 2mxc + c^2 &= a^2
\end{align*}

\(\text{Rearrange to standard quadratic form:}\)
\begin{align*}
(1 + m^2)x^2 + 2mxc + (c^2 – a^2) &= 0
\end{align*}

\(\text{For the roots to be equal, the discriminant must be zero:}\)
\begin{align*}
\Delta &= b^2 – 4ac
\end{align*}

\(\text{Here, } a = 1 + m^2, \, b = 2mc, \, \text{and } c = c^2 – a^2.\)

\(\text{Calculate the discriminant:}\)
\begin{align*}
\Delta &= (2mc)^2 – 4(1 + m^2)(c^2 – a^2) \\
&= 4m^2c^2 – 4(1 + m^2)(c^2 – a^2) \\
&= 4m^2c^2 – 4(c^2 + m^2c^2 – a^2 – m^2a^2) \\
&= 4m^2c^2 – 4c^2 – 4m^2c^2 + 4a^2 + 4m^2a^2 \\
&= -4c^2 + 4a^2 + 4m^2a^2 \\
&= 4(a^2 + m^2a^2 – c^2)
\end{align*}

\(\text{Set the discriminant to zero:}\)
\begin{align*}
4(a^2 + m^2a^2 – c^2) &= 0 \\
a^2 + m^2a^2 – c^2 &= 0 \\
c^2 &= a^2(1 + m^2)
\end{align*}

\(\text{Therefore, the equation } x^2 + (mx + c)^2 = a^2 \text{ has equal roots if } c^2 = a^2 (1 + m^2).\)

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Question 6 of Exercise 2.1 Class 10.
Find the condition that the roots of the equation \(\boldsymbol{ (mx + c)^2 – 4ax = 0} \) are equal.

Given the equation:
\[ (mx + c)^2 – 4ax = 0 \]

Expand \( (mx + c)^2 \):
\begin{align*}
(m^2x^2 + 2mxc + c^2) – 4ax &= 0 \\
m^2x^2 + 2mxc + c^2 – 4ax &= 0
\end{align*}

Rearrange to standard quadratic form:
\[ m^2x^2 + (2mc – 4a)x + c^2 = 0 \]

For the roots to be equal, the discriminant must be zero:
\[ \Delta = b^2 – 4ac \]

Here, \( a = m^2 \), \( b = 2mc – 4a \), and \( c = c^2 \).

Calculate the discriminant:
\begin{align*}
\Delta &= (2mc – 4a)^2 – 4(m^2)(c^2) \\
&= (2mc – 4a)^2 – 4m^2c^2 \\
&= 4m^2c^2 – 16mac + 16a^2 – 4m^2c^2 \\
&= 16a^2 – 16mac
\end{align*}

Set the discriminant to zero:
\[ 16a^2 – 16mac = 0 \]
\[ 16a(a – mc) = 0 \]

Therefore, the condition for the roots to be equal is:
\[ a = mc \]

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Question 7 of Exercise 2.1 Class 10. if the roots of the equation \(\boldsymbol{(c^2 -ab)x^2 -2(a^2 -bc)x + (b^2 – ac)=0}\) are equal then a=0 or \(\boldsymbol {a^3 +b^3 +c^3 =3abc}\).

Given the equation:
\[ (c^2 – ab)x^2 – 2(a^2 – bc)x + (b^2 – ac) = 0 \]

For the roots to be equal, the discriminant must be zero:
\[ \Delta = b^2 – 4ac \]

Here, \( a = c^2 – ab \), \( b = -2(a^2 – bc) \), and \( c = b^2 – ac \).

Calculate the discriminant:
\begin{align*}
\Delta &= [-2(a^2 – bc)]^2 – 4(c^2 – ab)(b^2 – ac) \\
&= 4(a^2 – bc)^2 – 4(c^2 – ab)(b^2 – ac) \\
&= 4(a^4 – 2a^2bc + b^2c^2) – 4(c^2b^2 – ac^3 – ab^3 + a^2bc) \\
&= 4a^4 – 8a^2bc + 4b^2c^2 – 4c^2b^2 + 4ac^3 + 4ab^3 – 4a^2bc \\
&= 4a^4 – 12a^2bc + 4ac^3 + 4ab^3
\end{align*}

Set the discriminant to zero:
\begin{align}
4a^4 – 12a^2bc + 4ac^3 + 4ab^3 &= 0 \\
a^4 – 3a^2bc + ac^3 + ab^3 &= 0 \\
a(a^3 – 3abc + c^3 + b^3) &= 0
\end{align}

Therefore, the condition for the roots to be equal is:
\[ a = 0 \quad \text{or} \quad a^3 + b^3 + c^3 = 3abc \]

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Question 8 of Exercise 2.1 Class 10. Show that the roots of the following equations are rational.

8(i) \(\boldsymbol{a(b-c)x^2 +b(c-a)x +c(a-b)=0}\)

Given the equation:
\[ a(b-c)x^2 + b(c-a)x + c(a-b) = 0 \]

For the roots to be rational, the discriminant must be a perfect square. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \Delta = b^2 – 4ac \]

Here, \( A = a(b-c) \), \( B = b(c-a) \), and \( C = c(a-b) \).

Calculate the discriminant:
\begin{align*}
\Delta &= [b(c-a)]^2 – 4[a(b-c)][c(a-b)] \\
&= b^2(c-a)^2 – 4a(b-c)c(a-b) \\
&= b^2(c^2 – 2ac + a^2) – 4ac(b^2 – ab – bc + ac) \\
&= b^2c^2 – 2ab^2c + b^2a^2 – 4acb^2 + 4a^2bc + 4a^2c^2 \\
&= b^2c^2 – 2ab^2c + b^2a^2 – 4acb^2 + 4a^2bc + 4a^2c^2 \\
&= (b^2c^2 + b^2a^2 – 2ab^2c) + (4a^2c^2 + 4a^2bc – 4acb^2) \\
&= b^2(c^2 + a^2 – 2ac) + 4a^2(c^2 + bc – bc) \\
&= b^2(c – a)^2 + 4a^2c^2
\end{align*}

Since \( \Delta \) is a sum of squares, it is always non-negative. For the roots to be rational, \( \Delta \) must be a perfect square. Given that \( b^2(c – a)^2 \) and \( 4a^2c^2 \) are both perfect squares, their sum is also a perfect square.

Therefore, the roots of the equation \( a(b-c)x^2 + b(c-a)x + c(a-b) = 0 \) are rational.

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8(ii) \(\boldsymbol{(a+2b)x^2 + 2(a+b+c)x + (a+2c) = 0} \)

Given the equation:
\[ (a+2b)x^2 + 2(a+b+c)x + (a+2c) = 0 \]

For the roots to be rational, the discriminant must be a perfect square. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \Delta = b^2 – 4ac \]

Here, \( a = a+2b \), \( b = 2(a+b+c) \), and \( c = a+2c \).

Calculate the discriminant:
\begin{align*}
\Delta &= [2(a+b+c)]^2 – 4(a+2b)(a+2c) \\
&= 4(a+b+c)^2 – 4(a+2b)(a+2c) \\
&= 4(a^2 + 2ab + 2ac + b^2 + 2bc + c^2) – 4(a^2 + 2ac + 2ab + 4bc) \\
&= 4a^2 + 8ab + 8ac + 4b^2 + 8bc + 4c^2 – 4a^2 – 8ac – 8ab – 16bc \\
&= 4b^2 + 4c^2 – 8bc \\
&= 4(b^2 – 2bc + c^2) \\
&= 4(b – c)^2
\end{align*}

Since \(\Delta = 4(b – c)^2\) is a perfect square, the roots of the equation are rational.

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Question 9 of Exercise 2.1 Class 10. for all values of k prove that root of the equation \(\boldsymbol{x^2 – 2(k+(1/k))x+3=0, k \neq 0}\) are real.

Given the equation:
\[ x^2 – 2\left(k + \frac{1}{k}\right)x + 3 = 0 \]

For the roots to be real, the discriminant must be non-negative. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \Delta = b^2 – 4ac \]

Here, \( a = 1 \), \( b = -2\left(k + \frac{1}{k}\right) \), and \( c = 3 \).

Calculate the discriminant:
\begin{align*}
\Delta &= \left[-2\left(k + \frac{1}{k}\right)\right]^2 – 4(1)(3) \\
&= 4\left(k + \frac{1}{k}\right)^2 – 12 \\
&= 4\left(k^2 + 2 \cdot k \cdot \frac{1}{k} + \frac{1}{k^2}\right) – 12 \\
&= 4\left(k^2 + 2 + \frac{1}{k^2}\right) – 12 \\
&= 4k^2 + 8 + \frac{4}{k^2} – 12 \\
&= 4k^2 + \frac{4}{k^2} – 4 \\
&= 4\left(k^2 + \frac{1}{k^2} – 1\right)
\end{align*}

Since \( k \neq 0 \), \( k^2 + \frac{1}{k^2} \geq 2 \) by the AM-GM inequality. Therefore:
\begin{align*}
k^2 + \frac{1}{k^2} – 1 &\geq 2 – 1 \\
&= 1
\end{align*}

Thus, the discriminant \(\Delta\) is:
\begin{align*}
\Delta &= 4\left(k^2 + \frac{1}{k^2} – 1\right) \\
&\geq 4 \cdot 1 \\
&= 4
\end{align*}

Since \(\Delta \geq 4\), the discriminant is always non-negative for all \( k \neq 0 \). Therefore, the roots of the equation \( x^2 – 2\left(k + \frac{1}{k}\right)x + 3 = 0 \) are real for all values of \( k \neq 0 \).

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Question 10 of Exercise 2.1 Class 10. Show that the roots of the equation \(\boldsymbol{ (b-c)x^2 + (c-a)x + (a-b) = 0} \) are real.

Given the equation:
\[ (b-c)x^2 + (c-a)x + (a-b) = 0 \]

For the roots to be real, the discriminant must be non-negative. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \Delta = b^2 – 4ac \]

Here, \( a = b-c \), \( b = c-a \), and \( c = a-b \).

Calculate the discriminant:
\begin{align*}
\Delta &= (c-a)^2 – 4(b-c)(a-b) \\
&= (c-a)^2 – 4[(b-c)(a-b)] \\
&= (c-a)^2 – 4[ba – b^2 – ca + bc] \\
&= (c-a)^2 – 4[ba – b^2 – ca + bc] \\
&= c^2 – 2ac + a^2 – 4ab + 4b^2 + 4ac – 4bc \\
&= a^2 + b^2 + c^2 – 2ab – 2bc + 2ac \\
&= (a – b)^2 + (b – c)^2 + (c – a)^2
\end{align*}

Since \(\Delta = (a – b)^2 + (b – c)^2 + (c – a)^2\) is a sum of squares, it is always non-negative.

Therefore, the roots of the equation \( (b-c)x^2 + (c-a)x + (a-b) = 0 \) are real.

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Exercise 2.2 solutions will take you further in in this unit.

Here is Video lecture of exercise 2.1 Class 10.

Explore a detailed guide on the discriminant of quadratic equations with Cuemath, where you’ll learn how to calculate and interpret it to classify roots effectively.