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Maths 3 Marks Question

Quadratic Equations · CBSE STD 10 (CBSE - English Medium). 18 questions.

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Question 13 Marks
Find whether the following equations have real roots. If real roots exist, find them.
$5x^2 - 2x - 10 = 0.$
Answer
Main concept used:
For real roots of quadratic equation.
$ax^2 + bx + c = 0, b^2 - 4ac > 0$
The Given equation is $5x^2 - 2x - 10 = 0$
Discriminant $(D) = b^2 - 4ac$
$\Rightarrow D = (-2)^2 - 4(5) (-10) (a = 5, b = -2, c = -10)$
$= 4 + 200$
$\Rightarrow D = 204 > 0$
So the roots of the given equation are real and distinct.
Now, $\sqrt{\text{D}}=\sqrt{204}\Rightarrow\ \sqrt{\text{D}}=2\sqrt{51}$
And, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{+2\pm2\sqrt{51}}{2\times5}$
$=\frac{2[1\pm\sqrt{51}]}{10}=\frac{1\pm\sqrt{51}}{5}$
$\Rightarrow\ \text{x}_1=\frac{1+\sqrt{51}}{5}$ and $\text{x}_2=\frac{1-\sqrt{51}}{5}$
Hence, the roots of the given equation are $\frac{1+\sqrt{51}}{5},\frac{1-\sqrt{51}}{5}$.
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Question 23 Marks
Find a natural number whose square diminished by $84$ is equal to thrice of 8 more than the given number.
Answer
Let the required number be x.
According to the question,
$x^2 - 84 = 3 \times (x + 8)$
$\Rightarrow x^2 - 84 = 3x + 24$
$\Rightarrow x^2 - 3x - 84 - 24 = 0$
$\Rightarrow x^2 - 3x - 108 = 0$
$\Rightarrow x^2 - 12x + 9x - 108 = 0$
$\Rightarrow x(x - 12) +9(x - 12) = 0$
$\Rightarrow (x - 12)(x + 9) = 0$
$\Rightarrow x - 12 = 0 or x + 9 = 0$
$\Rightarrow x = 12 or x = -9$
$(x = -9$ is not a natural number so it is rejected.)
Hence, the required number is $12$.
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Question 33 Marks
Find the roots of the following quadratic equations by the factorisation method:
$3\sqrt{2}\text{x}^2-5\text{x}-\sqrt{2}=0.$
Answer
Given equation is $3\sqrt{2}\text{x}^2-5\text{x}-\sqrt{2}=0$ $3\sqrt{2}\text{x}^2-(6\text{x}-\text{x})-\sqrt{2}=0$ [by splitting the middle term] $3\sqrt{2}\text{x}^2-(6\text{x}+\text{x})-\sqrt{2}=0$ $3\sqrt{2}\text{x}^2-3\sqrt{2}.\sqrt{2}\text{x}+\text{x}-\sqrt{2}=0$ $\Rightarrow\ 3\sqrt{2}\text{x}(\text{x}-\sqrt{2})+1(\text{x}-\sqrt{2})=0$ $\Rightarrow\ (\text{x}-\sqrt{2})(3\sqrt{2}\text{x}+1)=0$ Now, $\text{x}-\sqrt{2}=0\Rightarrow\ \text{x}=\sqrt{2}$ and $3\sqrt{2}\text{x}+1=0$ $\Rightarrow\ \text{x}=-\frac{1}{3\sqrt{2}}=\frac{-\sqrt{2}}{6}$Hence, the roots of the equation $3\sqrt{2}\text{x}^2-5\text{x}-\sqrt{2}=0$ are $-\frac{\sqrt{2}}{6}$ and $\sqrt{2}$.
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Question 43 Marks
Find whether the following equations have real roots. If real roots exist, find them.
$8x^2 + 2x - 3 = 0.$
Answer
Main concept used:
For real roots of quadratic equation.
$ax^2 + bx + c = 0, b^2 - 4ac > 0$
The Given equation is $8x^2 + 2x - 3 = 0$
Discriminant $(D) = b^2 - 4ac$
$\Rightarrow D = (2)^2 - 4(8) (-3) (a = 8, b = 2, c = -3)$
$\Rightarrow D = 4 + 96 \Rightarrow D = 100$
As $D > 0$, so, roots are real.
Now, Discriminant $\sqrt{\text{D}}=10$
So, roots are $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{-2\pm10}{2\times8}=\frac{-2\pm10}{16}$
$\Rightarrow\ \text{x}_1=\frac{-2+10}{16}$ and $\text{x}_2=\frac{-2-10}{16}$
$\Rightarrow\ \text{x}_1=\frac{8}{16}$ and $\text{x}_2=\frac{-12}{16}$
$\Rightarrow\ \text{x}_1=\frac{1}{2}$ and $\text{x}_2=\frac{-3}{4}$
So, the roots of the given equation are $\frac{1}{2}$ and $\frac{-3}{4}$.
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Question 53 Marks
Find whether the following equations have real roots. If real roots exist, find them.
$-2x^2 + 3x + 2 = 0.$
Answer
Main concept used:
For real roots of quadratic equation.
$ax^2 + bx + c = 0, b^2 - 4ac > 0$
The Given equation is $-2x^2 + 3x + 2 = 0$
Discriminant $(D) = b^2 - 4ac$
$\Rightarrow D = (3)^2 - 4(-2) (2) (a = -2, b = 3, c = 2)$
$\Rightarrow D = 9 + 16$
$\Rightarrow D = 25 > 0$
So, the given equation has real and distinct roots.
Now, $\sqrt{\text{D}}=5$
And, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{-3\pm5}{2(-2)}=\frac{-3\pm5}{-4}$
$\Rightarrow\ \text{x}_1=\frac{-3+5}{-4}$ and $\text{x}_2=\frac{-3-5}{-4}$
$\Rightarrow\ \text{x}_1=\frac{2}{-4}$ and $\text{x}_2=\frac{-8}{-4}$
$\Rightarrow\ \text{x}_1=\frac{-1}{2}$ and $\text{x}_2=2$
Hence, the roots of the given equation are 2 and $\frac{-1}{2}$.
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Question 63 Marks
Find the roots of the following quadratic equations by the factorisation method:
$\frac{2}{5}\text{x}^2-\text{x}-\frac{3}{5}=0.$
Answer
Given equation is $\frac{2}{5}\text{x}^2-\text{x}-\frac{3}{5}=0$
On multiplying by 5 on both sides, we get
$\Rightarrow 2x^2 - 5x - 3 = 0$
$2x^2 - (6x - x) - 3 = 0$ [by splitting the middle term]
$\Rightarrow 2x^2 - 6x + x - 3 = 0$
$\Rightarrow 2x(x - 3) + 1(x - 3) = 0$
$\Rightarrow (x - 3)(2x + 1) = 0$
Now, $x - 3 = 0$
$ \Rightarrow x = 3$
and $2x + 1 = 0$
$\Rightarrow\ \text{x}=-\frac{1}{2}$
Hence, the roots of the equation $2x^2 - 5x - 3 = 0$ are $-\frac{1}{2}$ and 3.
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Question 73 Marks
Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Answer
Yes, a quadratic equation having coefficients as rational number, has irrational roots.
For example, $2x^2 - 3x - 15 = 0$ has rational coefficients.
$D = b^2 - 4ac (a = 2, b = -3, c = -15)$
$\Rightarrow D = 129$
$\therefore\ \text{Roots are given by x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$\Rightarrow\ \text{x}=\frac{-(-3)\pm\sqrt{129}}{2\times2}$
$\Rightarrow\ \text{x}=\frac{3\pm\sqrt{129}}{4}$
The roots are irrational as $\sqrt{129}$ is irrational.
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Question 83 Marks
Find the roots of the following quadratic equations by the factorisation method:
$3\text{x}^2+5\sqrt{5}\text{x}-10=0.$
Answer
Given equation is $3\text{x}^2+5\sqrt{5}\text{x}-10=0$
$3\text{x}^2+6\sqrt{5}\text{x}-\sqrt{5}\text{x}-2\sqrt{5}.\sqrt{5}=0$ [by splitting the middle term]
$\Rightarrow\ 3\text{x}^2+6\sqrt{5}\text{x}-\sqrt{5}\text{x}-10$
$3\text{x}^2+6\sqrt{5}\text{x}-\sqrt{5}\text{x}-2\sqrt{5}.\sqrt{5}=0$
$\Rightarrow\ 3\text{x}(\text{x}+2\sqrt{5})-\sqrt{5}(\text{x}+2\sqrt{5})=0$
$\Rightarrow\ (\text{x}+2\sqrt{5})(3\text{x}-\sqrt{5})=0$
Now, $\text{x}+2\sqrt{5}=0$
$\Rightarrow\ \text{x}=-2\sqrt{5}$ and $3\text{x}-\sqrt{5}=0$
$\Rightarrow\ \text{x}=\frac{\sqrt{5}}{3}$
Hence, the roots of the equation $3\text{x}^2+5\sqrt{5}\text{x}-10=0$ are $-2\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.
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Question 93 Marks
At t minutes past $2$ pm, the time needed by the minutes hand of a clock to show $3$ pm was found to be $3$ minutes less than $\frac{\text{t}^2}{4}$ minutes. Find t.
Answer
Total time taken by min hand from $2$ pm to 3pm is $60$ min. After t min past $2$ pm the time needed by min. Hand of a clock to show $3$ pm is given by $3$ min less than $\frac{\text{t}^2}{4}$ min.
$\therefore\ \text{t}+\Big(\frac{\text{t}^2}{4}-3\Big)=60$
$\Rightarrow 4t + t^2 - 12 = 240$
$\Rightarrow t^2 + 4t - 252 = 0$
$\Rightarrow t^2 + 18t - 14t - 252 = 0$
$\Rightarrow t(t + 18) - 14(t + 18) = 0$
$\Rightarrow (t + 18)(t - 14) = 0$
$\Rightarrow t + 18 = 0 or t - 14 = 0$
$\Rightarrow t = -18 or t = 14min.$
Being, negative value, t = -18 is rejected.
Hence, $t = 14$min.
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Question 103 Marks
Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$2x^2 - 3x - 5 = 0$
Answer
Given equation is $2x^2 - 3x - 5 = 0$
On comparing with $ax^2 + bx + c = 0$, we get
$a = 2, b = -3$ and $c = -5$
By quadratic formula, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(-5)}}{2(2)}=\frac{3\pm\sqrt{9+40}}{4}$
$=\frac{3\pm\sqrt{49}}{4}=\frac{3\pm7}{4}=\frac{10}{4},\frac{-4}{4}=\frac{5}{2},-1$
So, $\frac{5}{2}$ and -1 are the roots of the given equation.
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Question 113 Marks
Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$5x^2 + 13x + 8 = 0.$
Answer
Given equation is $5x^2 + 13x + 8 = 0$
On comparing with $ax^2 + bx + c = 0$, we get
$a = 5, b = 13$ and $c = 8$
By quadratic formula, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{-(13)\pm\sqrt{(13)^2-4(5)(8)}}{2(5)}$
$=\frac{-13\pm\sqrt{169-160}}{10}=\frac{-13\pm\sqrt{9}}{10}$
$=\frac{-13\pm3}{10}=-\frac{10}{10},-\frac{16}{10}=-1,-\frac{8}{5}.$
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Question 123 Marks
A quadratic equation with integral coefficient has integral roots. Justify your answer.
Answer
No, a quadratic equation with integral coefficient $(0,\pm1,\pm2,\pm3\ .....)$ can have its roots in fraction, i.e., non integral.
For example, $5x^2 + 3x - 8 = 0$ has integral coefficients (coefficients $5, 3, -8$) are integrs).
Now, $5x + 3x - 8 = 0$
$\Rightarrow 5x^2 + 8x - 5x - 8 = 0$
$\Rightarrow x(5x + 8) -1(5x + 8) = 0$
$\Rightarrow (5x + 8)(x - 1) = 0$
$\Rightarrow 5x + 8 = 0 of (x - 1) = 0$
Therefore, the roots are given by $\text{x}=\frac{-8}{5}$ and $x = 1$
So, the given statement is false.
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Question 133 Marks
A natural number, when increased by $12$, equals $160$ times its reciprocal. Find the number.
Answer
Let the required number be x.
According to the question,
$\text{x}+12=\frac{1}{\text{x}}\times160$
$\Rightarrow x^2 + 12x = 160$
$\Rightarrow x^2 + 12x - 160 = 0$
$\Rightarrow x^2 + 20x - 8x - 160 = 0$
$\Rightarrow x(x + 20) -8(x + 20) = 0$
$\Rightarrow (x + 20)(x - 8) = 0$
$\Rightarrow x + 20 = 0 or x - 8 = 0$
$\Rightarrow x = -20 or x = 8$
But, $x = -20$ is not a natural number.
Hence, the required number is 8.
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Question 143 Marks
Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$\text{x}^2-3\sqrt{5}\text{x}+10=0.$
Answer
Given equation is $\text{x}^2-3\sqrt{5}\text{x}+10=0$
On comparing with $ax^2 + bx + c = 0$, we get
$a = 1$, $\text{b}=-3\sqrt{5}$ and $c = 10$
By quadratic formula, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{-(-3\sqrt{5})\pm\sqrt{(-3\sqrt{5})^2-4(1)(10)}}{2(1)}$
$=\frac{3\sqrt{5}\pm\sqrt{45-40}}{2}=\frac{3\sqrt{5}\pm\sqrt{5}}{2}$
$=\frac{3\sqrt{5}+\sqrt{5}}{2},\frac{3\sqrt{5}-\sqrt{5}}{2}=2\sqrt{5},\sqrt{5}$
So, $2\sqrt{5}$ and $\sqrt{5}$ are the roots of the given equation.
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Question 153 Marks
Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$-3x^2 + 5x + 12 = 0.$
Answer
Given equation is $-3x^2 + 5x + 12 = 0$On comparing with $ax^2 + bx + c = 0$, we get
$a = -3, b = 5$ and $c = 12$
By quadratic formula, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{-(5)\pm\sqrt{(5)^2-4(-3)(12)}}{2(-3)}$
$=\frac{-5\pm\sqrt{25+144}}{-6}=\frac{-5\pm\sqrt{169}}{-6}$
$=\frac{-5\pm13}{-6}=\frac{8}{-6},\frac{-18}{-6}=-\frac{4}{3},3.$
So, $-\frac{4}{3}$ and 3 are two roots of the given equation.
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Question 163 Marks
Find whether the following equations have real roots. If real roots exist, find them.
$\text{x}^2+5\sqrt{5}\text{x}-70=0.$
Answer
Main concept used:
For real roots of quadratic equation.
$ax^2 + bx + c = 0, b^2 - 4ac > 0$
The Given equation is $\text{x}^2+5\sqrt{5}\text{x}-70=0$
$D = b^2 - 4ac$
$=(5\sqrt{5})^2-4(1)(-70)\ \big(\text{a}=1,\text{b}=5\sqrt{5},\text{c}=-70\big)$
$= 25 \times 5 + 280 = 125 + 280$
$\Rightarrow D = 405 > 0$
So, the roots of the given equation are real and distinct.
For roots $\sqrt{\text{D}}=\sqrt{405}\Rightarrow\ \sqrt{\text{D}}=\sqrt{9\times9\times5}$
$\Rightarrow\ \sqrt{\text{D}}=9\sqrt{5}$
Now, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}=\frac{-5\sqrt{5}\pm9\sqrt{5}}{2\times1}=\frac{(-5\pm9)\sqrt{5}}{2}$
$\Rightarrow\ \text{x}_1=\frac{(-5+9)\sqrt{5}}{2}$ and $\Rightarrow\ \text{x}_2=\frac{(-5-9)\sqrt{5}}{2}$
$=\frac{4\sqrt{5}}{2}$ and $=\frac{-14\sqrt{5}}{2}$
$=2\sqrt{5}$ and $=-7\sqrt{5}$
Hence, the roots of the given quadratic equation are $2\sqrt{5}$ and $-7\sqrt{5}$.
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Question 173 Marks
Find the roots of the quadratic equations by using the quadratic formula in each of the following:
$-x^2 + 7x - 10 = 0.$
Answer
Given equation is $-x^2 + 7x - 10 = 0$On comparing with $ax^2 + bx + c = 0$, we get
$a = -1, b = 7$ and $c = -10$
By quadratic formula, $\text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$=\frac{-(7)\pm\sqrt{(7)^2-4(-1)(-10)}}{2(-1)}=\frac{-7\pm\sqrt{49-40}}{-2}$
$=\frac{-7\pm\sqrt{9}}{-2}=\frac{-7\pm3}{-2}=\frac{-4}{-2},\frac{-10}{-2}=2,5$
So, 2 and 5 are two roots of the given equation.
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Question 183 Marks
Find the roots of the following quadratic equations by the factorisation method:
$2\text{x}^2+\frac{5}{3}\text{x}-2=0.$
Answer
Given equation is $2\text{x}^2+\frac{5}{3}\text{x}-2=0$On multiplying by 3 on both sides, we get
$6x^2 + 5x - 6 = 0$
$\Rightarrow 6x^2 + (9x - 4x) - 6 = 0$ [by splitting the middle term]
$\Rightarrow 6x^2 + 9x - 4x - 6 = 0$
$\Rightarrow 3x(2x + 3) -2(2x + 3) = 0$
$\Rightarrow (2x + 3)(3x - 2) = 0$
Now, $2x + 3 = 0$
$\Rightarrow\ \text{x}=-\frac{3}{2}$
and $3\text{x} - 2 = 0$
$\Rightarrow\ \text{x}=\frac{2}{3}$
Hence, the roots of the equation $6x^2 + 5x - 6 = 0$ are $\frac{-3}{2}$ and $\frac{2}{3}$.
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