Exercise 5C
Q1 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q1 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 1
$2x^2 + x – 3 = 0$
Sol :
Given that $2x^2 + x – 3 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 2, b = 1, c = -3
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-1 \pm \sqrt{(1)^{2}-4(2)(-3)}}{2(2)}=\frac{-1 \pm \sqrt{1+24}}{4}=\frac{-1 \pm \sqrt{25}}{4}=\frac{-1 \pm 5}{4}=\frac{-1+5}{4}$ or $\frac{-1-5}{4}=\frac{4}{4}$ or $\frac{-6}{4}=1$ or $\frac{-3}{2}$
Thus, the roots of the given equation are x = 1 or $\frac{-3}{2}$.
Sol :
Given that $6x^2 + 7x – 20 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 6, b = 7, c = -20
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-7 \pm \sqrt{(7)^{2}-4(6)(-20)}}{2(6)}$
$=\frac{-7 \pm \sqrt{49+480}}{12}=\frac{-7 \pm \sqrt{529}}{12}=\frac{-7 \pm 23}{12}=\frac{-7+23}{12}$ or $\frac{-7-23}{12}=\frac{16}{12}$ or $-\frac{30}{12}=\frac{4}{3}$ or $-\frac{5}{2}$
Thus, the roots of the given equation are $x = \frac{4}{3}$ or $\frac{-5}{2}$.
Sol :
Given that $9x^2 + 6x = 35$
Sol :
Given that $2x^2 + x – 3 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 2, b = 1, c = -3
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-1 \pm \sqrt{(1)^{2}-4(2)(-3)}}{2(2)}=\frac{-1 \pm \sqrt{1+24}}{4}=\frac{-1 \pm \sqrt{25}}{4}=\frac{-1 \pm 5}{4}=\frac{-1+5}{4}$ or $\frac{-1-5}{4}=\frac{4}{4}$ or $\frac{-6}{4}=1$ or $\frac{-3}{2}$
Thus, the roots of the given equation are x = 1 or $\frac{-3}{2}$.
Q2 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q2 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 2
$6x^2 + 7x – 20 = 0$Sol :
Given that $6x^2 + 7x – 20 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 6, b = 7, c = -20
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-7 \pm \sqrt{(7)^{2}-4(6)(-20)}}{2(6)}$
$=\frac{-7 \pm \sqrt{49+480}}{12}=\frac{-7 \pm \sqrt{529}}{12}=\frac{-7 \pm 23}{12}=\frac{-7+23}{12}$ or $\frac{-7-23}{12}=\frac{16}{12}$ or $-\frac{30}{12}=\frac{4}{3}$ or $-\frac{5}{2}$
Thus, the roots of the given equation are $x = \frac{4}{3}$ or $\frac{-5}{2}$.
Q3 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q3 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 3
$9x^2 + 6x = 35$Sol :
Given that $9x^2 + 6x = 35$
⇒ $9x^2 + 6x – 35 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 9, b = 6, c = -35
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-6 \pm \sqrt{(6)^{2}-4(9)(-35)}}{2(9)}$
$=\frac{-6 \pm \sqrt{36+1260}}{18}=\frac{-6 \pm \sqrt{1296}}{18}=\frac{-6 \pm 36}{18}=\frac{-6+36}{18}$ or $\frac{-6-36}{18}=\frac{30}{18}$ or $-\frac{42}{18}=\frac{5}{3}$ or $-\frac{7}{3}$
Thus, the roots of the given equation are $x = \frac{5}{3}$ or $\frac{-7}{3}$.
Sol :
Given that $3x^2 + 7x – 6 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 3, b = 7, c = -6
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-7 \pm \sqrt{(7)^{2}-4(3)(-6)}}{2(3)}$
$=\frac{-7 \pm \sqrt{49+72}}{6}=\frac{-7 \pm \sqrt{121}}{6}=\frac{-7 \pm 11}{6}=\frac{-7+11}{6}$ or $\frac{-7-11}{6}=\frac{4}{6}$ or $-\frac{18}{6}=\frac{2}{3}$ or -3
Thus, the roots of the given equation are $x = \frac{2}{3}$ or -3
Sol :
Given that $x^2 – 66x + 189 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 1, b = -66, c = 189
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-66) \pm \sqrt{(-66)^{2}-4(1)(189)}}{2(1)}$
$=\frac{66 \pm \sqrt{4356-756}}{2}=\frac{66 \pm \sqrt{3600}}{2}=\frac{66 \pm 60}{2}=\frac{66+60}{2}$ or $\frac{66-60}{2}=\frac{126}{2}$ or $\frac{6}{2}=63$ or 3
Thus, the roots of the given equation are x = 63 or 3.
Sol :
Given that $√3x^2 + 11x + 6√3 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = √3, b = 11, c = 6√3
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-11 \pm \sqrt{(11)^{2}-4(\sqrt{3})(6 \sqrt{3})}}{2(\sqrt{3})}$
$=\frac{-11 \pm \sqrt{121-72}}{2 \sqrt{3}}=\frac{-11 \pm \sqrt{49}}{2 \sqrt{3}}=\frac{-11 \pm 7}{2 \sqrt{3}}=\frac{-11+7}{2 \sqrt{3}}$ or $\frac{-11-7}{2 \sqrt{3}}=\frac{-4}{2 \sqrt{3}}$ or $\frac{-18}{2 \sqrt{3}}=-\frac{2 \sqrt{3}}{3}$ or $-3 \sqrt{3}$
Thus, the roots of the given equation are $x = \frac{-2√3}{3}$ or -3√3.
Sol :
Given that $36x^2 + 23 = 60x$ ⇒ $36x^2 – 60x + 23 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 36, b = -60, c = 23
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-60) \pm \sqrt{(-60)^{2}-4(36)(23)}}{2(36)}$
$=\frac{60 \pm \sqrt{3600-3312}}{72}=\frac{60 \pm \sqrt{288}}{72}=\frac{60 \pm 12 \sqrt{2}}{72}=\frac{5 \pm \sqrt{2}}{6}=\frac{5+\sqrt{2}}{6}$ or $\frac{5-\sqrt{2}}{6}$
Thus, the roots of the given equation are x = (5 + √2)/6 or (5 – √2)/6.
Sol :
Given that $x^2 – 2x + 5 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 1, b = -2, c = 5
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4(1)(5)}}{2(1)}$
$=\frac{2 \pm \sqrt{4-20}}{2}=\frac{2 \pm \sqrt{-16}}{2}=\frac{2 \pm 4 \sqrt{-1}}{2}=1 \pm 2 \sqrt{-1}=1+2 \sqrt{-1}$ or $1-2 \sqrt{-1}$
Thus, the roots of the given equation are x = (1 + 2√(-1)) or (1 – 2√(-1)).
Sol :
Given that $3x^2 – 17x + 25 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 3, b = -17, c = 25
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-17) \pm \sqrt{(-17)^{2}-4(3)(25)}}{2(3)}=\frac{17 \pm \sqrt{-11}}{6}=\frac{17+\sqrt{-11}}{6}$ or $\frac{17-\sqrt{-11}}{6}$
Thus, the roots of the given equation are $x = \frac{(17 + √(-11))}{6}$ or \frac{(17 – √(-11))}{6}$.
Sol :
Given that $15x^2 – 28 = x$
⇒ $15x^2 – x – 28 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 15, b = -1, c = -28
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(15)(-28)}}{2(15)}$
$=\frac{1 \pm \sqrt{1+1680}}{30}=\frac{1 \pm \sqrt{1681}}{30}=\frac{1 \pm 41}{30}=\frac{1+41}{30}$ or $\frac{1-41}{30}=\frac{42}{30}$ or $-\frac{40}{30}=\frac{7}{5}$ or $-\frac{4}{3}$
Thus, the roots of the given equation are $x = \frac{7}{5}$ or $\frac{-4}{3}$
Sol :
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 9, b = 6, c = -35
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-6 \pm \sqrt{(6)^{2}-4(9)(-35)}}{2(9)}$
$=\frac{-6 \pm \sqrt{36+1260}}{18}=\frac{-6 \pm \sqrt{1296}}{18}=\frac{-6 \pm 36}{18}=\frac{-6+36}{18}$ or $\frac{-6-36}{18}=\frac{30}{18}$ or $-\frac{42}{18}=\frac{5}{3}$ or $-\frac{7}{3}$
Thus, the roots of the given equation are $x = \frac{5}{3}$ or $\frac{-7}{3}$.
Q4 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q4 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 4
$3x^2 + 7x – 6 = 0$Sol :
Given that $3x^2 + 7x – 6 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 3, b = 7, c = -6
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}}{2a}$.
$x=\frac{-7 \pm \sqrt{(7)^{2}-4(3)(-6)}}{2(3)}$
$=\frac{-7 \pm \sqrt{49+72}}{6}=\frac{-7 \pm \sqrt{121}}{6}=\frac{-7 \pm 11}{6}=\frac{-7+11}{6}$ or $\frac{-7-11}{6}=\frac{4}{6}$ or $-\frac{18}{6}=\frac{2}{3}$ or -3
Thus, the roots of the given equation are $x = \frac{2}{3}$ or -3
Q5 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 5
$x^2 – 66x + 189 = 0$Sol :
Given that $x^2 – 66x + 189 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 1, b = -66, c = 189
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-66) \pm \sqrt{(-66)^{2}-4(1)(189)}}{2(1)}$
$=\frac{66 \pm \sqrt{4356-756}}{2}=\frac{66 \pm \sqrt{3600}}{2}=\frac{66 \pm 60}{2}=\frac{66+60}{2}$ or $\frac{66-60}{2}=\frac{126}{2}$ or $\frac{6}{2}=63$ or 3
Thus, the roots of the given equation are x = 63 or 3.
Q6 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q6 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 6
$√3x^2 + 11x + 6√3 = 0$Sol :
Given that $√3x^2 + 11x + 6√3 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = √3, b = 11, c = 6√3
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-11 \pm \sqrt{(11)^{2}-4(\sqrt{3})(6 \sqrt{3})}}{2(\sqrt{3})}$
$=\frac{-11 \pm \sqrt{121-72}}{2 \sqrt{3}}=\frac{-11 \pm \sqrt{49}}{2 \sqrt{3}}=\frac{-11 \pm 7}{2 \sqrt{3}}=\frac{-11+7}{2 \sqrt{3}}$ or $\frac{-11-7}{2 \sqrt{3}}=\frac{-4}{2 \sqrt{3}}$ or $\frac{-18}{2 \sqrt{3}}=-\frac{2 \sqrt{3}}{3}$ or $-3 \sqrt{3}$
Thus, the roots of the given equation are $x = \frac{-2√3}{3}$ or -3√3.
Q7 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q7 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 7
$36x^2 + 23 = 60x$Sol :
Given that $36x^2 + 23 = 60x$ ⇒ $36x^2 – 60x + 23 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 36, b = -60, c = 23
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-60) \pm \sqrt{(-60)^{2}-4(36)(23)}}{2(36)}$
$=\frac{60 \pm \sqrt{3600-3312}}{72}=\frac{60 \pm \sqrt{288}}{72}=\frac{60 \pm 12 \sqrt{2}}{72}=\frac{5 \pm \sqrt{2}}{6}=\frac{5+\sqrt{2}}{6}$ or $\frac{5-\sqrt{2}}{6}$
Thus, the roots of the given equation are x = (5 + √2)/6 or (5 – √2)/6.
Q8 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 8
$x^2 – 2x + 5 = 0$Sol :
Given that $x^2 – 2x + 5 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 1, b = -2, c = 5
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4(1)(5)}}{2(1)}$
$=\frac{2 \pm \sqrt{4-20}}{2}=\frac{2 \pm \sqrt{-16}}{2}=\frac{2 \pm 4 \sqrt{-1}}{2}=1 \pm 2 \sqrt{-1}=1+2 \sqrt{-1}$ or $1-2 \sqrt{-1}$
Thus, the roots of the given equation are x = (1 + 2√(-1)) or (1 – 2√(-1)).
Q9 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q9 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 9
$3x^2 – 17x + 25 = 0$Sol :
Given that $3x^2 – 17x + 25 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 3, b = -17, c = 25
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-17) \pm \sqrt{(-17)^{2}-4(3)(25)}}{2(3)}=\frac{17 \pm \sqrt{-11}}{6}=\frac{17+\sqrt{-11}}{6}$ or $\frac{17-\sqrt{-11}}{6}$
Thus, the roots of the given equation are $x = \frac{(17 + √(-11))}{6}$ or \frac{(17 – √(-11))}{6}$.
Q10 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q10 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 10
$15x^2 – 28 = x$Sol :
Given that $15x^2 – 28 = x$
⇒ $15x^2 – x – 28 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 15, b = -1, c = -28
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(15)(-28)}}{2(15)}$
$=\frac{1 \pm \sqrt{1+1680}}{30}=\frac{1 \pm \sqrt{1681}}{30}=\frac{1 \pm 41}{30}=\frac{1+41}{30}$ or $\frac{1-41}{30}=\frac{42}{30}$ or $-\frac{40}{30}=\frac{7}{5}$ or $-\frac{4}{3}$
Thus, the roots of the given equation are $x = \frac{7}{5}$ or $\frac{-4}{3}$
Q11 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 11
$x^2 + 3x – 3 = 0$, giving your answer correct to two decimal places.Sol :
Given that $x^2 + 3x – 3 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 1, b = 3, c = -3
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-3 \pm \sqrt{(3)^{2}-4(1)(-3)}}{2(1)}$
$=\frac{-3 \pm \sqrt{9+12}}{2}=\frac{-3 \pm \sqrt{21}}{2}=\frac{-3 \pm 4.58}{2}=\frac{-3+4.58}{2}$ or $\frac{-3-4.58}{2}=\frac{1.58}{2}$ or $-\frac{7.58}{2}=0.79$ or -3.79
Thus, the roots of the given equation are x = 0.79 or -3.79.
Q12 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 12
$\left(\frac{2}{3}\right)x = \left(\frac{-1}{6}\right)x^2 –\frac{1}{3}$, giving your answer correct to two decimal places.
Sol :
Given that $\left(\frac{2}{3}\right)x = \left(\frac{-1}{6}\right)x^2 –\frac{1}{3}$
⇒ $x^2 + 4x + 2 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 1, b = 4, c = 2
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-4 \pm \sqrt{(4)^{2}-4(1)(2)}}{2(1)}$
$=\frac{-4 \pm \sqrt{16-8}}{2(1)}=\frac{-4 \pm \sqrt{8}}{2}=\frac{-4 \pm 2 \sqrt{2}}{2}=\frac{-4 \pm(2 \times 1.41)}{2}=\frac{-1.18}{2}$ or $\frac{6.82}{2}=-0.59$ or 3.41
Q13 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 13
$x^2 + 6x – 10 = 0$Sol :
Given that $x^2 + 6x – 10 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 1, b = 6, c = -10
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-6\pm\sqrt{6^2-4(1)(-10)}}{2(1)}$
$=\frac{-6 \pm \sqrt{36+40}}{2}=\frac{-6 \pm \sqrt{76}}{2}=\frac{-6 \pm 2 \sqrt{19}}{2}=-3 \pm \sqrt{19}$
Thus, the roots of the given equation are x = -3-√19 or -3+√19.
Q14 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 14
$\frac{(x^2 + 8)}{11} = 5x – x^2 – 5$Sol :
Given that $\frac{(x^2 + 8)}{11} = 5x – x^2 – 5$
⇒ $12x^2 – 55x + 63 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 12, b = -55, c = 63
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-55)\pm \sqrt{(-55)^2-4(12)(63)}}{2(12)}$
$=\frac{55\pm \sqrt{3025-3024}}{24}=\frac{54\pm \sqrt{1}}{24}=\frac{55 \pm 1}{24}$ $=\frac{55+1}{24}$ or $=\frac{55-1}{24}$
$=\frac{56}{24}$ or $=\frac{54}{24}$
$=\frac{7}{3}$ or $=\frac{9}{4}$
Thus, the roots of the given equation are $x =\frac{7}{3}$ or $\frac{9}{4}$
Q15 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q15 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 15
$y – \frac{3}{y} = \frac{1}{2}$Sol :
Given that $y –\frac{3}{y} = \frac{1}{2}$ ⇒ $2y^2 – y – 6 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 2, b = -1, c = -6
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-6)}}{2(2)}$
$=\frac{-1\pm \sqrt{1+48}}{4}=\frac{1\pm \sqrt{49}}{4}=\frac{1\pm 7}{4}$
$=\frac{1+7}{4}$ or $=\frac{1-7}{4}$
$=\frac{8}{4}$ or $=\frac{-6}{4}=2$ or $=\frac{-3}{2}$
Thus, the roots of the given equation are y = 2 or $\frac{-3}{2}$.
Q16 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q16 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 16
$2x + \frac{4}{x} = 9$Sol :
Given that $2x + \frac{4}{x} = 9 $
⇒ $2x^2 – 9x + 4 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 2, b = -9, c = 4
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-9) \pm \sqrt{(-9)^{2}-4(2)(4)}}{2(2)}$
$=\frac{9 \pm \sqrt{81-32}}{4}=\frac{9 \pm \sqrt{49}}{4}=\frac{9 \pm 7}{4}=\frac{9+7}{4}$ or $\frac{9-7}{4}=\frac{16}{4}$ or $\frac{2}{4}=4$ or $\frac{1}{2}$
Thus, the roots of the given equation are x = 4 or $\frac{1}{2}$.
Q17 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 17
$\frac{x}{(x + 1)}+ \frac{(x + 1)}{x}= \frac{34}{15}$, x ≠ 0, x ≠ -1.Sol :
Given that $\frac{x}{(x + 1)}+ \frac{(x + 1)}{x}= \frac{34}{15}$
Put $y = \frac{x}{(x + 1)}$, then
⇒ $y + \frac{1}{y} = \frac{34}{15}$
⇒ $15y^2 – 34y + 15 = 0$
Then, comparing this quadratic equation with $ay^2 + by + c = 0$,
we get a = 15, b = -34, c = 15
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$y=\frac{-(-34) \pm \sqrt{(-34)^{2}-4(15)(15)}}{2(15)}$
$=\frac{34 \pm \sqrt{1156-900}}{30}=\frac{34 \pm \sqrt{256}}{30}=\frac{34 \pm 16}{30}=\frac{34+16}{30}$ or $\frac{34-16}{30}=\frac{50}{30}$ or $\frac{18}{30}=\frac{5}{3}$ or $\frac{3}{5}$
If $y = \frac{5}{3}$, then
$\frac{x}{(x + 1) }= \frac{5}{3}$
⇒ 3x = 5x + 5
⇒ 2x = -5
⇒ $x =\frac{ -5}{2}$
If $y = \frac{3}{5}$, then
$\frac{x}{(x + 1) }= \frac{3}{5}$
⇒ 5x = 3x + 3
⇒ 2x = 3
⇒ $x = \frac{3}{2}$
Thus, the roots of the given equation are $x =\frac{ -5}{2} $or $\frac{3}{2}$
Q18 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q18 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 18
$\frac{2x}{(x – 4)} +\frac{ (2x – 5)}{(x – 3) }= 8\frac{1}{3}$Sol :
Given that $\frac{2x}{(x – 4) }+ \frac{(2x – 5)}{(x – 3)} = 8 \frac{1}{3} $
⇒ $\frac{[2x(x – 3) + (2x – 5)(x – 4)]}{(x – 4)(x – 3) }=\frac{25}{3}$
⇒ $3[2x^2 – 6x + 2x^2 – 13x + 20] = 25(x – 4)(x – 3)$
⇒ $3[4x^2 – 19x + 20] = 25(x^2 – 7x + 12)$
⇒ $12x^2 – 57x + 60 = 25x2 – 175x + 300$
⇒ $13x^2 – 118x + 240 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$, we get a = 13, b = -118, c = 240
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-118) \pm \sqrt{(-118)^{2}-4(13)(240)}}{2(13)}$
$=\frac{118 \pm \sqrt{13924-12480}}{26}=\frac{118 \pm \sqrt{1444}}{26}=\frac{118 \pm 38}{26}=\frac{118+38}{26}$ or $\frac{118-38}{26}=\frac{156}{26}$ or $\frac{80}{26}=6$ or $\frac{40}{13}$
Thus, the roots of the given equation are x = 6 or $\frac{40}{13}$.
Q19 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q19 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 19
$\frac{(x + 6)}{(x + 7) }–\frac{ (x + 1)}{(x + 2) }= \frac{1}{(3x + 1)}$Sol :
Given that $\frac{(x + 6)}{(x + 7) }– \frac{(x + 1)}{(x + 2) }= \frac{1}{(3x + 1)} $
⇒ $\frac{[(x + 6)(x + 2) – (x + 1)(x + 7)]}{(x + 7)(x + 2) }= \frac{1}{(3x + 1)}$
⇒ $(3x + 1)(x^2 + 8x + 12 – x^2 – 8x – 7) = (x^2 + 9x + 14)$
⇒ $5(3x + 1) = (x^2 + 9x + 14)$
⇒ $15x + 5 = x^2 + 9x + 14$
⇒ $x^2 – 6x + 9 = 0 $
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 1, b = -6, c = 9
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(1)(9)}}{2(1)}=\frac{6 \pm \sqrt{36-36}}{2}=\frac{6 \pm 0}{2}=3$ or 3
Thus, the roots of the given equation are x = 3 or 3.
Q20 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q20 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 20
$\frac{(x + 1)}{(2x + 5)} = \frac{(x + 3)}{(3x + 4)}$Sol :
Given that $\frac{(x + 1)}{(2x + 5) }= \frac{(x + 3)}{(3x + 4)}$
⇒ (x + 1)(3x + 4) = (x + 3)(2x + 5)
⇒ $3x^2 + 7x + 4 = 2x^2 + 11x + 15$
⇒ $x^2 – 4x – 11 = 0$
Then, comparing this quadratic equation with $ax^2 + bx + c = 0$,
we get a = 1, b = -4, c = -11
Putting the values of a, b and c quadratic formula $x = \frac{[- b ± √(b^2 – 4ac)]}{2a}$
$x=\frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-11)}}{2(1)}=\frac{4 \pm \sqrt{16+44}}{2}=\frac{4 \pm \sqrt{60}}{2}=\frac{4 \pm 2 \sqrt{15}}{2}=2 \pm \sqrt{15}$
Thus, the roots of the given equation are x = 2 + √15 or 2 – √15.
Q21 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Q21 | Ex-5C | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper
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Question 21
(i)
Solve using quadratic formula : $a^2x^2 – 3abx + 2b^2 = 0$
Sol :Given that $a^2x^2 – 3abx + 2b^2 = 0$
Then, comparing this quadratic equation with $Ax^2 + Bx + C = 0$,
we get $A = a2$, B = -3ab, $C = 2b^2$
Putting the values of A, B and C quadratic formula $x = \frac{[- B ± √(B^2 – 4AC)]}{2A}$.
$x=\frac{-(-3 a b) \pm \sqrt{(-3 a b)^{2}-4\left(a^{2}\right)\left(2 b^{2}\right)}}{2\left(a^{2}\right)}$
$=\frac{3 a b \pm \sqrt{9 a^{2} b^{2}-8 a^{2} b^{2}}}{2 a^{2}}=\frac{3 a b \pm \sqrt{a^{2} b^{2}}}{2 a^{2}}=\frac{3 a b \pm a b}{2 a^{2}}=\frac{3 a b+a b}{2 a^{2}}$ or $\frac{3 a b-a b}{2 a^{2}}=\frac{4 a b}{2 a^{2}}$ or $\frac{2 a b}{2 a^{2}}=\frac{2 b}{a}$ or $\frac{b}{a}$
Thus, the roots of the given equation are x = 2b/a or b/a.
(ii)
Solve using quadratic formula : $x^2 – x – a(a + 1) = 0$
Sol :
Given that $x^2 – x – a(a + 1) = 0$
Then, comparing this quadratic equation with $Ax^2 + Bx + C = 0$,
we get A = 1, B = -1, C = -a(a + 1)
Putting the values of A, B and C quadratic formula $x = \frac{[- B ± √(B^2 – 4AC)]}{2A}$.
$x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(-a(a+1))}}{2(1)}$
$=\frac{1 \pm \sqrt{1+4 a+4 a^{2}}}{2}=\frac{1 \pm \sqrt{(2 a+1)^{2}}}{2}=\frac{1 \pm(2 a+1)}{2}=\frac{1+2 a+1}{2}$ or $\frac{1-2 a-1}{2}=(a+1)$ or $-a$
Thus, the roots of the given equation are x = -a or (a + 1).
(iii)
Solve using quadratic formula : $10x^2 + 3bx + a^2 – 7ax – b^2 = 0$
Sol :
Given that $10x^2 + 3bx + a^2 – 7ax – b^2 = 0$
⇒ $10x^2 + (3b – 7a)x + (a^2 – b^2) = 0$
Then, comparing this quadratic equation with $Ax^2 + Bx + C = 0$, we get A = 10, B = (3b – 7a), $C = (a^2 – b^2)$.
Putting the values of A, B and C quadratic formula $x = \frac{[- B ± √(B^2 – 4AC)]}{2A}$.
$\begin{aligned} x &=\frac{-(3 b-7 a) \pm \sqrt{(3 b-7 a)^{2}-4(10)\left(a^{2}-b^{2}\right)}}{2(10)} \\ &=\frac{-(3 b-7 a) \pm \sqrt{9 b^{2}+49 a^{2}-42 a b-40 a^{2}+40 b^{2}}}{20} \\ &=\frac{-(3 b-7 a) \pm \sqrt{9 a^{2}+49 b^{2}-42 a b}}{20} \\ &=\frac{-(3 b-7 a) \pm \sqrt{(3 a-7 b)^{2}}}{20} \\ &=\frac{-(3 b-7 a) \pm(3 a-7 b)}{20} \\ &=\frac{-3 b+7 a+3 a-7 b}{20} \text { or } \frac{-3 b+7 a-3 a+7 b}{20} \end{aligned}$
$=\frac{10 a-10 b}{20}$ or $\frac{4 a+4 b}{20}$
$=\frac{a-b}{2}$ or $\frac{a+b}{5}$
Thus, the roots of the given equation are $x =\frac{(a – b)}{2}$ or $\frac{(a + b)}{5}$.
$=\frac{3 a b \pm \sqrt{9 a^{2} b^{2}-8 a^{2} b^{2}}}{2 a^{2}}=\frac{3 a b \pm \sqrt{a^{2} b^{2}}}{2 a^{2}}=\frac{3 a b \pm a b}{2 a^{2}}=\frac{3 a b+a b}{2 a^{2}}$ or $\frac{3 a b-a b}{2 a^{2}}=\frac{4 a b}{2 a^{2}}$ or $\frac{2 a b}{2 a^{2}}=\frac{2 b}{a}$ or $\frac{b}{a}$
Thus, the roots of the given equation are x = 2b/a or b/a.
(ii)
Solve using quadratic formula : $x^2 – x – a(a + 1) = 0$
Sol :
Given that $x^2 – x – a(a + 1) = 0$
Then, comparing this quadratic equation with $Ax^2 + Bx + C = 0$,
we get A = 1, B = -1, C = -a(a + 1)
Putting the values of A, B and C quadratic formula $x = \frac{[- B ± √(B^2 – 4AC)]}{2A}$.
$x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(-a(a+1))}}{2(1)}$
$=\frac{1 \pm \sqrt{1+4 a+4 a^{2}}}{2}=\frac{1 \pm \sqrt{(2 a+1)^{2}}}{2}=\frac{1 \pm(2 a+1)}{2}=\frac{1+2 a+1}{2}$ or $\frac{1-2 a-1}{2}=(a+1)$ or $-a$
Thus, the roots of the given equation are x = -a or (a + 1).
(iii)
Solve using quadratic formula : $10x^2 + 3bx + a^2 – 7ax – b^2 = 0$
Sol :
Given that $10x^2 + 3bx + a^2 – 7ax – b^2 = 0$
⇒ $10x^2 + (3b – 7a)x + (a^2 – b^2) = 0$
Then, comparing this quadratic equation with $Ax^2 + Bx + C = 0$, we get A = 10, B = (3b – 7a), $C = (a^2 – b^2)$.
Putting the values of A, B and C quadratic formula $x = \frac{[- B ± √(B^2 – 4AC)]}{2A}$.
$\begin{aligned} x &=\frac{-(3 b-7 a) \pm \sqrt{(3 b-7 a)^{2}-4(10)\left(a^{2}-b^{2}\right)}}{2(10)} \\ &=\frac{-(3 b-7 a) \pm \sqrt{9 b^{2}+49 a^{2}-42 a b-40 a^{2}+40 b^{2}}}{20} \\ &=\frac{-(3 b-7 a) \pm \sqrt{9 a^{2}+49 b^{2}-42 a b}}{20} \\ &=\frac{-(3 b-7 a) \pm \sqrt{(3 a-7 b)^{2}}}{20} \\ &=\frac{-(3 b-7 a) \pm(3 a-7 b)}{20} \\ &=\frac{-3 b+7 a+3 a-7 b}{20} \text { or } \frac{-3 b+7 a-3 a+7 b}{20} \end{aligned}$
$=\frac{10 a-10 b}{20}$ or $\frac{4 a+4 b}{20}$
$=\frac{a-b}{2}$ or $\frac{a+b}{5}$
Thus, the roots of the given equation are $x =\frac{(a – b)}{2}$ or $\frac{(a + b)}{5}$.
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