Maths Class X Question Bank

Question 1 of 38

1. Question
1 point(s)
Find the roots of the quadratic equation $x^2-3x=0 $.
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Hint

$x^2-3x=0\\ \\ x=\dfrac{-3\pm\sqrt{(-3)^2-4\times1\times0}}{2}\\ \\ =\dfrac{-(-3)\pm\sqrt9}{2}=\dfrac{3\pm3}{2}\\ \\ =\dfrac{3+3}{2},\dfrac{3-3}{2}\\ \\ =\dfrac{6}{2},\dfrac{0}{2}=3,0$
Question 2 of 38

2. Question
1 point(s)
Are $x=0,\ x=1 $ the solution of the equation $x^2+x+1=0 $?
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Hint

$x^2+x+1=0$
If $x=0$
$0^2+0+1\pm0$
If $x=1$
$1^2+1+1=3\pm0$
So $x=0,1$ are not the solution
Question 3 of 38

3. Question
1 point(s)
If $4a^2x^2-4abx+k=0 $ has equal roots, then find the value of $k $.
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Hint

$4a^2x^2-4abx+k=0\\ \\ b^2-4ac=0\\ \\ \Rightarrow (-4ab)^2-4\times(4a^2)k=0\\ \\ \Rightarrow 16a^2b^2-16a^2k=0\\ \\ \Rightarrow 16a^2(b^2-k)=0\\ \\ \Rightarrow k=b^2$
Question 4 of 38

4. Question
1 point(s)
Find the discriminant of the quadratic equation $bx^2-2\sqrt{ac}x+b=0 $
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Hint

$bx^2-2\sqrt{a}cx+b=0\\ \\ D=b^2-4ac\\ \\ =(-2\sqrt ac)^2-4\times b\times b\\ \\ =4ac-4b^2$
Question 5 of 38

5. Question
1 point(s)
If 2 is a root of the equation $x^2+bx+12=0 $ and the equation has equal roots, find the value of b,
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Hint

$x^2+bx+12=0\\ \\ (2)^2+b(2)+12=0\\ \\ \Rightarrow 2b=\ -16\Rightarrow b=-8$
Question 6 of 38

6. Question
1 point(s)
If one of the roots of the equation $3a^2x^2+8abx+4b^2=0 $ is $\dfrac{-2b}{a} $, then find the other root.
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Hint

$3a^2x^2+8abx+4b^2=0$
Let other root be $x$
$x+\left(\dfrac{-2b}{a}\right)=\dfrac{-8ab}{3a^2}\\ \\ \Rightarrow x=\dfrac{-8b}{3a}+\dfrac{2b}{a}=\dfrac{-8b+6b}{3a}\\ \\ x=\dfrac{-2b}{3a}$
Question 7 of 38

7. Question
1 point(s)
If $x=\dfrac{c}{b} $ is a root of $abx^2+(b^2-ac)\ x-k=0 $, then find the value of $ k$.
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Hint

$abx^2+(b^2-ac)x-k=0\\ \\ \Rightarrow ab\left(\dfrac{c}{b}\right)^2+b^2x-acx-k=0\\ \\ \Rightarrow ab\times \dfrac{c^2}{b^2}+b^2\times\dfrac{c}{b}-ac\times\dfrac{c}{b}-k=0\\ \\ \Rightarrow \dfrac{ac^2}{b}+bc-\dfrac{ac^2}{b}-k=0\\ \\ \Rightarrow bc=k\Rightarrow k=bc$
Question 8 of 38

8. Question
1 point(s)
State whether the quad ratic equation $(x-\sqrt2)^2-\sqrt2(x+1)=0 $ has two distinct real roots. Justify your answer.
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Hint

$\Rightarrow x^2-2\sqrt2x+2-\sqrt2x-\sqrt2=0\\ \\ \Rightarrow x^2-3\sqrt2x+(2-\sqrt2)=0\\ \\ D=b^2-4ac=(-3\sqrt2)^2-4(1)(2-\sqrt2)\\ \\ =18-8+4\sqrt2>0$
so the roots are real and distinct
Question 9 of 38

9. Question
1 point(s)
If $b=0,\ c<0 $, is it true that the roots of the quadratic equation $x^2+bx+c=0 $ are numerically equal and opposite in sign? Justify.
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Hint

Yes the roots are equal and opposite of the two roots are $x$ and $y$
$x+y=\dfrac{-b}{1}$ as $b=0$ then $x+y=0$ . This is only possible when $x$ and $y$ and equal and opposite
Question 10 of 38

10. Question
1 point(s)
Does $(x-1)^2+2(x+1)=0 $ have a real root? Justify your answer.
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Hint

$(x-1)^2+2(x+1)=0\\ \\ x^2+1-2x+2x+2=0\\ \\ x^2+3=0\ \Rightarrow x^2=-3$
This is not possible as under root of a negetive number leads to imaginary number has no real roots
Question 11 of 38

11. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$x^2-(\sqrt2+1)x+\sqrt2=0 $
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Hint

$x^2-(\sqrt2+1)x+\sqrt2=0\\ \\ x^2-\sqrt2x-x+\sqrt2=0\\ \\ x(x-\sqrt2)-1(x-\sqrt2)=0\\ \\ (x-\sqrt2)\ (x-1)=0\\ \\ x=1,\sqrt2$
Question 12 of 38

12. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$4x^2 + 4bx – (a^2 – b^2) = 0 $
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Hint

$4x^2+4bx-(a^2-b^2)=0\\ \\ 4x^2+4bx-a^2+b^2=0\\ \\ 4x^2+4bx+b^2-a^2=0\\ \\ (2x+b)^2-a^2=0\\ \\ (2x+b-a)\ (2x+b+a)=0\\ \\ 2x+b-a=0\Rightarrow 2x=a-b\Rightarrow x=\dfrac{a-b}{2}\\ \\ 2x+b+a=0\Rightarrow 2x=-(a+b)\Rightarrow x=\dfrac{-(a+b)}{2}$
Question 13 of 38

13. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$ax^2 + (4a^2 – 3b) x – 12ab = 0 $
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Hint

$ax^2+4a^2x-3bx-12ab=0\\ \\ ax(x+4a)-3b\ (x+4a)=0\\ \\ (ax-3b)\ (x+4a)=0\\ \\ x=\dfrac{3b}{a},-4a$
Question 14 of 38

14. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$\dfrac{m}{n}x^2+\dfrac{n}{m}=1-2x $
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Hint

$\dfrac{m}{n}x^2+\dfrac{n}{m}-1+2x=0\\ \\ \Rightarrow \dfrac{m^2x^2+n^2-nm+2xnm}{nm}=0\\ \\ \Rightarrow m^2x^2+n^2+2xnm-nm=0\\ \\ \Rightarrow (mx+n)^2-nm=0\\ \\ \Rightarrow (mx+n)^2-(\sqrt{nm})^2=0\\ \\ \Rightarrow (mx+n-\sqrt nm)\ (mx+n+\sqrt mn)=0\\ \\ x=\dfrac{\sqrt{ mn}-n}{m}\ \text{or}\ \dfrac{-\sqrt{mn}-n}{m}$
Question 15 of 38

15. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$(a+b)^2x^2-4abx-(a-b)^2=0 $
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Hint

$(a+b)^2x^2-4abx-(a-b)^2=0\\ \\ (a+b)^2x^2-(a+b)^2x+(a-b)^2x-(a-b)^2=0\\ \\ (a+b)^2x(x-1)+(a-b)^2(x-1)=0\\ \\ (a+b)^2x+(a-b)^2(x-1)=0\\ \\ x=-\left(\dfrac{a-b}{a+b}\right)^2,1$
Question 16 of 38

16. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$x^2 – x – a (a + 1) = 0 $
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Hint

$x^2-x-a(a+1)=0\\ \\ \Rightarrow x^2-a^2-x-a=0\\ \\ \Rightarrow (x+a)\ (x-a)-1(x+a)=0\\ \\ \Rightarrow (x+a)\ (x-a-1)=0\\ \\ \Rightarrow x=\ -a,a+1$
Question 17 of 38

17. Question
1 point(s)
Solve the following quadratic equations by factorisation:
$(x-3)\ (x-4)=\dfrac{34}{(33)^2} $
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Hint

$(x-3)\ (x-4)=\dfrac{34}{(33)^2}\\ \\ x^2-4x-3x+12=\dfrac{34}{1089}\\ \\ \Rightarrow 1089x^2-7623x+13034=0\\ \\ \Rightarrow 1089x^2-3234x-4389x+13034=0\\ \\ \Rightarrow 33x(33x-98)-133(33x-98)=0 \\ \\ \Rightarrow (33x-133)\ (33x-98)=0\\ \\ \Rightarrow x=\dfrac{133}{33},\dfrac{98}{33}$
Question 18 of 38

18. Question
1 point(s)
Solve the following quadratic equations by completing the squares: $4x^2+4\sqrt3x+3=0 $
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Hint

$4x^2+4\sqrt3x+3=0\\ \\ 4x^2+4\sqrt3x=-3$
Dividing each term by 4 we get
$x^2+\sqrt3x=\dfrac{-3}{4}\\ \\ \Rightarrow x^2+2\times x\times\dfrac{\sqrt3}{2}=\dfrac{-3}{4}\\ \\ \Rightarrow x^2+2\times x\times \dfrac{\sqrt3}{2}+\left(\dfrac{\sqrt 3}{2}\right)^2=\dfrac{-3}{4}+\left(\dfrac{\sqrt 3}{2}\right)^2\\ \\ \Rightarrow \left(x+\dfrac{\sqrt3}{2}\right)^2=\dfrac{-3}{4}+\dfrac{3}{4}=0\\ \\ \Rightarrow \left(x+\dfrac{\sqrt3}{2}\right)=0\Rightarrow x=\dfrac{-\sqrt3}{2}$
Question 19 of 38

19. Question
1 point(s)
Solve the following quadratic equations by completing the squares: $\sqrt3x^2+10x+7\sqrt3=0 $
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Hint

$\sqrt3 x^2+10x+7\sqrt3=0\\ \\ \Rightarrow x^2+\dfrac{10}{\sqrt3}x+\dfrac{7\sqrt3}{3}=0\\ \\ \Rightarrow x^2+2\times\dfrac{1}{2}\times\dfrac{10}{\sqrt3}x+7=0\\ \\ \Rightarrow x^2+2\times\dfrac{5}{\sqrt3}x+\left(\dfrac{5}{\sqrt3}\right)^2-\left(\dfrac{5}{\sqrt3}\right)^2+7=0\\ \\ \Rightarrow \left(x+\dfrac{5}{\sqrt3}\right)^2=\dfrac{25}{3}-7\\ \\ \Rightarrow \left(x+\dfrac{5}{\sqrt3}\right)^2=\dfrac{4}{3}\Rightarrow x+\dfrac{5}{\sqrt3}=\pm\dfrac{2}{\sqrt3}\\ \\ \Rightarrow x=-\sqrt3,\dfrac{-7}{\sqrt3}$
Question 20 of 38

20. Question
1 point(s)
Solve the following quadratic equations by completing the squares: $a^2x^2 – 3abx + 2b^2 = 0 $
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Hint

$a^2x^2-3abx+2b^2=0\\ \\ a^2x^2-3abx=-2b^2\\ \\ \Rightarrow x^2-\dfrac{3bx}{a}=\dfrac{2b^2}{a^2}\\ \\ \Rightarrow \left(x-\dfrac{3b}{2a}\right)^2=\dfrac{9b^2}{4a^2}-\dfrac{2b^2}{a^2}\\ \\ \Rightarrow \left(x-\dfrac{3b}{2a}\right)^2=\dfrac{b^2}{4a^2}\\ \\ \Rightarrow x-\dfrac{3b}{2a}=\pm\dfrac{b}{2a}\\ \\ \Rightarrow x=\left(\dfrac{3b}{2a}+\dfrac{b}{2a}\right),\left(\dfrac{3b}{2a}-\dfrac{b}{2a}\right)\\ \\ =\dfrac{4b}{2a},\dfrac{2b}{2a}\\ \\ x=\dfrac{2b}{a},\dfrac{b}{a}$
Question 21 of 38

21. Question
1 point(s)
Solve the following quadratic equations by completing the squares: $y^2+\dfrac{1}{2}y-1=0 $
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Hint

$y^2+\dfrac{1}{2}y-1=0\\ \\ y^2+\dfrac{1}{2}y=1\\ \\ \Rightarrow y^2+2\left(\dfrac{1}{4}\right)y+\left(\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2+1\\ \\ \Rightarrow \left(y+\dfrac{1}{4}\right)^2=\dfrac{17}{16}\\ \\ \Rightarrow y+\dfrac{1}{4}=\pm\dfrac{\sqrt{17}}{4}\\ \\ \Rightarrow y=\dfrac{-1}{4}\pm\dfrac{\sqrt{17}}{4}\\ \\ \Rightarrow y=\dfrac{-1}{4}+\dfrac{\sqrt{17}}{4}\ \text{and }\ \dfrac{-1-\sqrt{17}}{4}$
Question 22 of 38

22. Question
1 point(s)
Solve the following equations by using the formula: $x^2 – 2ax + 3x – 6a = 0 $
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Hint

$x^2-2ax+3x-6a=0\\ \\ x^2-(2a-3)x-6a=0\\ \\ D=b^2-4ac=\{-(2a-3)\}^2-4(1)\ (-6a)\\ \\ =(3-2a)^2+24a\\ \\ =(3+2a)^2\\ \\ x=\dfrac{(2a-3)\pm3+2a}{2}\\ \\ =\dfrac{2a-3+3+2a}{2},\dfrac{2a-3-3-2a}{2}\\ \\ =\dfrac{4a}{2},\dfrac{-6}{2}\\ \\ x=2a,-3$
Question 23 of 38

23. Question
1 point(s)
Solve the following equations by using the formula: $a^2b^2x^2 + b^2x – a^2x – 1 = 0 $
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Hint

$a^2b^2x^2+b^2x-a^2x-1=0\\ \\ b^2x(a^2x+1)-1(a^2x+1)=0\\ \\ (b^2x-1)\ (a^2x+1)=0\\ \\ x=\dfrac{1}{b^2},\dfrac{-1}{a^2}$
Question 24 of 38

24. Question
1 point(s)
Solve the following equations by using the formula: $a^2 – b^2 = 4ax – 4x^2 $
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Hint

$a^2-b^2=4ax-4x^2\\ \\ 4x^2-4ax+(a^2-b^2)=0\\ \\ D=b^2-4ac=(-4a)^2-4\times4\times(a^2-b^2)\\ \\ =16a^2-16a^2+16b^2=16b^2\\ \\ x=\dfrac{-b\pm\sqrt D}{2a}=\dfrac{-(-4a)\pm\sqrt{16}b^2}{2\times4}\\ \\ =\dfrac{4a+4b}{8},\dfrac{4a-4b}{8}\\ \\ =\dfrac{a+b}{2},\dfrac{a-b}{2}$
Question 25 of 38

25. Question
1 point(s)
Solve the following equations by using the formula: $(a – b)x^2 + (b – c)x + (c – a) = 0 $
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Hint

$(a-b)x^2+(b-c)x+(c-a)=0\\ \\ D=(b-c)^2-4(a-b)\ (c-a)\\ \\ =b^2+c^2-2bc-4ac+4a^2+4bc-4ab\\ \\ =(b+c-2a)^2\\ \\ x=\dfrac{-b\pm\sqrt D}{2a}=\dfrac{-(b-c)\pm(b+c-2a)}{2(a-b)}\\ \\ =\dfrac{-b+c+b+c-2a}{2a-2b}\\ \\ \dfrac{-b+c-b-c+2a}{2a-2b}\\ \\ =\dfrac{2c-2a}{2a-2b},\dfrac{-2b+2a}{2a-2b}=\dfrac{c-a}{a-b},\dfrac{-b+a}{a-b}=\dfrac{c-a}{a-b},1$
Question 26 of 38

26. Question
1 point(s)
Solve the following equations by using the formula: $(m^2+n^2)x^2+(m+n)x+\dfrac{1}{2}=0 $
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Hint

$x=\dfrac{-(m+n)\pm\sqrt{(m+n)^2-4(m^2+n^2)\left(\frac{1}{2}\right)}}{2(m^2+n^2)}\\ \\ =\dfrac{-(m+n)\pm\sqrt{-(m-n)^2}}{2(m^2+n^2)}\\ \\ =\dfrac{-(m+n)+m-n}{2(m^2+n^2)},\dfrac{-(m+n)-(m-n)}{2(m^2+n^2)}\\ \\ =\dfrac{-2n}{2m^2+2n^2},\dfrac{-2m}{2m^2+2n^2}$
Question 27 of 38

27. Question
1 point(s)
Find the values of $k $ for which the roots are real and equal in each of the following equations :
$(k + 1)x^2 + 2(k + 3)x + (k + 8) = 0 $
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Hint

$(k+1)x^2+2(k+3)x+(k+8)=0\\ \\ b^2-4ac=0\\ \\ [2(k+3)]^2-4\times(k+1)\ (k+8)=0\\ \\ =4k^2+24k+36-4k^2-36k-32\\ \\ \Rightarrow -12k+4=0\\ \\ \Rightarrow k=\dfrac{1}{3}$
Question 28 of 38

28. Question
1 point(s)
Find the values of $k $ for which the roots are real and equal in each of the following equations :
$5x^2 – 4x + 2 + k(4x^2 – 2x – 1) = 0 $
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Hint

$5x^2-4x+2+k(4x^2-2x-1)=0\\ \\ 5x^2-4x+2+4kx^2-2kx-k=0\\ \\ (5+4k)^2-(4+2k)x+(2-k)=0\\ \\ b^2-4ac=0\\ \\ \Rightarrow \{-(4+2k)\}^2-4\times(5+4k)\ (2-k)=0\\ \\ \Rightarrow 16+16k+4k^2-40-12k+16k^2=0\\ \\ \Rightarrow 20k^2+4k-24=0\\ \\ \Rightarrow 5k^2+k-6=0\\ \\ \Rightarrow 5k^2+6k-5k-6=0\\ \\ \Rightarrow (5k+6)=0,\ k-1=0\\ \\ \Rightarrow k=\dfrac{-6}{5},1$
Question 29 of 38

29. Question
1 point(s)
Find the values of $k $ for which the roots are real and equal in each of the following equations :
$(4 – k)x^2 + (2k + 4)x + (8k + 1) = 0 $
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Hint

$(4-k)x^2+(2k+4)x+(8k+1)=0\\ \\ (2k+4)^2-4(4-k)\ (8k+1)=0\\ \\ \Rightarrow 4k^2+16k+16-4(32k+4-8k^2-k)=0\\ \\ \Rightarrow 36k^2+16k-124k=0\\ \\ \Rightarrow 36k(k-3)=0\ \Rightarrow \ k=0,3$
Question 30 of 38

30. Question
1 point(s)
For what value of $(4 – k)x^2 + (2k + 4)x + (8k + 1) = 0, $ is a perfect square.
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Hint

$(4-k)x^2+(2k+4)x+(8k+1)=0\\ \\ (2k+4)^2-4(4-k)\ (8k+1)=0\\ \\ \Rightarrow 4k^2+16k+16-4(32k+4-8k^2-k)=0\\ \\ \Rightarrow 36k^2+16k-124k=0\\ \\ \Rightarrow 36k(k-3)=0\ \Rightarrow \ k=0,3$
Question 31 of 38

31. Question
1 point(s)
If the roots of the equations $ax^2+2bx+c=0 $ and $bx^2-2\sqrt{ac}x+b=0 $ are simultaneously real, then prove that $b^2=ac $.
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Hint

$ax^2+2bx+c=0\\ \\ D_1=b^2-4ac=(2b)^2-4ac=4b^2-4ac\\ \\ bx^2-2\sqrt{ac}x+b=0\\ \\ D_2=(-2\sqrt{ac})^2-4\times b\times b=4ac-4b^2$
$D_1\ge0$ and $D_2\ge0$
$4b^2-4ac\ge0$ and $4b^2\ge4ac\ \Rightarrow b^2\ge ac$ …(3)
$4ac-4b^2\ge0\ ac\ge b^2$ …(4)
From eqn (3) and (4)
$b^2=ac$ (Proved)
Question 32 of 38

32. Question
1 point(s)
For what value(s) of $k $, the equation $x^2-2kx-k=0 $ will have equal roots?
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Hint

$x^2-2kx-k=0\\ \\ D=0\\ \\ b^2-4ac=0\\ \\ (-2k)^2-4\times1\times(-k)=0\\ \\ 4k^2+4k=0\\ \\ 4k^2=-4k\\ \\ \Rightarrow k=-1$
Question 33 of 38

33. Question
1 point(s)
Solve the equation : $10ax^2+15ax-6x-9=0,\ a\ne0 $
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Hint

$10ax^2+15ax-6x-9\\ \\ 5ax(2x+3)-3(2x+3)=0\\ \\ (2x+3)\ (5ax-3)=0\\ \\ \Rightarrow x=\dfrac{-3}{2},\dfrac{3a}{5a}$
Question 34 of 38

34. Question
1 point(s)
Find the value of k for which the equation : kx (x – 2) + 6 = 0 has equal roots.
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Hint

$kx(x-2)+6=0\\ \\ \Rightarrow kx^2-2kx+6=0\\ \\ b^2-4ac=0\\ \\ \Rightarrow (-2k)^2-4(k)(6)=0\\ \\ \Rightarrow 4k^2-24k=0\\ \\ \Rightarrow 4k^2=24k\\ \\ \Rightarrow k=6$
Question 35 of 38

35. Question
1 point(s)
For what value of $k $ the equation $4x^2 – 2\ (k +1) x + (k + 1) = 0 $ has real and equal roots ?
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Hint

$4x^2-2(k+1)x+(k+1)=0\\ \\ b^2-4ac=0\\ \\ \Rightarrow [-2(k+1)]^2-4\times4\times(k+1)=0\\ \\ \Rightarrow 4k^2+8k+4-16k-16=0\\ \\ \Rightarrow 4k^2-8k-12=0\\ \\ \Rightarrow k^2-2k-3=0\\ \\ \Rightarrow k^2-3k+k-3=0\\ \\ \Rightarrow k(k-3)+1(k-3)=0\\ \\ \Rightarrow k=-1,3$
Question 36 of 38

36. Question
1 point(s)
Find the roots of the following quadratic equation $(x+3)\ (x+1)=3\ \left(x-\dfrac{1}{3}\right) $
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Hint

$(x+3)\ (x-1)=3\left(x-\dfrac{1}{3}\right)\\ \\ \Rightarrow x^2-x+3x-3=3x-1\\ \\ \Rightarrow x^2-x-2=0\\ \\ \Rightarrow x^2-2x+x-2=0\\ \\ \Rightarrow (x-2)\ (x+1)=0\\ \\ \Rightarrow x=2,-1$
Question 37 of 38

37. Question
1 point(s)
Find the value of $k $ such that the quadratic equation $x^2-2\ kx+(7k-12)=0 $ has real and equal roots.
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Hint

$x^2-2kx+(7k-12)=0\\ \\ b^2-4ac=0\\ \\ (-2k)^2-4\times 1\times(7k-12)=0\\ \\ 4k^2=4(7k-12)\\ \\ k^2=7k-12\\ \\ k^2-7k+12=0\\ \\ k^2-4k-3k+12=0\\ \\ k(k-4)-3(k-4)=0\\ \\ k=3,4$
Question 38 of 38

38. Question
1 point(s)
Find the nature of roots of the quadratic equation $\sqrt2x^2-\dfrac{3}{\sqrt2}x+\dfrac{1}{\sqrt2}=0 $
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Grading can be reviewed and adjusted.

Hint

$\sqrt2x^2-\dfrac{3}{\sqrt2}x+\dfrac{1}{\sqrt2}=0\\ \\ 2x^2-3x+1=0\\ \\ D=b^2-4ac\\ \\ =(-3)^2-4\times2\times1\\ \\ =9-8=1$
The roots are equal and distinct in nature