Download App

Maths 2 Marks Questions

Quadratic Equations · CBSE STD 10 (CBSE - English Medium). 11 questions.

Questions

2 Marks Questions

🎯

Test yourself on this topic

11 questions · timed · auto-graded

Question 12 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$x(1 - x) - 2 = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$x(1 - x) - 2 = 0$
$\Rightarrow x - x^2 - 2 = 0$
$\Rightarrow -x^2 + x - 2 = 0$
Now, $D = b^2 - 4ac$
$= (-12) - 4(-1) (-2)(a = -1, b = 1, c = -2)$
$= 1 - 8$
$\Rightarrow D = -7 < 0$
So, the given equation has no real roots.
View full question & answer→
Question 22 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$(x - 1)(x + 2) + 2 = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$(x - 1) (x + 2) + 2 = 0$
$\Rightarrow x^2 + 2x - x^2 + 2 = 0$
$\Rightarrow x^2 + x = 0$
$\Rightarrow x^2 + x + 0 = 0$
Now, $D = b^2 - 4ac (a = 1, b = 1, c = 0)$
$= (1)^2 - 4(1)(0) = 1$
$\Rightarrow D = 1 > 0$
So, the given equation has two distinct real roots.
View full question & answer→
Question 32 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$x2 - 3x + 4 = 0$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
Given quadratic equation is $x^2 - 3x + 4 = 0$
Now, $D = b^2 - 4ac$
$= (-3)^2 - 4(1)(4) (a = 1, b = -3, c = 4)$
$\Rightarrow D = 9 - 16$
$\Rightarrow D = -7 < 0$
$\therefore$ D < 0
So, the given equation has no real roots.
View full question & answer→
Question 42 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$(x + 1)(x - 2) + x = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$(x + 1)(x - 2) + x = 0$
$\Rightarrow x^2 - 2x + x - 2 + x = 0$
$\Rightarrow x^2 - 2 = 0$
$\Rightarrow x^2 + 0x - 2 = 0$
Now, $D = b^2 - 4ac$
$= (0)^2 - 4(1)(-2) (a = 1, b = 0, c = -2)$
$\Rightarrow D = 8 > 0$
So, the given equation has two distinct real roots.
View full question & answer→
Question 52 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$(x + 4)^2 - 8x = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$(x + 4)^2 - 8x = 0$
$\Rightarrow (x)^2 + (4)^2 + 2(x)(4) - 8x = 0$
$\Rightarrow x^2 + 16 + 8x - 8x = 0$
$\Rightarrow x^2 + 16 = 0$
$\Rightarrow x^2 + 0x + 16 = 0$
Now, $D = b^2 - 4ac$
$= (0)^2 - 4(1) (16) (a = 1, b = 0, c = 16)$
$\Rightarrow D = -64 < 0$
$As D < 0$, so the given equation has no real roots.
View full question & answer→
Question 62 Marks
Is $0.2$ a root of the equation $x^2 – 0.4 = 0$? Justify.
Answer
If 0.2 is a root of equation $x^2 - 0.4 = 0$, then 0.2 must satisfy the given equation.
$x^2 - 0.4 = 0$ [given]
$\Rightarrow (0.2)^2 - 0.4 = 0$
$\Rightarrow 0.04 - 0.4 = 0$
$\Rightarrow\ -0.36\neq0$
So, 0.2 is not a root of the given equation.
View full question & answer→
Question 72 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+\frac{1}{\sqrt{2}}=0$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+\frac{1}{\sqrt{2}}=0$
$\text{D}=\text{b}^2-4\text{ac}$
$\Rightarrow\ \text{D}=\Big(\frac{-3}{\sqrt{2}}\Big)^2-4(\sqrt{2})\Big(\frac{1}{\sqrt{2}}\Big)$
$\Big(\text{a}=\sqrt{2},\text{ b}=\frac{-3}{\sqrt{2}},\text{ c}=+\frac{1}{\sqrt{2}}\Big)$
$=\frac{9}{2}-\frac{4}{1}=\frac{9-8}{2}$
$\Rightarrow\ \text{D}=\frac{1}{2}>0$
$\therefore\ \text{D}>0$
Hence, the given quadratic equation has two real and distirict roots.
View full question & answer→
Question 82 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$2x^2 + x - 1 = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$2x^2 + x - 1 = 0$
Now, $D = b^2 - 4ac$
$= (1)^2 - 4(2)(-1) (a = 2, b = 1, c = -1)$
$= 1 + 8$
$\Rightarrow D = 9 > 0$
$\therefore\ \text{D}> 0$
So, the given equation has two distinct real roots.
View full question & answer→
Question 92 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$2\text{x}^2-6\text{x}+\frac{9}{2}=0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$2\text{x}^2-6\text{x}+\frac{9}{2}=0$
Now, $D = b^2 - 4ac$
$\Rightarrow\ \text{D}=(-6)^2-4(2)\Big(\frac{9}{2}\Big)$
$\Big(\text{a}1,\text{ b}=-6,\text{ c}=\frac{9}{2}\Big)$
$\Rightarrow\ \text{D}=36-36$
$\Rightarrow\ \text{D}=0$
So, the givne equation has twoi real and equaol roots.
View full question & answer→
Question 102 Marks
State whether the following quadratic equations have two distinct real roots. Justify your answer.
$3x^2 - 4x + 1 = 0.$
Answer
Main concept used:
Quadratic equation $ax^2 + bx + c = 0$ will have two distinct real roots if $D > 0 or b^2 - 4ac > 0.$
$3x^2 - 4x + 1 = 0$
$Now, D = b^2 - 4ac (a = 3, b = -4, c = 1)$
$= (-4)^2 - 4(3)(1) = 16 - 12$
$\Rightarrow D = 4 > 0$
$\therefore\ \text{D}>0$
So, the given equation has two distinct real roots.
View full question & answer→
Question 112 Marks
If $b = 0, c < 0$, is it true that the roots of $x^2 + bx + c = 0$ are numerically equal and opposite in sign? Justify.
Answer
Given equation is $x^2 + bx + c = 0$
b = 0 [Given]
$\therefore\ \text{x}^2+\text{c}=0$
$\Rightarrow\ \text{x}^2=-\text{c}$
$\Rightarrow\ \text{x}=\sqrt{-\text{c}}$
As c is negative so -c becomes positive or $\sqrt{-\text{c}}$ is real.
So, the roots of the given equation are
$\text{x}=\pm\sqrt{-\text{c}}$
or $\text{x}=\pm\sqrt{-\text{c}}$ and $-\sqrt{-\text{c}}$ [$\because$ (-c) is positive]
Hence, the roots of the given equation are real, equal and opposite in sign.
View full question & answer→
Generate Paper FreeAll Quadratic Equations topics
2 Marks Questions - Maths STD 10 Questions - Vidyadip