MCQ 11 Mark
If 3 is a root of the quadratic equation $x^2-p x+3=0$ then $p$ is equal to:
Answer(a) 4
Explanation:
$\because x=3$ is a root of the equation
$x^2-p x+3=0 \ldots$ (i)
then $x=3$ will satisfy equation (i)
$\begin{array}{l}\therefore(3)^2-3 p+3=0 \\ 9-3 p+3=0 \\ 3 p=12 \\ \Rightarrow p=\frac{12}{3}=4 Ans .\end{array}$
View full question & answer→MCQ 21 Mark
The roots of the quadratic equation $p x^2-q x+r=0$ are real and equal if:
- A
$p^2=4 q r$
- ✓
$q^2=4 p r$
- C
$-q^2=4 p r$
- D
$p^2 > 4 p r$
AnswerCorrect option: B. $q^2=4 p r$
(b) $q^2=4 p r$
Explanation:
Given, equation is $p x^2-q x+r=0$
On comparing it with $a x^2+b x+c=0$, we get
$a=p, b=-q, c=r$
For roots to be equal, $D =0$
i.e., $b^2-4 a c=0$
$\begin{array}{l}\Rightarrow(-q)^2-4 \times p \times r=0 \\ \Rightarrow q^2-4 p r=0 \\ \Rightarrow q^2=4 p r .\end{array}$
View full question & answer→MCQ 31 Mark
Which of the following is/are correct?
Statement (A): The equation $x^2-3 x-10=0$ has two roots i.e., -2 and 5 .
Statement (B): The discriminant of the quadratic equation $3 x^2-5 x-12=0$ is 144 .
Statement (C): $\sqrt{2}$ is solution of the quardratic equation. $x^2+\sqrt{2} x-4=0$
- A
- B
- ✓
- D
All A, B and C are correct
Answer(c) Only A and C are correct
Explanation
For statement A,
Given equation is $x^2-3 x-10=0$
$\begin{array}{l}x^2-5 x+2 x-10=0 \\ x(x-5)+2(x-5)=0 \\ (x-5)(x+2)=0 \\ x=-2,5\end{array}$
Statement A is correct
For statement B,
Given equation is $3 x^2-5 x-12=0$
a=3, b=-5, c=-12
$\therefore$ Discriminant (D) $=b^2-4 a c$
$\begin{array}{l}=25-4(3)(-12) \\ =25+144 \\ =169\end{array}$
Statement B is incorrect
For statement C,
$x^2+\sqrt{2} x-4=0$
Substituting $x=\sqrt{2}$ in the L.H.S.
L.H.S. $(\sqrt{2})^2-\sqrt{2} \times \sqrt{2}-4$
$2+2-4$
$4-4=0$ R.H.S.
L.H.S. $=$ R.H.S.
Hence, $\sqrt{2}$ is solution of the given quadratic equation statement $C$ is correct.
View full question & answer→MCQ 41 Mark
Which of the following is/are correct?
Statement (A): Two real and different roots if and if $b^2>4 a c$
Statement (B): Two real roots if and only if $b^2 \geq 4 a c$.
Statement (C): no real roots if and only if $b^2=4 a c$
- ✓
- B
- C
- D
All A, B and C are correct
Answer(a) Only A and B are correct
Explanation
Both A and B are correct, but statement C is incorrect because if $b^2<4 a c$, then equation has no real roots and if $b^2=$ $4 a c$, then two real and equal roots.
View full question & answer→MCQ 51 Mark
Statement (A): If one root of the quadratic equation $6 x^2-x-k=0$ is $\frac{2}{3}$, then the value of $k$ is 2 .
Statement (B): The quadratic equation $a x^2+b x+c=0, a \neq 0$ has almost two roots.
Which of the statement is valid?
Answer(c) Both A and B
Explanation
The given equation is $6 x^2-x-k=0$ ... one root is $\frac{2}{3}, x=\frac{2}{3}$
$\begin{array}{l}6 \times\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)-k=0 \\ \frac{6 \times 4}{9}-\frac{2}{3}-k=0 \\ k=\frac{8}{3}-\frac{2}{3} \\ k=\frac{6}{3} \\ k=2\end{array}$
The quadratic equation $a x^2+b x+c=0$ has two roots.
Hence, Both statements are correct.
View full question & answer→MCQ 61 Mark
Statement (A): The discriminant of the equation $4 x^2-12 x+9=0$ is 0 .
Statement (B): If $b^2-4 a c=0$ or $b^2=4 a c$ that means the equation has two real and equal roots.
Which of the statement is valid?
Answer(c) Both A and B
Explanation
The given equation, $4 x^2-12 x+9=0$. Comparing it with $a x^2+b x+c=0$, we get $a=4, b=-12, c=9$
$\begin{array}{l}\therefore \text { Discriminant }=b^2-4 a c=(-12)^2-4 \times 4 \times 9 \\ =144-144=0\end{array}$
Since, $b^2-4 a c=0$
Hence, the given equation has two equal real roots.
View full question & answer→MCQ 71 Mark
Statement (A): If roots of the equation $x^2-b x+c=0$ are two consecutive integers, then $b^2-4 a c=1$.
Statement (B): If $a, b, c$ are odd integers then the roots of the equation $4 a b c x^2+\left(b^2-4 a c\right) x-b=0$ are real and distinct.
Which of the statement is valid?
Answer(c) Both A and B
Explanation
Statment 1: Given equation, $x^2-b x+c=0$. Let $\alpha, \beta$ be two roots such that
$\begin{array}{l}|\alpha-\beta|=1 \\ \Rightarrow(\alpha+\beta)^2-4 \alpha \beta=1 \\ \Rightarrow b^2-4 c=1\end{array}$
Statement 2 : Given equation,
$4 a b c x^2+\left(b^2-4 a c\right) x-b=0$
$D =\left(b^2-4 a c\right)^2+16 a b^2 c$
$\Rightarrow D =\left(b^2+4 a c\right)^2>0$
Hence roots are real and unequal.
So, Both statements are true.
View full question & answer→MCQ 81 Mark
Which of the following equation has 2 as a root?
- A
$x^2-4 x+5=0$
- B
$x^2+3 x-12=0$
- ✓
$2 x^2-7 x+6=0$
- D
$3 x^2-6 x-2=0$
AnswerCorrect option: C. $2 x^2-7 x+6=0$
(c) $2 x^2-7 x+6=0$
Explanation :
On substituting $x=2$ in $2 x^2-7 x+6$, we get
$\begin{array}{l}2(2)^2-7(2)+6=2(4)-14+6 \\ =8-14+6=14-14=0\end{array}$
So, $x=2$ is root of the equation $2 x^2-7 x+6=0$.
View full question & answer→MCQ 91 Mark
If the roots of equation $x^2-6 x+k=0$ are real and distinct, then value of $k$ is:
Answer(d) $<9$
Explanation:
Equation: $x^2-6 x+k=0$, comparing it with $a x^2+b x+c=0$, we get
a=1, b=-6, c=k
Roots are real and distinct,
D>0
$\begin{array}{l}\Rightarrow b^2-4 a c>0 \\ \Rightarrow(-6)^2-4 \times 1 \times k>0 \\ \Rightarrow 36-4 k=0 \\ \Rightarrow 36>4 k \\ \Rightarrow 9>k \Rightarrow k<9\end{array}$
View full question & answer→MCQ 101 Mark
If $x^2+k x+6=(x-2)(x-3)$ for all values of $x$, then the value of $k$ is:
Answer(a) -5
Explanation:
Given,
$\begin{array}{l}x^2+k x+6=(x-2)(x-3) \\ \Rightarrow x^2+k x+6=x^2-5 x+6\end{array}$
Compare coefficient of $x$ both sides.
k=-5.
View full question & answer→MCQ 111 Mark
The condition for the sum and the product of the roots of the quadratic equation $a x^2-b x+c=0$ to be equal, is:
- A
$b+c=0$
- ✓
$b-c=0$
- C
$a+c=0$
- D
$a+b+c=0$
AnswerCorrect option: B. $b-c=0$
(b) $b-c=0$
Explanation:
... Sum of roots = Product of roots
$\begin{array}{l}-\left(\frac{-b}{a}\right)=\frac{c}{a} \\ \Rightarrow b=c \\ \Rightarrow b-c=0\end{array}$
View full question & answer→MCQ 121 Mark
The quadratic equation whose roots are $a, \frac{1}{a}$ is:
- ✓
$a x^2-\left(a^2+1\right) x+a=0$
- B
$a x^2-\left(a^2-1\right) x+a=0$
- C
$a x^2-\left(a^2-1\right) x-a=0$
- D
AnswerCorrect option: A. $a x^2-\left(a^2+1\right) x+a=0$
(a) $a x^2-\left(a^2+1\right) x+a=0$
Explanation:
Quadratic equation whose roots are $a, \frac{1}{a}$ be
$\begin{array}{l}x^2-\left(a+\frac{1}{a}\right) x+1=0 \\ \Rightarrow x^2-\left(\frac{a^2+1}{a}\right) x+1=0 \\ \Rightarrow a x^2-\left(a^2+1\right) x+a=0\end{array}$
View full question & answer→MCQ 131 Mark
The roots of the equation $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7$ are:
Answer(d) 1, 2
Explanation:
We have,
$\begin{array}{l}\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30} \\ \Rightarrow \frac{x-7-(x+4)}{(x+4)(x-7)}=\frac{11}{30} \\ \Rightarrow \frac{-11}{x^2-3 x-28}=\frac{11}{30} \\ \Rightarrow-30=x^2-3 x-28 \\ \Rightarrow x^2-3 x+2=0 \\ \Rightarrow x^2-2 x-x+2=0 \\ \Rightarrow x(x-2)-1(x-2)=0 \\ \Rightarrow(x-1)(x-2)=0 \\ \Rightarrow x=1,2\end{array}$
View full question & answer→MCQ 141 Mark
If the sum of squares of two consecutive odd natural numbers is 290 , then the two numbers are:
Answer(a) 11,13
Explanation:
Let the two consecutive odd natural numbers be $x$ and $x+2$.
Then, according to the question,
$\begin{array}{l}x^2+(x+2)^2=290 \\ \Rightarrow 2 x^2+4 x+4=290 \\ \Rightarrow 2 x^2+4 x-286=0 \\ \Rightarrow x^2+2 x-143=0 \\ \Rightarrow x^2+13 x-11 x-143=0 \\ \Rightarrow(x+13)(x-11)=0 \\ \Rightarrow(x+13)=0 \text { or } x-11=0 \\ \Rightarrow x=-13 \text { or } 11\end{array}$
But $x$ is a natural number.
$\therefore x=11$
So, the two numbers are 11 and 13.
View full question & answer→MCQ 151 Mark
The solution of the equation $x^2+x+1=1$ is:
- ✓
$x=0,-1$
- B
$x=-1,2$
- C
$x=0,1$
- D
AnswerCorrect option: A. $x=0,-1$
(a) $x=0,-1$
Explanation:
Given equation is
$\begin{array}{l}x^2+x+1=1 \\ \Rightarrow x^2+x=0 \\ \Rightarrow x(x+1)=0 \\ \Rightarrow x=0 \text { or } x+1=0 \\ x=0 \text { or } x=-1 .\end{array}$
View full question & answer→MCQ 161 Mark
The roots of $x^2-(a+1) x+b^2=0$ are equal. Then choose the correct value of $a$ and $b$ from the following options:
AnswerCorrect option: C. $5,-3$
(c) $5,-3$
Explanation:
The roots of $x^2-(a+1) x+b^2=0$ are equal.
$\begin{array}{l} D = B ^2-4 AC =0 \\ \Rightarrow(a+1)^2-4 b^2=0 \\ \Rightarrow a+1= \pm 2 b\end{array}$
From the options
$a=5, b=-3$ satifes the above relation.
View full question & answer→MCQ 171 Mark
The sum of a number and its reciprocal is $\frac{10}{3}$. The number is:
Answer(c) both (a) and (b)
Explanation:
Let the number be $x$.
$x+\frac{1}{x}=\frac{10}{3}$
$\begin{array}{l}\Rightarrow \frac{x^2+1}{x}=\frac{10}{3} \\ \Rightarrow 3 x^2+3=10 x \\ \Rightarrow 3 x^2-10 x+3=0 \\ \Rightarrow 3 x^2-9 x-x+3=0 \\ \Rightarrow 3 x(x-3)-1(x-3)=0 \\ \Rightarrow(3 x-1)(x-3)=0 \\ \Rightarrow 3 x-1=\text { or } x-3=0\end{array}$
\x=3 or $x=\frac{1}{3}$
View full question & answer→MCQ 181 Mark
The age of a father is 25 years more than his son's age. The product of their ages is 84 in years. The son's age after 10 years is:
Answer(c) 13
Explanation:
Let age of son be $x$ years.
So, age of father be $(x+25)$ years.
According to the question,
$\begin{array}{l}x(x+25)=84 \\ \Rightarrow x^2+25 x-84=0 \\ \Rightarrow x^2+28 x-3 x+84=0 \\ \Rightarrow(x+28)(x-3)=0 \\ \Rightarrow x=-28 \text { or } x=3 \\ \backslash x=3\end{array}$
( $\because$ Age can not be negative)
Son age after 10 years $(3+10)=13$ years.
View full question & answer→MCQ 191 Mark
If $\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}$, then the value(s) of $x$ is/are:
AnswerCorrect option: D. $3, \frac{4}{3}$
(d) $3, \frac{4}{3}$
Explanation:
We have,
$\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}$
$\begin{array}{l}\Rightarrow \frac{6(x-1)-2 x}{x(x-1)}=\frac{1}{x-2} \\ \Rightarrow \frac{4 x-6}{x(x-1)}=\frac{1}{x-2} \\ \Rightarrow(4 x-6)(x-2)=x(x-1) \\ \Rightarrow 4 x^2-14 x+12=x^2-x \\ \Rightarrow 3 x^2-13 x+12=0 \\ \Rightarrow 3 x^2-9 x-4 x+12=0 \\ \Rightarrow 3 x(x-3)-4(x-3)=0 \\ \Rightarrow(3 x-4)(x-3)=0 \\ \Rightarrow x=\frac{4}{3}, 3\end{array}$
View full question & answer→MCQ 201 Mark
The roots of the equation $2 x^2+2 x=3$ are:
- ✓
$-1.8225,0.8225$
- B
$-0.7125,1.4225$
- C
$-1 \cdot 1115,1 \cdot 2225$
- D
$-1 \cdot 6445,0 \cdot 6295$
AnswerCorrect option: A. $-1.8225,0.8225$
(a) $-1.8225,0.8225$
Explanation:
We have,
$\begin{array}{l}2 x^2+2 x=3 \\ \Rightarrow 2 x^2+2 x-3=0\end{array}$
Using quadratic formula,
$\begin{array}{l}x=\frac{-2 \pm \sqrt{(2)^2-4 \times 2 \times(-3)}}{2 \times 2} \\ =\frac{-2 \pm \sqrt{4+24}}{4} \\ =\frac{-2 \pm \sqrt{25}}{4} \\ =\frac{-2 \pm 5.29}{4}=\frac{-2+5.29}{4}, \frac{-2-5.29}{4} \\ =0.8225,-1.8225\end{array}$
View full question & answer→MCQ 211 Mark
If $(x+1)(2 x+8)=(x+7)(x+3)$, then using factorisation method, the values of $x$ are:
- A
$\sqrt{12}, \sqrt{13}$
- B
$-\sqrt{13},-\sqrt{13}$
- ✓
$\pm \sqrt{13}$
- D
$\sqrt{13}, \sqrt{13}$
AnswerCorrect option: C. $\pm \sqrt{13}$
(c) $\pm \sqrt{13}$
Explanation:
We have,
$\begin{array}{l}(x+1)(2 x+8)=(x+7)(x+3) \\ \Rightarrow 2 x^2+10 x+8=x^2+10 x+21 \\ \Rightarrow x^2-13=0 \\ \Rightarrow(x)^2-(\sqrt{13})^2=0 \\ \Rightarrow(x-\sqrt{13})(x+\sqrt{13})=0 \\ \Rightarrow x=-\sqrt{13}, \sqrt{13} .\end{array}$
View full question & answer→MCQ 221 Mark
If $x=2$ and $x=3$ are the roots of the equation $3 x^2-2 m x+2 n=0$, then the values of $m, n$ respectively are:
AnswerCorrect option: A. $\frac{15}{2}, 9$
(a) $\frac{15}{2}, 9$
Explanation:
Since $x=2, x=3$ are the roots of
$3 x^2-2 m x+2 n=0$
$\therefore$ For $x=2$,
$\begin{array}{l}3(2)^2-2 m(2)+2 n=0 \\ \Rightarrow 12-4 m+2 n=0 \\ \Rightarrow 2 m-n=6 \ldots \text { (i) }\end{array}$
For $x=3$,
$3(3)^2-2 m(3)+2 n=0$
and $27-6 m+2 n=0$
and $6 m-2 n=27 \ldots$ (ii)
Solving the equations (i) and (ii) for $m$ and $n$, we get
$m=\frac{15}{2}, n=9$
View full question & answer→MCQ 231 Mark
If $a , b$ are the roots of $x^2+p x+q=0$, then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is:
- ✓
$\frac{p^2-2 q}{q}$
- B
$\frac{2 q-p^2}{q}$
- C
$\frac{p^2+2 q}{q}$
- D
AnswerCorrect option: A. $\frac{p^2-2 q}{q}$
(a) $\frac{p^2-2 q}{q}$
Explanation:
Here, $a+b=-p$
and $a b=q$
$\begin{array}{l}\backslash \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta} \\ =\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}=\frac{p^2-2 q}{q}\end{array}$
View full question & answer→MCQ 241 Mark
The value of $k$ for which the equation $x^2+4 k x+\left(k^2-k+2\right)=0$ has equal roots, is:
- A
$-1,-\frac{2}{3}$
- B
$-\frac{2}{3}, 1$
- ✓
$\frac{2}{3},-1$
- D
$1, \frac{2}{3}$
AnswerCorrect option: C. $\frac{2}{3},-1$
(c) $\frac{2}{3}, \quad 1$
Explanation :
Since, the given equation has equal roots,
$\therefore$ Discriminant $=0$
$\begin{array}{l}\Rightarrow b^2-4 a c=0 \\ \Rightarrow(4 k)^2-4 \times 1 \times\left(k^2-k+2\right)=0 \\ \Rightarrow 16 k^2-4 k^2+4 k-8=0 \\ \Rightarrow 12 k^2+4 k-8=0 \\ \Rightarrow 3 k^2+k-2=0 \\ \Rightarrow 3 k^2+3 k-2 k-2=0 \\ \Rightarrow 3 k(k+1)-2(k+1)=0 \\ \Rightarrow(3 k-2)(k+1)=0 \\ \Rightarrow 3 k-2=0 \text { or } k+1=0 \\ \Rightarrow k=\frac{2}{3} \text { or }-1\end{array}$
View full question & answer→MCQ 251 Mark
The roots of the equation $21 x^2-8 x-4=0$ are:
- A
$-\frac{1}{2}, \frac{1}{7}$
- B
$\frac{1}{3}, \frac{4}{9}$
- ✓
$\frac{2}{3},-\frac{2}{7}$
- D
$\frac{4}{7},-\frac{2}{3}$
AnswerCorrect option: C. $\frac{2}{3},-\frac{2}{7}$
(c) $\frac{2}{3},-\frac{2}{7}$
Explanation:
We have,
$21 x^2-8 x-4=0$
Using quadratic formula,
$\begin{array}{l}x=\frac{-(-8) \pm \sqrt{(-8)^2-4 \times 21 \times(-4)}}{2 \times 21} \\ =\frac{8 \pm \sqrt{64+336}}{42} \\ =\frac{8 \pm \sqrt{400}}{42}=\frac{8 \pm 20}{42} \\ =\frac{8+20}{42} \text { or } \frac{8-20}{42}=\frac{28}{42} \text { or } \frac{-12}{42} \\ =\frac{2}{3} \text { or } \frac{-2}{7} .\end{array}$
View full question & answer→MCQ 261 Mark
The roots of the equation $\sqrt{2 x+9}+x=13$ are:
- A
$-8,-20$
- B
$-8,20$
- ✓
- D
$8,-20$
Answer(c)8,20
Explanation:
$\begin{array}{l}\sqrt{2 x+9}+x=13 \\ \Rightarrow \sqrt{2 x+9}=13-x\end{array}$
On squaring both sides, we get
$\begin{array}{l}2 x+9=(13-x)^2 \\ 2 x+9=169+x^2-26 x \\ \Rightarrow x^2-26 x-2 x-9+169=0 \\ \Rightarrow x^2-28 x+160=0 \\ \Rightarrow x^2-20 x-8 x+160=0 \\ \Rightarrow x(x-20)-8(x-20)=0 \\ \Rightarrow(x-8)(x-20)=0 \\ \Rightarrow x=8 \text { or } 20 .\end{array}$
View full question & answer→MCQ 271 Mark
The roots of the quadratic equation $2 x^2-5 x-4=0$ are:
- A
- B
$4.84,-1.98$
- C
$7,-2$
- ✓
$3 \cdot 14,-0 \cdot 64$
AnswerCorrect option: D. $3 \cdot 14,-0 \cdot 64$
(d) X$3 \cdot 14,-0.64$
Explanation:
We have,
$2 x^2-5 x-4=0$
Using quadratic formula,
$\begin{array}{l}x=\frac{-(-5) \pm \sqrt{(-5)^2-4 \times 2 \times(-4)}}{2 \times 2} \\ =\frac{5 \pm \sqrt{25+32}}{4} \\ =\frac{5 \pm \sqrt{57}}{4}=\frac{5 \pm 7.55}{4} \\ =\frac{5+7.55}{4}, \frac{5-7.55}{4} \\ =3.14,-0.64\end{array}$
View full question & answer→MCQ 281 Mark
If $a$ is a natural number and one of the roots of the equation $3 x^2-14 x+8=0$, then the value of $a$ is:
Answer(a)4
Explanation:
We have,
$\begin{array}{l}3 x^2-14 x+8=0 \\ \Rightarrow 3 x^2-12 x-2 x+8=0 \\ \Rightarrow 3 x(x-4)-2(x-4)=0 \\ \Rightarrow(3 x-2)(x-4)=0 \\ \Rightarrow x=\frac{2}{3}, 4\end{array}$
Since, $a$ is a natural number.
$\therefore a=4 \text {. }$
View full question & answer→MCQ 291 Mark
One root of the quadratic equation $3 x^2-4 x-4=0$ is:
- A
$\frac{3}{2}$
- ✓
- C
$\frac{2}{3}$
- D
Answer(b)2
Explanation:
We have,
$\begin{array}{l}3 x^2-4 x-4=0 \\ \Rightarrow 3 x^2-6 x+2 x-4=0 \\ \Rightarrow 3 x(x-2)+2(x-2)=0 \\ \Rightarrow(3 x+2)(x-2)=0 \\ \Rightarrow x=-\frac{2}{3}, 2\end{array}$
View full question & answer→MCQ 301 Mark
If the equation $3 x^2-6 x+k=0$ has real and distinct roots, then the value of $k$ is:
- A
$k \leq 3$
- B
$k=3$
- C
$k>3$
- ✓
$k<3$
Answer(d) $k<3$
Explanation:
Since, the given equation has real and distinct (unequal) roots,
$\therefore D >0$
$\begin{array}{l}\Rightarrow(-6)^2-4 \times 3 \times k>0 \\ \Rightarrow 36-12 k>0 \\ \Rightarrow 36>12 k \\ \Rightarrow 3>k \text { or } k<3 .\end{array}$
View full question & answer→MCQ 311 Mark
If $x=2$ is a solution of the quadratic equation $k x^2+2 x-3=0$, then the value of $k$ is:
- A
- B
$\frac{1}{4}$
- C
- ✓
$-\frac{1}{4}$
AnswerCorrect option: D. $-\frac{1}{4}$
(d) $-\frac{1}{4}$
Explanation:
Since, $x=2$ is a solution of $k x^2+2 x-3=0$,
$\begin{array}{l}\therefore k(2)^2+2(2)-3=0 \\ \Rightarrow 4 k+4-3=0 \\ \Rightarrow 4 k+1=0 \\ \Rightarrow k=-\frac{1}{4} .\end{array}$
View full question & answer→MCQ 321 Mark
The equation $x^2+2 x+1=(4-k x)^2+3$ will be quadratic, if the value of $k$ is:
AnswerCorrect option: B. $k \neq \pm 1$
(b) $k \neq \pm 1$
Explanation:
We have,
$\begin{array}{l}x^2+2 x+1=(4-k x)^2+3 \\ \Rightarrow x^2+2 x+1=16-8 k x+k^2 x^2+3 \\ \Rightarrow x^2+2 x+1=k^2 x^2-8 k x+19 \\ \Rightarrow\left(k^2-1\right) x^2-2(4 k+1) x+18=0\end{array}$
Now, for this equation to be quadratic, coefficient of $x^2$ should not be zero.
i.e., $k^2-1 \neq 0$
$\Rightarrow k \neq \pm 1 .$
View full question & answer→MCQ 331 Mark
The number of possible solutions of a quadratic equation are:
Answer(b) at most two
Explanation:
A quadratic equation cannot have more than 2 solution, i.e., either 2,1 or 0 .
View full question & answer→MCQ 341 Mark
The discriminant (D) of a quadratic equation $a x^2+b x+c=0$ is given by the formula:
- A
$D =b-4 a c$
- B
$D =b^2+4 a c$
- ✓
$D =b^2-4 a c$
- D
$D =b+4 a c$
AnswerCorrect option: C. $D =b^2-4 a c$
(c)$D =b^2-4 a c$
View full question & answer→