Question 11 Mark
Find the nature of the roots of the following quadratic equations: $x^2-2 \sqrt{3} x-1=0$ If real roots exist, find them.
Answer$
x^2-2 \sqrt{3} x-1=0
$
Here $a=1, b=-2 \sqrt{3}, c=-1$
$
\therefore D = b ^2-4 ac
$
$
=(-2 \sqrt{3})^2-4 \times 1 \times(-1)
$
$
=12+4
$
$
=16
$
$\therefore D >0$
$\therefore$ Roots are real and unequal.
View full question & answer→Question 21 Mark
Find the nature of the roots of the following quadratic equations: $x^2-\frac{1}{2} x-\frac{1}{2}=0$
Answer$
x^2-\frac{1}{2} x-\frac{1}{2}=0
$
Here $a=1, b=-\frac{1}{2}, c=-\frac{1}{2}$
$
\therefore D = b ^2-4 ac
$
$
\begin{aligned}
& =\left(\frac{-1}{2}\right)^2-4 \times 1 \times\left(\frac{-1}{2}\right) \\
& =\frac{1}{4}+2 \\
& =\frac{9}{4}
\end{aligned}
$
$
\therefore D =\frac{9}{4}>0
$
$\therefore$ Roots are real and unequal.
View full question & answer→Question 31 Mark
Discuss the nature of the roots of quadratic equations :$2 \sqrt{3} x^2-5 x+\sqrt{3}=0$
Answer$
\begin{aligned}
& 2 \sqrt{3} x^2-5 x+\sqrt{3}=0 \\
& \text { Here } a=2 \sqrt{3}, b=-5, c=\sqrt{3} \\
& \therefore D=b^2-4 a c \\
& =(-5)^2-4 \times 2 \sqrt{3} \times \sqrt{3} \\
& =25-24 \\
& =1 \\
& \because D>0
\end{aligned}
$
$\therefore$ Roots are real and distinct.
View full question & answer→Question 41 Mark
Discuss the nature of the roots of quadratic equations : $-2x^2 + x + 1 = 0$
Answer$-2x^2 + x + 1 = 0$
Here $a = -2, b = 1, c = 1$
$\therefore D = b^2 - 4ac$
$= (1)^2 - 4 x (-2) x 1$
$= 1 + 8$
$= 9$
$\because D > 0$
$\therefore $ Roots are real and distinct.
View full question & answer→Question 51 Mark
Discuss the nature of the roots of quadratic equations :$x^2-\frac{1}{2} x+4=0$
Answer$
x^2-\frac{1}{2} x+4=0
$
Here $a=1, b=-\frac{1}{2}, c=1$
$
\therefore D = b ^2-4 ac
$
$=\left(-\frac{1}{2}\right)-4 \times 1 \times 4$
$=\frac{1}{4}-16$
$=-\frac{63}{4}$
$\because D<0$
$\therefore$ Roots are not real.
View full question & answer→Question 61 Mark
Discuss the nature of the roots of quadratic equations : $3 x^2-4 \sqrt{3} x+4=0$
Answer$
\begin{aligned}
& 3 x^2-4 \sqrt{3} x+4=0 \\
& \text { Here } a=3, b=-4 \sqrt{3}, c=4 \\
& \therefore D=b^2-4 a c \\
& =(-4 \sqrt{3})^2-4 \times 3 \times 4 \\
& =48-48 \\
& =0 \\
& \because D=0
\end{aligned}
$
$\therefore$ Roots are real and equal.
View full question & answer→Question 71 Mark
Discuss the nature of the roots of quadratic equations : $3 x^2-2 x+\frac{1}{3}=0$
Answer$
3 x^2-2 x+\frac{1}{3}=0
$
Here $a=3, b=-2, c=\frac{1}{3}$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(-2)^2-4 \times 3 \times \frac{1}{3} \\
& =4-4 \\
& =0 \\
& \because D=0
\end{aligned}
$
$\therefore$ Roots are real and equal.
View full question & answer→Question 81 Mark
Discuss the nature of the roots of quadratic equations : $x^2 – 4x – 1 = 0$
Answer$x^2 – 4x – 1 = 0$
Here $a = 1, b = -4, c = -1$
$\therefore D = b^2 - 4ac$
$= (-4)^2 - 4 x 1 x (-1)$
$= 16 + 4$
$= 20$
$\because D > 0$
Roots are real and distinct.
View full question & answer→Question 91 Mark
Find the discriminant of the following equations and hence find the nature of roots: $2x^2 + 15x + 30 = 0$
Answer$2x^2 + 15x + 30 = 0$
Here $a = 2, b = 15, c = 30$
$\therefore D = b^2 - 4ac$
$= (15)^2 - 4 \times 2 \times 30$
$= 225 - 240$
$= -15$
$∴$ Discriminant $= -15$
$\because D < 0$
$\therefore $ Root are not real.
View full question & answer→Question 101 Mark
Find the discriminant of the following equations and hence find the nature of roots: $16x^2 - 40x + 25 = 0$
Answer$16x^2 - 40x + 25 = 0$
$a = 16, b = -40, c = 25$
$\therefore D = b^2 - 4ac$
$= (-40)^2 - 4 \times 16 \times 25$
$= 1600 - 1600$
$= 0$
$\therefore $ Discriminant $= 0$
$\because $ D $= 0$
$\therefore $ Roots are real and equal.
View full question & answer→Question 111 Mark
Find the discriminant of the following equations and hence find the nature of roots: $3x^2 + 2x - 1 = 0$
Answer$3x^2 + 2x - 1 = 0$
Here $a = 3, b = 2, c = -1$
$\therefore D = b^2 - 4ac$
$= (2)^2 - 4 x 3 x (-1)$
$= 4 + 12$
$= 16$
$\therefore $ Discriminant $= 16$
$\because D > 0$
$\therefore $ Roots are real and distinct.
View full question & answer→Question 121 Mark
Find the discriminant of the following equations and hence find the nature of roots: $7x^2 + 8x + 2 = 0$
Answer$7x^2 + 8x + 2 = 0$
Here $a = 7, b = 8, c = 2$
$\therefore D = b^2 - 4ac$
$= (8)^2 - 4 x 7 x 2$
$= 64 - 56$
$= 8$
$\therefore $ Discriminant $= 8$
$\because D > 0$
$\therefore $ Roots are real and distinct.
View full question & answer→Question 131 Mark
Find the discriminant of the following equations and hence find the nature of roots: $2x^2– 3x + 5 = 0$
Answer$2x^2– 3x + 5 = 0$
Here $a = 2, b = -3, c = 5$
$\therefore D - b^2 - 4ac$
$= (-3)^2 - 4 x 2 x 5$
$= 9 - 40$
$= -31$
$\therefore $ Discriminant $= -31$
$\because D < 0,$
$\therefore $ Roots are not real.
View full question & answer→Question 141 Mark
Find the discriminant of the following equations and hence find the nature of roots: $3x^2 – 5x – 2 = 0$
Answer$3x^2 – 5x – 2 = 0$
Here $a = 3, b = -5, c = -2$
$\therefore D = b^2 - 4ac$
$= (-5)^2- 4 x 3 x (-2)$
$= 25 + 24$
$= 49$
$\therefore $ Discriminant $= 49$
$\therefore D > 0$
$\because $ Roots are real and distinct.
View full question & answer→Question 151 Mark
Solve equation by factorization
(x – 3) (2x + 5) = 0
Answer$
(x-3)(2 x+5)=0
$
Either $x-3=0$,
Then $x=3$
or
$
\begin{aligned}
& 2 x+5=0 \text { then } 2 x=-5 \\
& \Rightarrow x=\frac{-5}{2}
\end{aligned}
$
Hence $x=3,=\frac{-5}{2}$.
View full question & answer→Question 161 Mark
Solve equation by factorization$x^2-(1+\sqrt{2}) x+\sqrt{2}=0$
Answer$
\begin{aligned}
& x^2-(1+\sqrt{2}) x+\sqrt{2}=0 \\
& \Rightarrow x^2-x-\sqrt{2} x+\sqrt{2}=0 \\
& \Rightarrow(x-1)-\sqrt{2}(x-1)=0 \\
& \Rightarrow(x-1)(x-\sqrt{2})=0
\end{aligned}
$
Either $x-1=0$,
then $x=1$
or
$
x-\sqrt{2}=0 \text {, }
$
then $x =\sqrt{2}$
Hence $x=1, \sqrt{2}$.
View full question & answer→Question 171 Mark
Solve equation by factorization$\frac{x^2-5 x}{2}=0$
Answer$
\begin{aligned}
& \frac{x^2-5 x}{2}=0 \\
& x^2-5 x=0 \\
& \Rightarrow x(x-5)=0
\end{aligned}
$
Either $x =0$ or $x -5=0$,
then $x=5$
Hence $x=0,5$.
View full question & answer→Question 181 Mark
Solve equation by factorization
$4x^2 = 3x$
Answer$
\begin{aligned}
& 4 x^2=3 x \\
& x(4 x-3)=0
\end{aligned}
$
Either $x=0$,
or
$
\begin{aligned}
& 4 x-3=0 \\
& \text { then } 4 x=3
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow x=\frac{3}{4} \\
& \therefore x=0, \frac{3}{4} .
\end{aligned}
$
View full question & answer→Question 191 Mark
If a is a root of the equation $x^2 – (a + b)x + k = 0$, find the value of $k$.
Answer$x^2 – x(a + b)x + k = 0, x = a$
$\therefore x = a$ is its solution
$\therefore (a)^2- a(a + b) + k = 0$
$\Rightarrow a^2 - a^2 - ab + k = 0$
$\Rightarrow -ab + k = 0$
$\therefore k = ab.$
View full question & answer→Question 201 Mark
If $\sqrt{2}$ is a root of the equation $k x^2+\sqrt{2}-4=0$, find the value of $k$.
Answer$
k x^2+\sqrt{2}-4=0, x=\sqrt{2}
$
$x =\sqrt{2}$ is its solution
$
\begin{aligned}
& \therefore k (\sqrt{2})^2+\sqrt{2} \times \sqrt{2}-4=0 \\
& \Rightarrow 2 k +2-4=0 \\
& \Rightarrow 2 k -2=0 \\
& \Rightarrow 2 k =2 \\
& \Rightarrow k =\frac{2}{2}=1 \\
& \therefore k =1 .
\end{aligned}
$
View full question & answer→Question 211 Mark
In each of the following, determine whether the given numbers are roots of the given equations or not; $x^2 – x + 1 = 0; 1, – 1$
Answer$x^2 – x + 1 = 0; 1, – 1$
Where $x = 1$, then
$(1)^2 – 1 + 1 = 1 – 1 + 1 = 1 \neq 0$
$\therefore x = 1$ does not satisfy it
and $(-1)^2 - (-1) + 1 = 0$
$1 + 1 + 1 \Rightarrow 3 \neq 0$
$\therefore x = -1$, does not satisfy it
$\therefore x = 1, -1$ are not roots of the equation.
View full question & answer→Question 221 Mark
Check whether the following are quadratic equations:$x^2+\frac{1}{x^2}=3, x \neq 0$
Answer$
\begin{aligned}
& x^2+\frac{1}{x^2}=3, x \neq 0 \\
& x^4+6=3 x^2 \\
& x^4-3 x^2+6=0
\end{aligned}
$
It is not a quadratic equation.
View full question & answer→Question 231 Mark
Check whether the following are quadratic equations: $x+\frac{2}{x}=x^2, x \neq 0$
Answer$
\begin{aligned}
& x+\frac{2}{x}=x^2, x \neq 0 \\
& x^2+2=x^3 \Rightarrow x^3-x^2-2=0
\end{aligned}
$
It is not a quadratic equation.
View full question & answer→Question 241 Mark
Check whether the following are quadratic equations: $x-\frac{3}{x}=2, x \neq 0$
Answer$
\begin{aligned}
& x-\frac{3}{x}=2, x \neq 0 \\
& x^2-3=2 x \Rightarrow x^2-2 x-3=0
\end{aligned}
$
It is a quadratic equation.
View full question & answer→Question 251 Mark
Check whether the following are quadratic equations:$(x-3)^3+5=x^3+7 x^2-1$
Answer$
\begin{aligned}
& (x-3)^3+5=x^3+7 x^2-1 \\
& x^3-3 x^2 \times 3+3 x \times 9-27+5=x^3+7 x^2-1 \\
& -9 x^2+27 x-22-7 x^2+1=0 \\
& -16 x^2+27 x-21=0 \\
& \Rightarrow 16 x^2-27 x+21=0
\end{aligned}
$
It is a quadratic equation.
View full question & answer→Question 261 Mark
Check whether the following are quadratic equations: $(2x + 1) (3x – 2) = 6(x + 1) (x – 2)$
Answer$(2x + 1) (3x – 2) = 6(x + 1) (x – 2)$
$6x^2– 4x + 3x – 2 = 6(x^2 – 2x + x – 2)$
$6x^2 – x – 2 = 6x^2 – 12x + 6x – 12$
$12x - 6x - x = -12 + 2$
$5x = -10.$
View full question & answer→Question 271 Mark
Check whether the following are quadratic equations:$\sqrt{3} x^2-2 x+\frac{3}{5}=0$
Answer$
\sqrt{3} x^2-2 x+\frac{3}{5}=0
$
It is a quadratic equation as it is power of 2 .
View full question & answer→Question 281 Mark
Solve the following equation by factorisation :
$2x^2 + ax – a^2= 0$
Answer$
\begin{aligned}
& 2 x^2+a x-a^2=0 \\
& \Rightarrow 2 x^2+2 a x-a x-a^2=0 \\
& \Rightarrow 2 x(x+a)-a(x+a)=0 \\
& \Rightarrow(x+a)(2 x-a)=0
\end{aligned}
$
Either $x+a=0$,
then $x=-a$
or
$2 x-a=0 \text {, }$
then $2 x = a$
$\Rightarrow x =\frac{a}{2}$
Hence $x =- a, \frac{a}{2}$.
View full question & answer→Question 291 Mark
Solve the following equation by factorisation :
$3x^2+ 11x + 10 = 0$
Answer$
\begin{aligned}
& 3 x^2+11 x+10=0 \\
& \Rightarrow 3 x^2+6 x+5 x+10=0 \\
& 3 x(x+2)+5(x+2)=0 \\
& \Rightarrow(x+2)(3 x+5)=0
\end{aligned}
$
Either $x+2=0$,
then $x=-2$
or
$
3 x+5=0 \text {, }
$
then $3 x=-5$
$
\Rightarrow x =\frac{-5}{3}
$
Hence $x=-2, \frac{-5}{3}$.
View full question & answer→Question 301 Mark
Solve the following equation by factorisation :
$x^2 + 6x – 16 = 0$
Answer$x^2 + 6x – 16 = 0$
$\Rightarrow x^2 + 8x – 2x – 16 = 0$
$\Rightarrow x(x + 8) – 2(x + 8) = 0$
$\Rightarrow (x + 8)(x - 2) = 0$
Either $x + 8 = 0,$
then $x = –8$
or
$x – 2 = 0,$
then $x = 2$
Hence $x = –8, 2.$
View full question & answer→Question 311 Mark
Discuss the nature of the roots of equation: $\sqrt{3} x^2-2 x-\sqrt{3}=0$
Answer$
\begin{aligned}
& \sqrt{3} x^2-2 x-\sqrt{3}=0 \\
& \text { Here } a=\sqrt{3}, b=-2, c=-\sqrt{3} \\
& \therefore D=b^2-4 a c \\
& =(-2)^2-4 \times \sqrt{3} \times(-\sqrt{3}) \\
& =4+12 \\
& =16 \\
& \because D>0
\end{aligned}
$
$\therefore$ Roots are real and distinct.
View full question & answer→Question 321 Mark
Discuss the nature of the roots of equation: $5 x^2-6 \sqrt{5} x+9=0$
Answer$
\begin{aligned}
& 5 x^2-6 \sqrt{5} x+9=0 \\
& \text { Here } a=5, b=-6 \sqrt{5}, c=9 \\
& \therefore D=b^2-4 a c \\
& =(-6 \sqrt{5})^2-4 \times 5 \times 9 \\
& =180-180 \\
& =0 \\
& \therefore D=0
\end{aligned}
$
$\therefore$ Roots are real and equal.
View full question & answer→Question 331 Mark
Discuss the nature of the roots of equation: $3x^2 – 7x + 8 = 0$
Answer$3x^2 – 7x + 8 = 0$
Here $a = 3, b = –7, c = 8$
$\therefore D = b^2 – 4ac$
$= (–7)^2 – 4 \times 3 \times 8$
$= 49 – 96$
$= –47$
$\because D < 0$
$\therefore$ Roots are not real.
View full question & answer→