Question 12 Marks
If $p-15=0$ and $2 x^2+p x+25=0$;find the values of $x$.
AnswerGiven $p - 15 = 0$ i.e $p - 15$
So, the given quadratic equation becomes
$2 x^2+15 x+25=0 $
$ \Rightarrow 2 x^2+10 x+5 x+25=0$
$ \Rightarrow 2 x(x+5)+5(x+5)=0$
$ \Rightarrow(x+5)(2 x+5)=0$
$ \Rightarrow x=-5,-\frac{5}{2}$
View full question & answer→Question 22 Marks
Solve $3 \sqrt{2 x^2}-5 x-\sqrt{2}=0$
AnswerGiven: $3 \sqrt{2 x^2}-5 x-\sqrt{2}=0$
$\Rightarrow 3 \sqrt{2 x^2}-6 x+x-\sqrt{2}=0 $
$\Rightarrow(3 \sqrt{2} x+1)(x-\sqrt{2})=0 $
$ \Rightarrow x=-\frac{1}{3} \sqrt{2} \text { or } x=\sqrt{2}$
View full question & answer→Question 32 Marks
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
$x^2+4 x+5=0$
AnswerGiven quadratic equation is
$x^2+4 x+5=0$
$D=b^2-4 a c={ }^`(4)^{\wedge} 2-4(1)(5)=16-20=-4$
Since D < 0, the roots of the given quadratic equation does not exist.
View full question & answer→Question 42 Marks
Find the value of k for which equation $4 x^2+8 x-k=0$ has real roots.
AnswerGiven quadratic equation is $4 x^2+8 x-k=0 \ldots$ (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
$\Rightarrow D=b^2-4 a c \geq 0 $
$ \Rightarrow 8^2-4(4)(-k) \geq 0 $
$ \Rightarrow 64+16 k \geq 0$
$ \Rightarrow 16 k \geq-64$
$ \Rightarrow k \geq-4$
Hence, the given quadratic equation has real roots for.
View full question & answer→Question 52 Marks
Solve :
$(x+5)(x-5)=24$
AnswerGiven $: (x+5)(x-5)=24$
$\Rightarrow x^2-5^2=24 \quad$............. since $(a-b)(a+b)=a^2-b^2$
$\Rightarrow x^2-25=24$
$ \Rightarrow x^2=49 $
$ \Rightarrow x= \pm 7$
View full question & answer→Question 62 Marks
One root of the quadratic equation $x^2+(3-2 a) x-6 a=0$ is $-3,$ find its other root.
AnswerGiven quadratic equation is $x^2+(3-2 a) x-6 a=0 \ldots (i)$
One of the roots of (i) is -3, so it satisfies $(i)$
$\Rightarrow x^2+3 x-2 a x-6 a=0$
$ \Rightarrow x(x+3)-2 a(x+3)=0$
$ \Rightarrow(x-2 a)(x+3)=0$
$ \Rightarrow x =-3,2 a $
Hence, the other root is $2a.$
View full question & answer→Question 72 Marks
Solve $a^2 x^2-b^2=0$
Answer$a^2 x^2-b^2=0$
$\Rightarrow(a x)^2-b^2=0 $
$\Rightarrow(a x+b)(a x-b)=0$
If $a x+b=0$ and $a x-b=0$
then $x=\frac{-b}{a}$ and $x=\frac{b}{a}$
View full question & answer→Question 82 Marks
Solve: $(2x+3)^2= 81$
Answer$(2 x+3)^2=81$
$\Rightarrow 2 x-3= \pm 9$
$\Rightarrow2 x+3=9$ and $2 x+3=-9$
$\Rightarrow2 x+3=9$ and $2 x=-12$
$\Rightarrow x=3$ and $x=-6$
View full question & answer→Question 92 Marks
Without solving the following quadratic equation Find the value of p for which the roots are equal
$p x^2-4 x+3=0$
Answer$p x^2-4 x+3=0$
Here $a = p, b = -4$ and $c = 3$
Given equation has equal roots then $D = 0$
$\Rightarrow b^2-4 a c=0 $
$ \Rightarrow[-4]^2-4(p)(3)=0$
$ \Rightarrow 16-12 p=0 $
$ \Rightarrow-12 p=-16 $
$ \Rightarrow p=\frac{-16}{-12}=\frac{4}{3}$
View full question & answer→Question 102 Marks
Solve $: (a+b)^2 x^2-(a+b) x-6=0 ; a+b \neq 0$
Answer$(a+b)^2 x^2-(a+b) x-6=0 ; a+b \neq 0$
$\Rightarrow(a+b)^2 x^2-3(a+b) x+2(a+b) x-6=0$
$ \Rightarrow(a+b) x[(a+b) x-3]+2[(a+b) x-3]=0$
$\Rightarrow[(a+b) x-3][(a+b) x+2]=0 $
$ \Rightarrow (a+b) x-3=0 \text { or }(a+b) x+2=0$
$\Rightarrow x=\frac{3}{a+b}$ or $x=\frac{-2}{a+b}$
View full question & answer→Question 112 Marks
Solve $2x^2 – 9x + 10 =0$; when $x \in Q$
Answer$2 x^2-9 x+10=0$
$\Rightarrow 2 x^2-5 x-4 x-10=0 $
$\Rightarrow x(2 x-5)-2(2 x-5)=0 $
$\Rightarrow(2 x-5)(x-2)=0$
then $x = 5/2$ and $x = 2$
Since $x \in Q$ then $x=\frac{5}{2}$ and $2$
View full question & answer→Question 122 Marks
Solve $x^2 – 4x – 12 =0$; when $x \in I$
Answer$x^2-4 x-12=0$
$\Rightarrow x^2-6 x+2 x-12=0$
$ \Rightarrow x(x-6)+2(x-6)=0 $
$ \Rightarrow ( x -6)( x +2)=0$
then $x = 6$ and $x = -2$
Since $x \in I$ then $x =6$ and $-2$
View full question & answer→Question 132 Marks
Solve : $x^2 – 11x – 12 =0$; when $x \in N$
Answer$x^2-11 x-12=0$
$\Rightarrow x^2-12 x+x-12=0 $
$\Rightarrow x(x-12)+1(x-12)=0 $
$ \Rightarrow(x-12)(x+1)=0$
then $x=12$ and $x=-1$
Since $x \in N$ then $x =12$
View full question & answer→Question 142 Marks
Solve equation using factorisation method:$(2x - 3)^2 = 49$
Answer$(2x - 3)^2 = 49$ Taking square root on both sides
$2x - 3 = ± 7$
When $2x - 3 = 7 \Rightarrow 2x = 10 \Rightarrow x = 5$
and, when $2x - 3 = -7 \Rightarrow 2x = -4 \Rightarrow x = -2$
View full question & answer→Question 152 Marks
Solve equation using factorisation method:
$x=\frac{3 x+1}{4 x}$
Answer$x=\frac{3 x+1}{4 x}$
⇒ 4x2 = 3x + 1
⇒ 4x2 - 3x - 1 = 0
⇒ 4x2 - 4x + x - 1 = 0
⇒ 4x( x -1) +1 (x -1) = 0
⇒ (x -1)(4x + 1) = 0
since x -1=0 or 4x + 1= 0
then $x=1$ or $x=\frac{-1}{4}$
View full question & answer→Question 162 Marks
Solve equation using factorisation method:
$\frac{6}{x}=1+x$
Answer$\frac{6}{x}=1+x$
$\Rightarrow 6 = x + x^2$
$\Rightarrow x^2 + x - 6 = 0$
$\Rightarrow x2 + 3x - 2x - 6 = 0$
$\Rightarrow x(x +3) -2 (x + 3) = 0$
$\Rightarrow (x + 3) (x -2) = 0$
Since $x + 3 = 0$ or $x -2 = 0$
then $x = -3$ or $x = 2$
View full question & answer→Question 172 Marks
Solve equation using factorisation method:
$2 x^2=\frac{1}{2} x=0$
Answer$2 x^2-\frac{1}{2} x=0$
$ \Rightarrow x\left(2 x-\frac{1}{2}\right)=0$
Since $x=0$ or $2 x-\frac{1}{2}=0$
then $x =0$ or $x =\frac{1}{4}$
View full question & answer→Question 182 Marks
Solve equation using factorisation method:
$x^2 - 16 = 0$
Answer$x^2 - 16 = 0$
$\Rightarrow x^2 - 4^2 = 0$
$\Rightarrow (x + 4) (x - 4) = 0$
Since $x + 4 = 0$ or $x - 4 = 0$
then $x = -4$ or $x = 4$
View full question & answer→Question 192 Marks
Determine whether $x = -1$ is a root of the equation $x^2 - 3x +2=0$ or not.
Answer$x^2 - 3x +2=0$
Put $x = -1$ in L.H.S.
$L.H.S. = (-1)^2 - 3(-1) +2$
$= 1 +3 +2=6 \neq R.H.S.$
Then $x = -1$ is not the solution of the given equation.
View full question & answer→Question 202 Marks
Find the value of $x$, if $a + 1=0$ and $x^2 + ax - 6 =0$.
AnswerIf $a+1=0,$ then $a = -1$
Put this value in the given equation $x^2 + ax - 6 =0$
$x2 - x - 6 = 0$
$\Rightarrow x2 - 3x + 2x - 6 = 0$
$\Rightarrow x( x - 3) + 2 (x - 3) = 0$
$\Rightarrow ( x -3) (x + 2) = 0$
If $x -3 = 0$ or $x + 2=0$
then $x = 3$ or $x = -2$
View full question & answer→Question 212 Marks
Find the quadratic equation, whose solution set is :
(-2,3}
AnswerSince solution set is $\{-2,3)$
$\Rightarrow x = -2$ or $x = 3$
$\Rightarrow x + 2 = 0$ or $x - 3 = 0$
$\Rightarrow (x + 2)(x -3) = 0$
$\Rightarrow x^2 - 3x + 2x - 6 = 0$
$\Rightarrow x^2 - x - 6 = 0$
which is the required equation.
View full question & answer→Question 222 Marks
Find the quadratic equation, whose solution set is :
${3,5}$
AnswerSince solution set is
$\Rightarrow x = 3$ or $x = 5$
$\Rightarrow x -3 = 0$ or $x -5 = 0$
$\Rightarrow (x - 3) (x - 5) = 0$
$\Rightarrow x^2 - 5x - 3x + 15 = 0$
$\Rightarrow x^2 - 8x + 15 = 0$ which is the required equation.
View full question & answer→Question 232 Marks
Solve equation using factorisation method:
$x^2 - 10x - 24 = 0$
Answer$x^2 - 10x - 24 = 0$
$\Rightarrow x^2 - 12x + 2x - 24 = 0$
$\Rightarrow x(x - 12) + 2(x - 12) = 0$
$\Rightarrow (x -12) (x + 2) = 0$
Since $x - 12 = 0$ or $x + 2 = 0$
then $x = 12$ or $x = -2$
View full question & answer→Question 242 Marks
Solve equation using factorisation method:$2x^2 - 9x + 10 = 0$, When
(i) $x\in N$
(ii) $x\in Q$
Answer$2x^2 - 9 + 10 = 0$
$\Rightarrow 2 x^2-5 x-4 x+10=0$
$\Rightarrow x(2 x-5)-2(2 x-5)=0$
$\Rightarrow(2 x-5)(x-2)=0$
$\Rightarrow 2 x-5=0 \text { or } x-2=0$
$\Rightarrow x=\frac{5}{2} \text { or } x=2$
(i) when $x \in N$ we have $x =2$
(ii) When $x \in Q$ we have $x =2, \frac{5}{2}$
View full question & answer→Question 252 Marks
Solve equation using factorisation method:
$x^2 - (a + b) x + ab = 0$
Answer$x^2 - (a + b) x + ab = 0$
$\Rightarrow x^2 - ax - bx + ab = 0$
$\Rightarrow x (x -a) -b(x -a) = 0$
$\Rightarrow (x -a) (x -b) = 0$
since $x -a = 0$ or $x -b = 0$
then $x = a$ or $x = b$
View full question & answer→Question 262 Marks
Solve equation using factorisation method $: 2\left(x^2-6\right)=3(x-4)$
Answer$2\left(x^2-6\right)=3(x-4)$
$\Rightarrow 2 x^2-12=3 x-12$
$ \Rightarrow 2 x^2-3 x=0 $
$ \Rightarrow x(2 x-3)=0$
Since $x=0$ or $2 x-3=0$
then $x =0$ or $x =\frac{3}{2}$
View full question & answer→Question 272 Marks
Find the value of k for which the equation $3x^2- 6x + k = 0$ has distinct and real roots.
Answer$3x^2- 6x + k = 0$
Here, $a = 3, b = -6$ and $c = k$
Since the roots are distinct and real,
$b2 - 4ac > 0$
$\Rightarrow (-6)^2- 4 x 3 x k > 0$
$\Rightarrow 36 > 12k$
$\Rightarrow 3 > k$
$\Rightarrow K< 3$
View full question & answer→Question 282 Marks
The equation $3 x^2-12 x+(n-5)=0$ has equal roots. Find the value of $n$.
Answer$3 x^2-12 x+(n-5)=0$
Here $a = 3, b = − 12$ and $c = n – 5$
Given: equation has equal roots
Then $D = 0$
$\Rightarrow b^2-4 a c=0 $
$ \Rightarrow[-12]^2-4(3)(n-5)=0$
$ \Rightarrow 144-12 n +60=0 $
$ \Rightarrow-12 n =-204 $
$ \Rightarrow n =\frac{-204}{-12}=17$
View full question & answer→Question 292 Marks
Without solving comment upon the nature of roots of each of the following equation:
$2x^2 + 8x + 9 = 0$
Answer$2 x^2+8 x+9=0$
$a=2, b=8$ and $c=9$
$\therefore \text { Discriminant }= b ^2-4 ac$
$=(8)^2-4(2)(9) $
$ =64-72 $
$=-8=\text { a negative value }$
Since $D < 0,$ then the equation has imaginary roots.
View full question & answer→Question 302 Marks
Without solving comment upon the nature of roots of each of the following equations:
$x ^2- ax - b ^2=0$
Answer$x ^2- ax - b ^2=0$
$a =1, b =- a$ and $c =- b ^2$
$\therefore \text { Discriminant }=b^2-4 a c$
$ =(-a)^2-4(1)(-b) 2$
$\ =a^2+4 b^2=\text { a positive value }$
Since $a ^2+4 b ^2$ is not a perfect square, roots are irrational
Since $D>0$, then equation has two real and unequal roots.
View full question & answer→Question 312 Marks
Without solving comment upon the nature of roots of each of the following equations : $x ^2+2 \sqrt{3} x -9=0$
Answer$a = 1, b = 2\sqrt3$ and $c = − 9$
$\therefore \text { Discriminant }= b ^2-4 ac $
$=(2 \sqrt{3}) 2-4(1)(-9) $
$ =12+36=48$
Since $48$ is not a perfect square, roots are irrational
Since $D >0,$ then equation has two real and unequal roots
View full question & answer→Question 322 Marks
Without solving comment upon the nature of roots of each of the following equation:
$25 x^2-10 x+1=0$
Answer$25 x^2-10 x+1=0$
$a=25, b=-10$ and $c=1$
$\therefore$ Discriminant $= b ^2-4 ac$
$=(-10)^2-4(25)(1)$
$ \begin{aligned} & =100-100 \\ & =0 \end{aligned} $
Since $D=0$, then equation has real and equal roots.
View full question & answer→Question 332 Marks
Without solving comment upon the nature of roots of each of the following equations $:6x^2 – 13x + 4 = 0$
Answer$6 x^2-13 x+4=0$
$a=6, b=-13 \text { and } c=4$
$\therefore \text { Discriminant }=b 2-4 a c$
$=(-13)^2-4(6)(4)$
$=169-96=73$
$= 169 – 96 = 73$
Since $73$ is not a perfect square, roots are irrational
Since $D >0,$ then equation has two real and unequal roots.
View full question & answer→Question 342 Marks
Without solving comment upon the nature of roots of each of the following equations $:7x^2 – 9x + 2 = 0$
Answer$a = 7, b = − 9$ and $c = 2$
$\therefore \text { Discriminant }= b ^2-4 ac$
$=(-9)^2-4(7)(2)$
$= 81 – 56 = 25$
Since $D >0,$ then equation has two real and unequal roots.
View full question & answer→Question 352 Marks
Is $x = -3$ a solution of the quadratic equation $2x^2 - 7x + 9 = 0$?
Answer$2x^2 - 7x + 9 = 0$
For $x = -3$ to be solution of the given quadratic equation it should satisfy the equation
So, substituting $x = 5$ in the given equation, we get
$L.H.S=2(-3)^2 - 7(-3) + 9$
$= 18 + 21 + 9$
$= 48$
$\neq R.H.S$
Hence, $x = -3$ is not a solution of the quadratic equation $2x^2 - 7x + 9 = 0.$
View full question & answer→Question 362 Marks
Is $x = 5$ a solution of the quadratic equation $x^2 - 2x - 15 = 0$?
Answer$x^2 - 2x - 15 = 0$
For $x = 5$ to be solution of the given quadratic equation it should satisfy the equation.
So, substituting $x = 5$ in the given equation, we get
$L.H.S = (5)^2 - 2(5) - 15$
$= 25 - 10 - 15$
$= 0$
$= R.H.S$
Hence, $x = 5$ is a solution of the quadratic equation $x^2 - 2x - 15 = 0$.
View full question & answer→Question 372 Marks
Find which of the following equation are quadratic:
$(x - 1)^2 + (x + 2)^2 + 3(x +1) = 0$
Answer$(x - 1)^2 + (x + 2)^2 + 3(x +1) = 0$
$\Rightarrow x^2 - 2x + 1 + x^2 + 4x + 4 + 3x + 3 = 0$
$\Rightarrow 2x^2 + 5x + 8 = 0$;
which is of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is a quadratic equation.
View full question & answer→Question 382 Marks
Find which of the following equations are quadratic:
$7x^3 - 2x^2 + 10 = (2x - 5)^2$
Answer$7x^3 - 2x^2 + 10 = (2x - 5)^2$
$\Rightarrow 7x^3 - 2x^2 + 10 = 4x^2 - 20x + 25$
$\Rightarrow 7x^3 - 6x^2 + 20x - 15 = 0$;
which is not of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is not a quadratic equation.
View full question & answer→Question 392 Marks
Find which of the following equation are quadratic:
$x^2 + 5x - 5 = (x - 3)^2$
Answer$x^2 + 5x - 5 = (x - 3)^2$
$\Rightarrow x^2 + 5x - 5 = x^2 - 6x + 9$
$\Rightarrow 11x - 14 =0$; which is not of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is not a quadratic equation.
View full question & answer→Question 402 Marks
Find which of the following equation are quadratic:
$(x - 4)(3x + 1) = (3x - 1)(x +2)$
Answer$(x - 4)(3x + 1) = (3x - 1)(x +2)$
$\Rightarrow 3x^2 + x - 12x - 4 = 3x^2 + 6x - x - 2$
$\Rightarrow 16x + 2 =0$; which is not of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is not a quadratic equation.
View full question & answer→Question 412 Marks
Find which of the following equation are quadratic:
$5x^2 - 8x = -3(7 - 2x)$
Answer$5x^2 - 8x = -3(7 - 2x)$
$\Rightarrow 5x^2 - 8x = 6x - 21$
$\Rightarrow 5x^2 - 14x + 21 =0;$ which is of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is a quadratic equation.
View full question & answer→Question 422 Marks
Find which of the following equation are quadratic:
$(3x - 1)^2 = 5(x + 8)$
Answer$(3x - 1)^2 = 5(x + 8)$
$\Rightarrow (9x^2 - 6x + 1) = 5x + 40$
$\Rightarrow 9x^2 - 11x - 39 =0;$ which is of the form $ax^2 + bx + c = 0.$
$\therefore $ Given equation is a quadratic equation.
View full question & answer→