SELINA Solutions for Class 10 Maths Chapter 5 - Quadratic Equations
Aim for top marks in the prelim and board exams with the Selina Solutions for ICSE Class 10 Mathematics Chapter 5 Quadratic Equations. Improve your analytical abilities to prove whether the given equation is a quadratic equation or not using our textbook solutions. Also, learn the concept of two equal roots and two distinct real roots through Maths practice.
Revise the quadratic formula and the factorisation method to prove quadratic equations with the support of our ICSE Class 10 Maths textbook solutions. To revisit the problem-solving methods, you can watch our concept videos or go through our well-written revision notes.
Chapter 5 - Quadratic Equations Exercise Ex. 5(A)
x2 + 5x - 5 = (x - 3)2
x2 + 5x - 5 = (x - 3)2
⇒ x2 + 5x - 5 = x2 - 6x + 9
⇒ 11x - 14 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
7x3 - 2x2 + 10 = (2x - 5)2
7x3 - 2x2 + 10 = (2x - 5)2
⇒ 7x3 - 2x2 + 10 = 4x2 - 20x + 25
⇒ 7x3 - 6x2 + 20x - 15 = 0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
(x - 1)2 + (x + 2)2 + 3(x +1) = 0
(x - 1)2 + (x + 2)2 + 3(x +1) = 0
⇒ x2 - 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0
⇒ 2x2 + 5x + 8 = 0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
Find which of the following equations are quadratic:
(3x - 1)2 = 5(x + 8)
(3x - 1)2 = 5(x + 8)
⇒ (9x2 - 6x + 1) = 5x + 40
⇒ 9x2 - 11x - 39 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
5x2 - 8x = -3(7 - 2x)
5x2 - 8x = -3(7 - 2x)
⇒ 5x2 - 8x = 6x - 21
⇒ 5x2 - 14x + 21 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
(x - 4)(3x + 1) = (3x - 1)(x +2)
(x - 4)(3x + 1) = (3x - 1)(x +2)
⇒ 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2
⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Is x = 5 a solution of the quadratic equation x2 - 2x - 15 = 0?
x2 - 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 x2 - 2x - 15 = 0.
Is x = -3 a solution of the quadratic equation 2x2 - 7x + 9 = 0?
2x2 - 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
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0.
If is a solution of
equation 3x2 + mx + 2 = 0, find the value of m.
For x = to be solution of
the given quadratic equation it should satisfy the equation
So, substituting x = in the given
equation, we get
and
1 are the solutions of equation mx2 + nx
+ 6 = 0. Find the values of m and n.
For x = and x = 1 to be
solutions of the given quadratic equation it should satisfy the equation
So, substituting x = and x = 1 in the
given equation, we get
Solving equations (1) and (2) simultaneously,
If 3 and -3 are the solutions of equation ax2 + bx - 9 = 0. Find the values of a and b.
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 3 and x = -3 in the given equation, we get
Solving equations (1) and (2) simultaneously,
Chapter 5 - Quadratic Equations Exercise Ex. 5(B)
Without solving, comment upon the nature of roots of each of the following equations :
(i)7x2 - 9x +2 =0 (ii)6x2 - 13x +4 =0
(iii)25x2 - 10x +1=0 (iv)
(v)x2 - ax - b2 =0 (vi)2x2 +8x +9=0
Find the value of p, if the following quadratic equation has equal roots : 4x2 - (p - 2)x + 1 = 0
Find the value of 'p', if the following quadratic equations have equal roots :
x2 + (p - 3)x + p = 0
x2 + (p - 3)x + p = 0
Here, a = 1, b = (p - 3), c = p
Since, the roots are equal,
⇒ b2- 4ac = 0
⇒ (p - 3)2- 4(1)(p) = 0
⇒p2 + 9 - 6p - 4p = 0
⇒ p2- 10p + 9 = 0
⇒p2-9p - p + 9 = 0
⇒p(p - 9) - 1(p - 9) = 0
⇒ (p -9)(p - 1) = 0
⇒ p - 9 = 0 or p - 1 = 0
⇒ p = 9 or p = 1
The equation 3x2 - 12x + (n - 5)=0 has equal roots. Find the value of n.
Find the value of m, if the following equation has equal roots : (m - 2)x2 - (5+m)x +16 =0
Find the value of k for which the equation 3x2- 6x + k = 0 has distinct and real roots.
Chapter 5 - Quadratic Equations Exercise Ex. 5(C)
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
Solve :
2x2 - 9x + 10 = 0, When
(i) x∈ N
(ii) x∈ Q
Solve :
Solve :
Solve :
Solve :
Find the quadratic equation, whose solution set is :
(i) (ii)
Find the value of x, if a + 1=0 and x2 + ax - 6 =0.
If a+1=0, then a = -1
Put this value in the given equation x2 + ax - 6 =0
Find the value of x, if a + 7=0; b + 10=0 and 12x2 = ax - b.
If a + 7 =0, then a = -7
and b + 10 =0, then b = - 10
Put these values of a and b in the given equation
Use the substitution y= 2x +3 to solve for x, if 4(2x+3)2 - (2x+3) - 14 =0.
4(2x+3)2 - (2x+3) - 14 =0
Put 2x+3 = y
Without solving the quadratic equation 6x2 - x - 2=0, find whether is a solution of this equation or not.
Consider the equation, 6x2 - x - 2=0
Put in L.H.S.
Since L.H.S.= R.H.S., then is a solution of the given equation.
Determine whether x = -1 is a root of the equation x2 - 3x +2=0
or not.
x2 - 3x +2=0
Put x = -1 in L.H.S.
L.H.S. = (-1)2 - 3(-1) +2
= 1 +3 +2=6 R.H.S.
Then x = -1 is not the solution of the given equation.
If x = is a solution of the quadratic equation 7x2+mx - 3=0;
Find the value of m.
7x2+mx - 3=0
Given x = is the solution of the given equation.
Put given value of x in the given equation
If x = -3 and x = are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.
If quadratic equation x2 - (m + 1) x + 6=0 has one root as x =3;
find the value of m and the root of the equation.
Given that 2 is a root of the equation 3x2 - p(x + 1) = 0 and that the equation px2 - qx + 9 = 0 has equal roots, find the values of p and q.
or x = -(a + b)
If -1 and 3 are the roots of x2+px+q=0
then find the values of p and q
Chapter 5 - Quadratic Equations Exercise Ex. 5(D)
Solve each of the following equations using the formula :
(i)x2 - 6x =27 (ii)x2 - 10x +21=0
(iii)x2 +6x - 10 =0 (iv)x2 +2x - 6=0
(v)3x2+ 2x - 1=0 (vi)2x2 + 7x +5 =0
(vii) (viii)
(ix) (x)
(xi) (xii)
(xiii) (xiv)
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :
(i)x2 - 8x+5=0
(ii)5x2 +10x - 3 =0
x2 - 5x - 10 = 0
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
x2 - 3x - 9 =0
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
(i)2x2 - 10x +5=0
Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :
(i)3x2 - 12x - 1 =0
(ii)x2 - 16 x +6= 0
(iii)2x2 + 11x + 4= 0
Solve:
(i)x4 - 2x2 - 3 =0
(ii)x4 - 10x2 +9 =0
Solve :
(i)(x2 - x)2 + 5(x2 - x)+ 4=0
(ii)(x2 - 3x)2 - 16(x2 - 3x) - 36 =0
Solve :
(i)
(ii)
(iii)
Solve the equation . Write your answer correct to two decimal places.
Solve the following equation and give your answer correct to 3 significant figures:
Consider the given equation:
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x - 1)2 - 3x + 4 = 0
Solve the quadratic equation x2 - 3(x + 3) = 0; Give your answer correct to two significant figures.
x2 - 3(x + 3) = 0
Chapter 5 - Quadratic Equations Exercise Ex. 5(E)
Solve:
Solve: (2x+3)2=81
Solve each of the following equations, giving answer upto two decimal places.(i)x2 - 5x -10=0(ii) 3x2 - x - 7 =0
Solve :
(i)x2 - 11x - 12 =0; when x N
(ii)x2 - 4x - 12 =0; when x I
(iii)2x2 - 9x + 10 =0; when x Q
Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
Consider the given equation:
Chapter 5 - Quadratic Equations Exercise Ex. 5(F)
Solve : (x+5)(x-5)=24
Given: (x+5)(x-5)=24
Solve :
Given:
Solve :
Given:
or
One root of the quadratic equation is
. Find the value of m. Also, find the other root of the
equation.
Given quadratic equation is …. (i)
One of the roots of (i) is
, so it satisfies (i)
So, the equation (i)
becomes
Hence, the other root is.
One root of the quadratic equation is -3, find its
other root.
Given quadratic equation is …. (i)
One of the roots of (i) is -3, so it satisfies (i)
Hence, the other root is 2a.
If and
;find the values of x.
Given i.e
So, the given quadratic equation becomes
Hence, the values of x are and
.
Find the solution of the equation; if
and
.
Given quadratic equation is ….. (i)
Also, given and
and
So, the equation (i) becomes
Hence, the solution of given quadratic equation are and
.
If m and n are roots of the equation where x ≠ 0 and x ≠ 2; find m × n.
Given quadratic equation is
Since, m and n are roots of the equation, we have
and
Hence, .
Solve, using formula :
Given quadratic equation is
Using quadratic formula,
⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Solve the quadratic equation
(i) When
(integers)
(ii) When (rational numbers)
Given quadratic equation is
(i) When the equation
has no roots
(ii) When the roots of
are
or
Find the value of m for which the equation has real and equal
roots.
Given quadratic equation is
The quadratic equation has real and equal roots if its discriminant is zero.
or
Find the values of m for which equation has equal roots.
Also, find the roots of the given equation.
Given quadratic equation is …. (i)
The quadratic equation has equal roots if its discriminant is zero
When , equation (i) becomes
When , equation (i) becomes
∴ x =
Find the value of k for which equation has real roots.
Given quadratic equation is …. (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
Hence, the given quadratic equation has real roots for.
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i)
(ii)
(i) Given quadratic
equation is
D = b2 - 4ac = = 25 - 24 = 1
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have
or
(ii) Given quadratic equation is
D = b2 - 4ac = = 16 - 20 = - 4
Since D < 0, the roots of the given quadratic equation does not exist.
Solve :
(i) and
x > 0.
(ii) and x < 0.
(i) Given quadratic
equation is
or
But as x > 0, so x can't be negative.
Hence, x = 6.
(ii) Given quadratic equation is
or
But as x < 0, so x can't be positive.
Hence,
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