10th Grade > Mathematics
QUADRATIC EQUATIONS MCQs
:
B
Step 1:- For x2–2x–3=0, value of discriminant
D=(−2)2–4(−3)(1)=4+12=16.
Step 2:- Since, D is a perfect square, roots are rational and unequal.
Step 3:- Solving the equation,
x=−(−2)+√(16)2 or −(−2)−√(16)2
x=3,−1
Thus, both of them are integers.
:
D
Given, the roots of quadratic equation 9x2+6kx+4=0 are real and equal.
Discriminant, Δ=b2−4acΔ=36k2–4(4)(9)=36k2–4(36)
The roots of quadratic equation are real and equal then Δ=036k2–4(36)=0k2−4=0k2=4k=√4∴k=± 2
:
D
Let, Length = x
Breadth = y
Given,
Area of Rectangle = 634 m2
Length : x
Thrice the breadth : 3y
2 more that thrice : 3y+2
So, x=3y+2
Area of the rectangle=length×breadth
634=xy634=(2+3y)y634=2y+3y2∴3y2+2y−634=0
This equation resembles the general form of quadratic equation ax2+bx+c=0.
Lets find the values of y satisfying the equation.(Roots of the equation)
y=−b ±√b2− 4ac2a=−2 ±√22 − 4×3(−634)2 × 3y=−2 ±√4 + 76086=−2 ±√76126y=−2 ± 87.2466y=−2+87.2466 or −2−87.2466y=14.20 or −14.87
Length is always positive.
∴y=14.20 m
x=2+3y∴x=2+3(14.20)=44.6 m
Length =44.6 m
Breadth =14.2 m
:
A
x+51=2x+10x−6
⇒(x + 5)(x - 6) = 2x + 10
⇒x2−6x+5x−30=2x+10
⇒x2−3x−40=0 is of the form of ax2+bx+c
Hence, is a quadratic equation.
:
B
Let the numbers be x & x+1
Square of consecutive number: x2 & (x+1)2
Sum of square: x2+(x+1)2
∴x2+(x+1)2=365
Lets solve
x2+x2+2x+1=3652x2+2x−364=0x2+x−182=0
(Dividing by 2)
Factorise,
x2+14x−13x−182=0x(x+14)−13(x+14)=0(x+14)(x−13)=0∴x=13 & −14
Since numbers are positive integers, x=13
So, required consecutive numbers are 13 & 14.
:
C
By using the method of factorisation,
x2−3x−10=0
⇒ x2−5x+2x−10=0
⇒x(x−5)+2(x−5)=0
⇒ (x−5)(x+2)=0
⇒x=5 and x=−2
:
C and D
Let the number be x.
Square of number: x2
Lesser by 15: x2−15
of two times the number: x2−15=2x
We get, x2−15=2x
Lets solve, x2−2x−15=0x2−5x+3x−15=0x(x−5)+3(x−5)=0(x−5)(x+3)=0∴x=5 & x=−3
The number satisfying the given statement are 5 & −3.
:
B
Multiply the equation by 5.
we get 25x2–30x–10=0
(5x)2–[2×(5x)×3]+32–32–10=0
(5x–3)2–9–10=0(5x–3)2–19=0(5x–3)2=195x−3=±√19x=3±√195
Roots of the equation are 3+√195 & 3−√195
:
C
Given: x=23 is a solution of the quadratic equation 7x2+mx−3=0.
Substituting x=23 in the given quadratic equation we get
⇒ 7(23)2+m(23)−3=0
⇒ 7(49)+m(23)−3=0
⇒ 289+2m3−3=0
⇒ 28+6m−27=0
⇒ m=−16
:
B
Step 1 : For, x2+2x+m=0,
⇒ a=1, b=2, c=m
We know that
D=b2–4ac
⇒D=(2)2–4m
=4−4m
Step 2 : The roots of a quadratic equation are real and distinct only when D>0
Step 3 : 4−4m>0
4>4m
m<1