Chapter-4, QUADRATIC EQUATIONS Extra Questions

1. Which of the following is a quadratic equation? 

(A) x2 + 2x + 1 = (4 – x)2 + 3 (B) –2x2  = (5 – x)  2 2 /5

(C) (k + 1)x2 +3/2 x = 7, where k = –1 (D) x3 – x2  = (x – 1)3

Explanation:


The quadratic formula needs to be located.

A quadratic formula has the conventional form ax² + bx + c = 0 in the variable x.

where a 0 and a, b, and c are real numbers.

We must determine whether the formulas provided have a degree of two.

As a result of the choices, A) x² + 2x + 1 = (4 - x)² + 3

Algebraic identity yields the following result: (a - b)² = a² - 2ab + b² x²+ 2x + 1 = 16 + x² - 8x + 3

Combining results in x2 - x2 + 2x + 8x + 1 - 16 - 3 = 0.

10x - 18 = 0

The equation has a degree of 1.

Consequently, (4 - x) = x² + 2x + 1.The equation 2 + 3 is not quadratic.

B) -2x² = (5 - x)(2x - 2/5)

2x² = (5-x)(10-x)/5

-10x² = (5-x)(10-x)

According to the additive and distribution assets,

-10x² = 50x - 10- 10 - 10x² + 2x

As a result of the arrangement, -10x²+ 10x² + 52x - 10 = 0.

52x - 10 = 0

The formula has a degree of 1.

As a result, the equation -2x² = (5 - x)(2x - 2/5) is not a quadratic one.

C) Where k = -1 (-1 + 1)x2 + (3/2)x = 7, where k + 1x2 + (3/2)x = 7.

3/2x = 7

The formula has a degree of 1.

Since k = -1, the formula (k + 1)x² + (3/2)x = 7 is not quadratic.

D) x³ - x² = (x - 1)³

Algebraic identity reveals that (a - b)³ = a³ - b³ - 3a2b + 3ab² and (x3 - x²) = x³ - 1 - 3x²+ 3x.

Grouping results in x³ - x³ - x² + 3x² - 3x + 1 = 0.

2x² - 3x + 1 = 0

The equation has a degree of two.

Consequently, x3 - x² = (x - 1) equation 3 is quadratic.

2. Which of the following is not a quadratic equation? 

(A) 2(x – 1)2  = 4x2 – 2x + 1 (B) 2x – x2  = x2 + 5 

(C) 22 2 ( 2 3) 3 5 x x xx + += − (D) (x2 + 2x)  = x4 + 3 + 4x3

Explanation: 


The polynomial problems of degree 2 in a single factor of form ax2+bx+c, where a, b, c, ∈, and R are all positive, are known as quadratic equations.

Expansion results in (x+1)2=2(x-3)x²+1+2x=2x-6

.Note:(a+b)²=a²+b²+2abx²+2x-2x+1+6=0x²+7=0x²+0x+7=0

The a=1, b=0, and c=7 in the formula above make it in form ax²+bx+c=0.

Since (x+1)²=2(x-3) is given, it follows that it is a quadratic equation.

3. Which of the following equations has 2 as a root? 

(A) x2 – 4x + 5 = 0 (B) x2 + 3x – 12 = 0 

(C) 2x2 – 7x + 6 = 0 (D) 3x2 – 6x – 2 = 0

Explanation:

Formula A with x=2 inserted are as follows: x²+4x5(22)+4254+85=7

B: x²+3x+12⇒(22)+3×2+12⇒4+6+12=22

C: 2x²−7x+6⇒2(22)−7×2+6⇒8−14+6=0

D: 3x²−6x+2⇒3(22)−6×2+2⇒12−12+2=2

Then, the root of equation (C) is 2.

4. If  ½ 2 is a root of the equation x2 + kx – 5 /4 = 0, then the value of k is 

(A) 2 (B) – 2 (C)  ¼ (D) ½

Explanation:

The formula is as follows: x2 + kx - 5/4 = 0.

The solution to the problem is 1/2.

The value of k must be determined.

Put x = 1/2, (1/2)² + k(1/2) - 5/4 = 0

k/2 = 5/4 - 1/4 

k/2 = (5 - 1)/4

k/2 = 4/4

k/2 = 1

k = 2(1)

k = 2

1/4 + k/2 - 5/4 = 0

Consequently, k has a value of 2.

5. Which of the following equations has the sum of its roots as 3? 

(A) 2x2 – 3x + 6 = 0 (B) –x2 + 3x – 3 = 0 

(C) 2 3/ 2 1 0 x x (D) 3x2 – 3x + 3 = 0

Explanation:

B is the right answer when x2+3x-3=0.

6. Values of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is (A) 0 only (B) 4 (C) 8 only (D) 0, 8

Explanation:

Since 2x² - kx + k = 0 is the formula,

The k numbers for which the equation has equal roots must be determined.

When the discriminant of the equation is zero, the roots of the quadratic equation ax² + bx + c = 0 are equivalent.

Discriminatory = b² – 4 ac

In this case, a=2, b=-k, and c=-k.(k - 8) = 0 

k = 0 k - 8 = 0

k = 2 - 4(2)(k) = 0 

k² - 8k = 0

k

As a result, the equation's roots are 0 and 8.

7. Which constant must be added and subtracted to solve the quadratic equation 

9x2 + ¾ x – 2 = 0 by the method of completing the square?

Explanation:

Since 9x2 + (3/4)x - 2 = 0, the quadratic formula is zero.

To solve the quadratic problem using the method of completing the square, we must identify the constants that must be added and subtracted.

Let y = 3x

Now, (3x)² + (3x)/4 - √2 = 0

y² + (1/4)y - √2 = 0 --------------------- (1)

(A + B)² = A² + 2AB + B2 is an algebraic identity. --------------- (2)

A² = 1 a = 1 2ab equals 1/4 of 2(1) when (1) and (2) are compared.b = 1/4 b = 1/8 b² = (1/8)² = 1/64

So, y² + (1/4)y - √2 + 1/64 - 1/64 = 0

Rearranging results in the equation y2 + (1/4)y + 1/64 - 2 - 1/64 = 0 (y + 1/8).² = √2 + 1/64

Now, (3x + 1/8)² = √2 + 1/64

As a result, 1/64 is the constant that needs to be added and removed.

8. The quadratic equation 2x2  – 5x + 1 = 0 has 

(A) two distinct real roots (B) two equal real roots (C) no real roots (D) more than 2 real roots

Explanation:

The quadratic formula is 2x² - √5x + 1 = 0.

The type of the equation's roots must be determined.

Discriminatory = b² – 4 ac

In this case, a = 2, b = - √5, 

and c = 1 b2 = (- √5).4(2)(1) = 8 b2 - 4ac = 5 - 8 = -3 b² - 4ac<  0 2 = 5 4ac

When the formula's discriminant is smaller than zero, the quadratic equation ax² + bx + c = 0 does not have any real roots.

As a result, the equation lacks any true roots.

9. Which of the following equations has two distinct real roots? 

(A) 2x2 – 9 3 2 /4=0  (B) x2 + x – 5 = 0 

(C) x2 + 3x + 2 2 = 0 (D) 5x2  – 3x + 1 = 0

Explanation:

The right answer is D x2 + x - 5 = 0.

If D = b2 - 4ac > 0, a quadratic equation has two unique roots.

Take into account the quadratic formula x2 + x - 5 = 0.

Here, D = (1)2 - 4(1)(-5) = 1 + 20 = 21 > 0

Thus, there are two distinct real roots in this equation.

10. Which of the following equations has no real roots? 

(A) x2  – 4x + 3 2 = 0 (B) x2  + 4x – 3 2 = 0 

(C) x2  – 4x – 3 2 = 0 (D) 3x2  + 4 3 x + 4 = 0

Explanation:

We need to identify an equation without any actual roots.

When the equation's discriminant is smaller than zero, the quadratic equation ax² + bx + c = 0 does not have any real roots.

Discriminatory = b² – 4 ac

As a result of the choices, A) x² - 4x + 3√2 = 0

In this instance, a = 1, b = 4, and c = 32 b² - 4ac = (-4)² - 4(1)(3√2) = 16 - 12√2 = 16 - 12(1.414) = 16 - 16.968 = -0.968 < 0

As a result, the equation lacks any true roots.

B) x² + 4x - 3√2 = 0

In this example, a = 1; b = 4; and c = -3√2 b² - 4ac = (4)² - 4(1)(-3√2) = 16 + 122 > 0

The equation has two unique real roots as a result.

C) x² - 4x - 3√2 = 0

Here, b = -4, c = -3√2, and b² - 4ac = (-4)² - 4(1)(-3√2) = 16 + 12√2 > 0

Thus, the equation has two unique real solutions.

origins.

D) 3x² + 4√3x + 4 = 0

In this case, a = 3, b = 4√3,

and c = 4 b2 - 4 ac = (4√3).² - 4(3)(4)

 = 48 - 48 

= 0

The formula has two equal roots as a result.

11. (x2  + 1)2  – x2  = 0 has 

(A) four real roots (B) two real roots 

(C) no real roots (D) one real root.

Explanation:


Choice (C)

Since (x²+1)²=x² and (x²+1)²=0

The two sides of the square root. We obtain ²2+1=x, 

x²+1=x,

=>x²x+1=0, 

or x²+x+1=0.

The discriminant for the aforementioned equations is either 1-4 = -3 0 or 1-4=-3 0.

In light of this, (x²+1)²x² = 0 has no actual roots.

12. State whether the following quadratic Justifies your answer. the equation has two distinct real roots. 

(i) x2  – 3x + 4 = 0 (ii) 2x2 + x – 1 = 0 (iii) 2x2  – 6x +  9/2 =0

 (iv) 3x2 – 4x + 1 = 0(v) (x + 4)2  – 8x = 0 (vi) (x – 2 )2  – 2(x + 1) = 0 

(vii) 2 3 / 2x +1/2 (viii) x (1 – x) – 2 = 0

(ix) (x – 1) (x + 2) + 2 = 0 (x) (x + 1) (x – 2) + x = 0  

Explanation:


Given that 2x² - 6x + 9/2 = 0 is true

If the equation has two unique real roots, we must identify them.

Discriminatory = b² – 4 ac

In this case, a = 2, b = 6, 

and c = 9/2

b² - 4ac = (-6)² - 4(2)(9/2) 

= 36 - 36 

= 0

When the discriminant of the equation is 0. we know that the quadratic formula ax² + bx + c = 0 has equal roots.

The formula has two equal roots as a result.

13. Write whether the following statements are true or false. Justify your answers. 

(i) Every quadratic equation has exactly one root. 

(ii) Every quadratic equation has at least one real root. 

(iii) Every quadratic equation has at least two roots. 

(iv) Every quadratic equation has at most two roots. 

(v) If the coefficient of x2  and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots. 

(vi) If the coefficient of x2  and the constant term have the same sign and if the 

coefficient of x term is zero, then the quadratic equation has no real roots.

Explanation:

1. Untrue.

There won't be any genuine roots for a quadratic equation if b2-4ac is negative.


2. Untrue.

Take into account the positive value of b2 in the equation x2+0x-1, which is 4ac=4. The quadratic equation, therefore, has true roots.


3. True.

The value of ac will be negative if the term for the constant (c) and x2 coefficient (a) have the opposite signs. -4ac will therefore be good. As a result, b2-4ac is going to have a positive sign, and the quadratic equation will only have real roots.


4. False.

Take the given Quadratic formula, for instance.

By dividing the middle term, 8x² - 2x -1 = 0

becomes 8x² -4x + 2x -1 = 0.

4x(2x-1)+1(2x-1)=0

 (4x+1)(2x-1)=0

if 4x+1 =0

THEN , x = -1/4

If 2x-1 = 0, then x = 1/2 and x = -1/4, respectively.

14. A quadratic equation with an integral coefficient has integral roots. Justify your answer.

Explanation:

The total of the values of the quadratic formula is given.

Checking for total roots in the formula is necessary.

Second-degree expressions in algebra are quadratic equations.

A quadratic equation is, in terms of an "equation of degree 2" that has two solutions for x, known as the quadratic equations' roots and denoted by (α, β, ).

Both integral and non-integral roots are possible.

15. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Explanation:

Absolutely take into account the quadratic equation 2x2 + x - 4 = 0 with a reasonable variable; its roots are 1+√33/4 and 1-√334, respectively, which are irrational.

16. Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

Explanation:


We must assess whether a quadratic equation has rational coefficients but irrational bases on each side.

A quadratic equation has the conventional form ax² + bx + c = 0 in the variable x.

where a 0 and a, b, and c are real numbers.

A quadratic equation is an "equation of degree 2" with two solutions for x that are referred to as the quadratic equations' roots and are represented by the symbols (α, β ).

17. Is 0.2 a root of the equation x2  – 0.4 = 0? Justify.

Explanation:

Not because the quadratic formula (0.2)² - 0.4 = 0.04 - 0.4≠  0. requires that 0.2 fulfill.

18. If b = 0, c < 0, is it true that the roots of x2  + bx + c = 0 are numerically equal and opposite in sign? Justify.

Explanation:

Assuming that b = 0 & c 0, as well as the quadratic solution x2 + bx + c = 0, we can substitute b = 0 into Eq. (i) to obtain x² + 0 + c = 0 x²=c [where c > 0 c > 0].

Since x=±√c, every root of x² + bx + c = 0 is identical in number and sign.

19. Find the roots of the quadratic equations by using the quadratic formula in each of the following: 

(i) 2x2  – 3x – 5 = 0 (ii) 5x2  + 13x + 8 = 0 

(iii) –3x2  + 5x + 12 = 0 (iv) –x2  + 7x – 10 = 0 

(v) x2  + 2 2 x – 6 = 0 (vi) x2  – 3 5 x + 10 = 0 

(vii)  ½ x2 – 11 x + 1 = 0

Explanation:

The quadratic equation is 2x² - 3x - 5 = 0, so there.

The formula's roots must be located.

The quadratic method is given as x = [-b±  √b² - 4ac]/2a.

Here, a=2, b=-3, and c=-5; hence, b2 - 4ac = (-3)² - 4(2)(-5) = 9 + 40 = 49

x = [- (-3) ± √49]/2(2) x = [3 ± 7]/4

Now, x = (3+7)/4 = 10/4 = 5/2 x = (3-7)/4 = -4/4 = -1

As a result, the roots of the formula are 5/2 and -1.

20. Find the roots of the following quadratic equations by the factorization method: 

(i) 2x2 +5/3 x-2=0 (ii) ⅖ x2- ⅗ =0

(iii) 2 3 2x – 5x – 2 = 0 (iv) 3x2  + 5 5x – 10 = 0 

(v) 21x2  – 2x +  1/21=0

Explanation:


provided, 2x²+3/5x−2=0

⇒6x²+5x−6=0

6x²−4x+9x−6=0

2x(3x−2)+3(3x−2)=0

 (3x−2)(2x+3)=0 

(x−3/2)(x+2/3)=0

 x=3/2,−2/3

Therefore, that's the response.

21. Find whether the following equations have real roots. If real roots exist, find them

(i) 8x2  + 2x – 3 = 0 (ii) –2x2  + 3x + 2 = 0 

(iii) 5x2  – 2x – 10 = 0 (iv)  1/2x-3 +1/x-5=1 x#3/2.5

(v) x2  + 5 5 x – 70 = 0

Explanation:


The quadratic formula is 8x² + 2x - 3 = 0, as is known.

Checking to see if the equation has actual roots is necessary.

When the equation's discriminant is bigger than zero, a quadratic equation with the form ax² + bx + c = 0 has two unique real roots.

Discriminatory = b2 – 4 ac

In this case, a = 8, b = 2, and c = -3.

So, b² - 4ac = (2)² - 4(8)(-3) = 4 + 96 = 100 > 0

The equation thus has two unique real roots.

The quadratic formula is x = [-b ± √  b² - 4ac]/2a, x = (-2 ± √  100)/2(8), and x = (-2  ±10)/16.

Now, x = (-2 + 10)/16 = 8/16 = 1/2 x = (-2 - 10)/16 = -12/16 = -3/4

As a result, the roots of the formula are 1/2 and -3/4.

22. Find a natural number whose square diminished by 84 is equal to thrice 8 more than the given number.

Explanation:


A number that is natural that is provided as a squared value of 84, which is three times the supplied number.

Finding the natural number is necessary.

Let x be the natural number.

So, x² - 84 = 3(x + 8)

3x + 24 x2 - 84 = x² - 3x - 24 - 84 = 0 x² - 3x - 108 = 0

After factoring, x² - 12x + 9x - 108 = 0 

and x(x - 12) + 9(x - 12) = 0

(x + 9)(x - 12) respectively.

Now, x + 9 = 0 x = -9

In addition, x - 12 = 0; x = 12.

x = -9 is disregarded since a natural number cannot be negative.

Consequently, x = 12 is the natural number.

23. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Explanation:


Provided, a natural integer multiplied by 12 has a reciprocal that is 160 times larger.

Finding the natural number is necessary.

Let x be the natural number.

So, x + 12 = 160(1/x)

Rearranging results in x² + 12x = 160

and x² + 12x - 160 = 0.

Factoring reveals that x² + 20x - 8x - 160 = 0 

for x(x + 20). - 8(x + 20) = 0

 (x - 8)(x + 20) = 0

Now, x - 8 = 0 x = 8

Additionally, x + 20 = 0 x = -20

x = -20 is disregarded since a negative number cannot be negative.

Consequently, x = 8 is the natural number.

24. A train, traveling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Explanation:

Suppose x= Actual Rate.

We are aware that time = speed/distance 

Thus, the following equation can be created in response to the query.

360/x− 360/x+5 =48/60     [60 minutes in an hour]

⇒1/x−1/x+5 =48/60 . 1/360

⇒5/x²+5x=1/450

⇒x²+5x−2250=0

⇒x²−45x+50x−2250=0

⇒x(x−45)+50(x−45)=0

⇒(x−45)(x+50)=0

x=45,−50

Since the speed cannot be zero,

x=45.

25. If Zeba were younger by 5 years than what she really is, then the square of her 

age (in years) would have been 11 more than five times her actual age. What is her age now?

Explanation:

Make Zeba's true age five years.

She was x years older when the was 5 years younger.

Given the circumstance, her age squared now equals 11, which is more than five times her actual age (x5).3. =5x+11⇒(x−5)²=5x+11

⇒x²+25−10x=5x+11

⇒x²−15x+14=0

⇒x²−14x−x+14=0    dividing the middle word⇒x(x−14)−1(x−14)=0

⇒(x−1)(x−14)=0

⇒x=14

[In this case, x=1 because she is x=5, hence x=5=15=4, proving that age cannot be negative]

Zeba is now 14 years old as a result.

26. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times her present age of Nisha. Find the present ages of both Asha and Nisha.

Explanation:


27 and 5 years old

Should Nisha be x years old at this time?

Asha's current age then equals x² + 2.            [by the stated condition]

Now, after (x²+2)x years, Nisha will be the same age as her mother. Following that, Asha's age likewise rose by [(x²+2)x] years.

Again, according to the provided condition, Asha's age is equal to one year less than ten times Nisha's current age (x²+2)+(x²+2)x=10x1

⇒2x²-x+4=10x1 

2x²-11x+5=0 

2x²-10xx+5=0 

2x(x5)1(x5)=0).

⇒(x−5)(2x−1)=0 ∴x=5

[In this case, x=12 is not feasible since Asha would be 214 years old at x=12, which is never feasible]

Nisha must therefore be at least 5 years old.

So Asha's needed age is x²+2=(5)²+2=25+2=27 years.

27. In the center of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2  [see Fig. 4.1]. Find the length and breadth of the pond.

Explanation:


Because the circular grass measures 50 m x 40 m.

There are 1184 m² of grass bordering the pond.

The pond's length and width need to be measured.

Let EFGH represent the rectangle pond and ABCD represent the rectangular lawn.

The size of the grassy space surrounding the pond should be x m.

Thus, the grass is fifty meters long.

forty meters is the lawn's width.

Pond length = (50 - 2x) m

Pond width equals (40 - 2x)m

Lawn area minus pond area equals the area of grass.

1184 = (50 × 40) - [(50 - 2x) × (40 - 2x)]

1184 = 2000 - [2000 - 100x - 80x + 4x²]

1184 = 2000 - 2000 + 100x + 80x - 4x²

1184 = 180x - 4x²

So, 4x² - 180x + 1184 = 0

Using a factor of 4, x² - 45x + 296 = 0

The factoring results are: x² - 37x - 8x + 296 = 0

x(x - 37) - 8(x - 37) = 0

(x - 8)(x - 37) = 0.

Now, x - 8 = 0 x = 8

Also, x = 37 and x - 37 = 0.

If x equals 37, the

Lake length = 50 – 2(37) = 50 – 74 = -24

x = 37 is therefore not conceivable.

Consequently, the grassy space is 8 meters wide.

The pond's length is now 50 - 2(8) = 50 - 16 = 34 m.

The pond width is equal to 40 - 2(8) = 40 - 16 = 24m.

The lake is therefore 34 and 24 meters long and wide, respectively.

28. At t minutes past 2 pm, the time needed by the minute's hand of a clock to show 3 pm was found to be 3 minutes less than  t²/4  minutes. Find it.

Explanation:


Response: Based on the circumstances, we have allocated the time required by the minute's hand show (t²/4). – 3

Consequently, the quadratic formula is (t²/4). – 3 = 60 – t Since it will be 60 minutes from 2:00 to 3:30 PM.

(t² – 12)/4 = 60 – t 

(t² – 12) = 240 – 4 t

t(t + 18) - 14(t + 18) = 0

(t + 18) t² - 12 - 240 + 4 t = 0

t² + 4t - 252 = 0

t² + 18t - 14t - 252 = 0t = - 18 or t = 14 or (t - 14) = 0 or (t + 18) = 0 We shall ignore t=-18, leaving us with a time constraint of t=14.

Chapter-4, QUADRATIC EQUATIONS