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Let α and β be the roots of a quadratic equation ax² + bx + c = 0, then

Sum of the roots is given by α+β = −b/a​

and product of the roots is given by αβ = c/a​

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(ax^{2}+(4a^{2}-3b)x-12ab=0\)

⇒ ax² + 4a²x – 3bx – 12ab = 0 

⇒ ax(x + 4a) -3b(x + 4a) = 0 

⇒ (ax – 3b)(x + 4a) = 0 

⇒ x = 3b/a, -4a

Given equation is ax² + (4a² – 3b)x – 12ab = 0

⇒ ax² + 4a²x – 3bx – 12ab = 0

⇒ ax(x + 4a) – 3b(x + 4a) = 0

⇒ (x + 4a)(ax – 3b) = 0

Now, either x + 4a = 0 ⇒ x = -4a

Or, ax – 3b = 0 ⇒ x = \(\frac{3b}{a}\)

Thus, the roots of the given quadratic equation are x = \(\frac{3b}{a}\) and -4a respectively.

A real number α is called a root of the quadratic equation

ax² + bx + c = 0, a ≠ 0 if

aα² + bα + c = 0.

x = α is a solution of the quadratic equation, or α satisfies the quadratic equation.

The zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic. Equation ax² + bx + c = 0 are the same.

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a , b, c are real numbers, a ≠ 0.

Correct answer is (D) x³ – x² = (x – 1)³

(b) 5, -2

a - 3 = \(\frac{10}{a}\) ⇒ a² - 3a = 10

⇒ a² - 3a - 10 = 0 ⇒ a² - 5a + 2a - 10 = 0

⇒ a (a - 5) + 2(a - 5 ) = 0

⇒ (a - 5)(a + 2) = 0

⇒ a - 5 = 0 or a + 2 = 0

⇒ a = 5, -2.

(d) 0 and 3

5^2x - 5^x+3 + 125 = 5^x

⇒ 5^2x - 5^x x 5³ + 125 = 5^x

⇒ 5^2x - 5^x(125 + 1) + 125 = 0

⇒ 5^2x - 126 x 5^x + 125 = 0

Let 5^x = y

Then, the given equation reduces to

y²- 126y + 125 = 0

⇒ y² - 125y -y + 125 = 0

⇒ y(y - 125) - 1(y - 125 ) = 0

⇒ (y - 125) (y - 1) = 0

⇒ y - 125 = 0 or  y - 1 = 0 

⇒ y = 125 or 1 ⇒ 5^x = 125 or 5^x = 1

⇒ 5^x = 5³ or 5^x = 5⁰ ⇒ x = 3, 0.

Correct answer is

(C) Roots are not real

Correct answer is

(D) 3x² + 15x + 3 = 0

Correct answer is

(B) x² – 8x + 15 = 0

(c) x² + 13x - 30 = 0

• x² - 2x + 15 = 0

⇒ (x - 5)(x + 2) = 0 ⇒ x = 5, -2

• x² + 15x - 2 = 0. Solving x ≠ 2, -15

• x² + 13x - 30 = 0

⇒ (x + 15)(x - 2) = 0 ⇒ x = -15, 2

• x² - 30 = 0 ⇒ x² = 30 ⇒ x = ±\(\sqrt{30}.\)

Correct answer is (d) 3

Given:

3x² - 10x + 3 = 0

One root of the equation is 1/3.

Let the other root be a.

Product of the roots = c/a

⇒ 1/3 x a = 3/3

⇒ a = 3

(d) 2

x² - 7x + 10 = 0

⇒ (x - 5)(x - 2) = 0 ⇒ x = 5,2

x² - 10 + 16 = 0 ⇒ (x - 8)(x - 2) = 0

⇒ x = 8, 2

∴ Common root = 2.

Correct option is (B) 4x² – 17x + 4 = 0

\(2x-3\sqrt{x}=2\)

\(\Rightarrow2x-2=3\sqrt x\)

\(\Rightarrow(2x-2)^2=9x\)     (By squaring both sides)

\(\Rightarrow4x^2-8x+4=9x\)

\(\Rightarrow4x^2-17x+4=0\)

The roots of the given equation can be found by solving the equation \(4x^2-17x+4=0.\)

Correct option is B) 4x² – 17x + 4 = 0

Correct option is (A) (p² - 2q)/q

Given that \(\alpha\) and \(\beta\) are roots of \(x^2-px+q=0\)

Then sum of roots \(=\frac{-b}a=\frac{-(-p)}1=p\)

\(\therefore\) \(\alpha+\beta=p\)     _____________(1)

And product of roots \(=\frac{c}a=\frac q1=q\)

\(\therefore\) \(\alpha\beta=q\)          _____________(2)

Now, \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) \(=\frac{\alpha^2+\beta^2}{\alpha\beta}\)

\(=\frac{\alpha^2+\beta^2+2\alpha\beta-2\alpha\beta}{\alpha\beta}\)

\(=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\)

= \(\frac{p^2-2q}{q}\)

Correct option is D) (p² + q)/q

Correct option is (C) (x + 1)² = 2(x – 3)

Standard form of quadratic equation is \(ax^2+bx+c=0,a\neq0.\)

(A) \((x-2)\,(x+1)=(x-1)\,(x+3)\)

\(\Rightarrow x^2-2x+x-2=x^2-x+3x-3\)

\(\Rightarrow x^2-x-2=x^2+2x-3\)

\(\Rightarrow 2x-3+x+2=0\)

\(\Rightarrow3x-1=0\) which is a linear equation.

(B) \(x^4-1=0\) which is a quartic equation.

(C) \((x+1)^2=2(x-3)\)

\(\Rightarrow x^2+2x+1=2x-6\)

\(\Rightarrow x^2+2x+1-2x+6=0\)

\(\Rightarrow x^2+7=0\) which is a quadratic equation.

(D) \(x^2+3x+1=(x-2)^2\)

\(\Rightarrow x^2+3x+1=x^2-4x+4\)

\(\Rightarrow7x-3=0\) which is a linear equation.

Correct option is C) (x + 1)² = 2(x – 3)

(c) \(\frac{9}{13}.\)

Let \(\sqrt{\frac{x}{1-x}}\) = y. Then, the given equation reduces to

y + \(\frac{1}{y}=\frac{13}{6}\) ⇒ 6 (y² + 1) = 13 y 

⇒ 6y² – 13y + 6 = 0 ⇒ 6y² – 9y – 4y + 6 = 0 

⇒ 3y (2y – 3) – 2(2y – 3) = 0 

⇒ (3y – 2) (2y – 3) = 0 ⇒ y = \(\frac{2}{3}\) and \(\frac{3}{2}\)

when y = \(\frac{2}{3}\), \(\sqrt{\frac{x}{1-x}}\) = \(\frac{2}{3}\) ⇒ \(\frac{x}{1-x}\) = \(\frac{4}{9}\)

⇒ 9x = 4 – 4x ⇒ 13x = 4 ⇒ x = \(\frac{4}{13}\)

when y = \(\frac{3}{2}\), \(\sqrt{\frac{x}{1-x}}\) = \(\frac{3}{2}\) ⇒ \(\frac{x}{1-x}\) = \(\frac{9}{4}\)

⇒ 4x = 9 – 9x ⇒ 13x = 9 ⇒ x = \(\frac{9}{13}.\)

Let α and β be the two roots of the given equation 

4x² – 2x + (λ – 4) = 0 

According to the given condition 

α = 1/β 

Here a = 4, b = – 2 and c = (λ – 4) 

α + β = 2/4 = 1/2 

1/β + β = 1/2

α β = \(\frac{k-4}{4}\)

\(\frac{1}{\beta}\beta=\frac{k-4}{4}\)

K = 8

Correct option is (B) \(\pm\sqrt{a^2+b^2}\)

We have \(\frac{b}{x-a}=\frac{x+a}{b}\)

\(\Rightarrow b^2=(x+a)(x-a)\)      (By cross multiplication)

\(\Rightarrow x^2-a^2=b^2\)           \((\because(x+a)(x-a)=x^2-a^2)\)

\(\Rightarrow x^2=a^2+b^2\)

\(\Rightarrow x=\pm\sqrt{a^2+b^2}\)

Correct option is B) ± \(\sqrt{a^2+b^2}\)

Correct option is (B) ax² + bx + c = 0, a ≠ 0

The standard form of a quadratic equation is \(ax^2+bx+c=0,\) \(a\neq0.\)

Correct option is B) ax² + bx + c = 0, a ≠ 0

(c) 11, – 5

Given, log₁₀ (x² – 6x + 45) = 2 ⇒ x² – 6x + 45 = 10² = 100 

⇒ x² – 6x – 55 = 0 ⇒ x² – 11x + 5x – 55 = 0 

⇒ x(x – 11) + 5(x – 11) = 0 ⇒ (x + 5) (x – 11) = 0 

⇒ x = – 5 or 11.

Let the given equation has roots α and β 

α = 1 

Here a = 1, b = a and c = 3 

Sum of the roots 

α + β = – b/a = – a 

product of the roots 

α β = c/a = 3 

1 β = 3 

β = 3

Correct option is (A) no roots

Given equation is \(x-\frac{7}{x-3}=3-\frac{7}{x-3}\)

\(\Rightarrow\) \(x=3-\frac{7}{x-3}+\frac{7}{x-3}\)

\(\Rightarrow\) x = 3

But \(x\neq3\) as \(\frac{7}{x-3}\) is a rational term present in given equation.

\(\therefore\) Given equation has no roots.

Correct option is A) no roots

Correct option is (C) no root

Given equation is

\(\sqrt{x+4}-\sqrt{x-3}+1=0\)         ______________(1)

\(\Rightarrow\) \(\sqrt{x+4}+1=\sqrt{x-3}\)

\(\Rightarrow x+4+1+2\sqrt{x+4}=x-3\)     (By squaring both sides)

\(\Rightarrow2\sqrt{x+4}=x-3-x-5\)

\(\Rightarrow2\sqrt{x+4}=-8\)

\(\Rightarrow\sqrt{x+4}=-4\)

\(\Rightarrow x+4=(-4)^2=16\)     (By squaring both sides)

\(\Rightarrow x=16-4=12\)

Put x = 12 in equation (1), we get

\(\sqrt{12+4}-\sqrt{12-3}+1=0\)

\(\Rightarrow4-3+1=0\)

\(\Rightarrow2=0\)                 (Not satisfy)

\(\therefore\) x = 12 is not a root of given equation.

It implies given equation has no root.

Correct option is C) no root

Correct option is (A) 2

Given equation is

\(\sqrt{5-x}=x\sqrt{5-x}\)      _______________(1)

\(\Rightarrow5-x=x^2(5-x)\)   (By squaring both sides)

\(\Rightarrow x^2(5-x)-(5-x)=0\)

\(\Rightarrow(5-x)(x^2-1)=0\)

\(\Rightarrow5-x=0\;or\;x^2-1=0\)

\(\Rightarrow x=5\;or\;x=1\;or\;x=-1\)

Put x = -1 in equation (1), we get

\(\sqrt{5-(-1)}=-1\sqrt{5-(-1)}\)

\(\sqrt6=-\sqrt6\)         (Not satisfied)

Hence, x = -1 is not a solution of given equation.

But x = 1 & x = 5 are roots of given equation.

\(\therefore\) Number of roots satisfying the given equation is 2.

Correct option is A) 2

Correct option is (C) one real root

Given equation is \(x+\sqrt{x-2}=4\)      _______________(1)

\(\Rightarrow\) \(\sqrt{x-2}=4-x\)

\(\Rightarrow x-2=(4-x)^2\)

\(\Rightarrow x-2=x^2-8x+16\)

\(\Rightarrow x^2-9x+18=0\)

\(\Rightarrow x^2-6x-3x+18=0\)

\(\Rightarrow x(x-6)-3(x-6)=0\)

\(\Rightarrow(x-6)(x-3)=0\)

\(\Rightarrow x-6=0\) or \(x-3=0\)

\(\Rightarrow x=6\) or \(x=3\)

Put x = 3 in equation (1), we have

\(3+\sqrt{3-2}=4\)

\(\Rightarrow\) 3+1 = 4

\(\Rightarrow\) 4 = 4                 (Satisfies)

Hence, x = 3 is a root of given equation.

Put x = 6 in equation (1), we have

\(6+\sqrt{6-2}=4\)

\(\Rightarrow\) 6+2 = 4

\(\Rightarrow\) 8 = 4               (Not satisfies)

\(\therefore\) x = 6 is not a root of given equation.

Hence, x = 3 is only real root of given equation.

Thus, given equation has only one real root.

Correct option is C) one real root

Correct option is (A) \(\sqrt{85}\)

Let \(\alpha\;and\;\beta\) are roots of \(x^2-7x-9=0\)

\(\therefore\) Sum of roots \(=\frac{-(-7)}1=7\)

\(\Rightarrow\) \(\alpha+\beta=7\)  _____________(1)

And product of roots \(=\frac{-9}1=-9\)

\(\Rightarrow\) \(\alpha\beta=-9\)   _____________(2)

Now, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\)

\(=7^2-4\times-9\)

= 49+36 = 85

\(\therefore\) \(\alpha-\beta=\sqrt{85}\)

Hence, difference of roots of given equation is \(\sqrt{85}.\)

Correct option is A)\(\sqrt{85}\)

Correct option is (A) 4 or -1

Given equation is

\(log_{10}(x^2-3x+6)=1\)

\(\Rightarrow x^2-3x+6=10^1=10\)    (By taking anti-log)

\(\Rightarrow x^2-3x+6-10=0\)

\(\Rightarrow x^2-4x+x-4=0\)

\(\Rightarrow x(x-4)+1(x-4)=0\)

\(\Rightarrow(x-4)(x+1)=0\)

\(\Rightarrow x+1=0\) \(or\;x-4=0\)

\(\Rightarrow\) x = -1 or x = 4

Correct option is A) 4 or -1

In the given equation 2x² + kx + 4 = 0 

Let α and β be the two roots 

α = 2 (given) 

here a = 2, b = k and c = 4 

sum of the roots

⇒ α + β = \(-b/a\)

⇒ 2 + β = \(-k/2\)

β = – k 

α β = 2 

β = 2/2 = 1(putting value of α = 2)

Correct option is (C) an extraneous root between -10 and -6

Given equation is

\(\sqrt{x+10}-\frac{6}{\sqrt{x+10}}=5\)      _____________(1)

\(\Rightarrow\) x + 10 - 6 \(=5\sqrt{x+10}\)

\(\Rightarrow\) x+4 \(=5\sqrt{x+10}\)

\(\Rightarrow(x+4)^2=25(x+10)\)      (By squaring both sides)

\(\Rightarrow x^2+8x+16=25x+250\)

\(\Rightarrow x^2-17x-234=0\)

\(\Rightarrow x^2-26x+9x-234=0\)

\(\Rightarrow x(x-26)+9(x-26)=0\)

\(\Rightarrow(x-26)(x+9)=0\)

\(\Rightarrow\) x - 26 = 0 or x+9 = 0

\(\Rightarrow\) x = 26 or x = -9

Hence, roots of transformed equation are -9 and 26.

Put x = 26 in equation (1), we obtain

\(\sqrt{26+10}-\frac{6}{\sqrt{26+10}}=5\)

\(\Rightarrow\) \(\sqrt{36}-\frac{6}{\sqrt{36}}=5\)

\(\Rightarrow6-\frac66=5\)

\(\Rightarrow\) 6 - 1 = 5

\(\Rightarrow\) 5 = 5    (Satisfied)

Hence, x = 26 is a root of given equation.

Put x = -9 in equation (1), we obtain

\(\sqrt{-9+10}-\frac{6}{\sqrt{-9+10}}=5\)

\(\Rightarrow\) \(\sqrt{1}-\frac{6}{\sqrt{1}}=5\)

\(\Rightarrow\) 1 - 6 = 5

\(\Rightarrow\) - 5 = 5    (Not satisfied)

\(\therefore\) x = -9 is not a root of given equation but x = -9 is a root of its transformed equation.

\(\therefore\) x = -9 is an extraneous root of given which lies between -10 and -6.    \((\because-10<-9<-6)\)

Hence, an extraneous root of given equation lies between -10 and -6.

Correct option is C) an extraneous root between -10 and -6

In the given equation ax² + bx + c = 0 

Let α and β be the two roots 

Given α = 3β ……..1 

α β = (product of the roots) 

3 β² = c / a ……….. by using 1 

β² = …………….2

α + β = \(-b/a\)

4β = \(-b/a\)

Squaring both the sides

16 β² = \(b^2/a^2\)

By using 2

\(16\times\frac{c}{3a}=b^2/a^2\)

\(16/3=b^2/ac\)

x² + 2x – 5 = 0 

If x² + 2x + k = (x + a)², then 

x² + 2x + k = x² + 2ax + a² 

Comparing the coefficients, we get

2 = 2a and k = a² 

∴ a = 1 and k = (1)² = 1

Now, x² + 2x – 5 = 0

∴ x² + 2x + 1 – 1 – 5 = 0 

∴ (x + 1)² – 6 = 0 

∴ (x + 1)² = 6

Taking square root of both sides, we get

x + 1 = ± √6 

∴ x + 1 √6 or x + 1 = √6 

∴ x = √6 – 1 or x = -√6 – 1

∴ The roots of the given quadratic equation are √6 -1 and – √6 -1.

Given

9x² + 6x + 1 = 0

⇒ 9x² + 3x + 3x + 1 = 0

⇒ 3x(3x + 1) + 1(3x + 1) = 0

⇒ (3x + 1) (3x + 1) = 0

⇒ 3x + 1 = 0 or 3x + 1 = 0

⇒ x = -1/3 or x = -1/3

Hence -1/3 is the root of the equation 9x² + 6x + 1 = 0

If the equation does not possess real roots then 

d = b² – 4ac = 0 

Here a = 1, b = – b, c = 1 

d = b² – 4 < 0 

b² < 4 

b < \(\pm\) 2 

– 2 < b < 2

x = 1 is a solution of x² + kx + 3 = 0, which means it must satisfy the equation.

(1)² + k(1) + 3 = 0

k = -4

Hence the required value of k = -4

Find other root:

We have equation, x² – 4x + 3 = 0

x² – x – 3x + 3 = 0

x(x – 1) – 3(x – 1) = 0

(x – 1)(x – 3) = 0

either x – 1 = 0 or x – 3 = 3

x = 1 or x = 3

Other root is 3.

Given equation is ax² + bx – 6 = 0

As 3/4 is its root, then must satisfy the equation

a(3/4)² + b(3/4) – 6 = 0

9a +12b – 96 = 0 ….(1)

Again, x = -2 is its root

a(-2)² + b(-2) – 6 = 0

4a – 2b – 6 = 0 …(2)

Solving (1) and (2), we get

a = 4 and b = 5

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

\(x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\)

⇒ x² –x -√3x + √3 = 0 

⇒ x(x – 1) - √3(x – 1) = 0 

⇒ (x - √3)(x – 1) = 0 

⇒ x = √3, 1

False. For example, a quadratic equation x² – 9 = 0 has two distinct roots – 3 and 3.

(i) False, for example : x² = 1 is a quadratic equation with two roots.

(ii) False, for example x² + 1 = 0 has no real root.

(iii) False, for example : x²+1 = 0 is a quadratic equation which has no real roots.

(iv) True, because every quadratic polynomial has almost two zeroes.

(v) True, because if in ax²+bx+c = 0, a and c have opposite signs, then ac<0

and so b²–4ac > 0.

(vi) True, because if in ax²+bx+c = 0, a and c have same sign and b = 0, then

b²–4ac = –4ac < 0.

Correct answer is (A) 2

Correct answer is (C) no real roots

Correct answer is (D) 0, 8

Correct answer is (A) x² – 4x + 3√2 = 0

Correct answer is (C) (x + 2) (x – 1) = x² – 2x – 3

Equating the expression with 0,

x² - 6x + 8 = 0

On factorising it further,

x² - 4x - 2x + 8 = 0

x(x - 4) – 2(x - 4) = 0

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

∴ x = 4 or x = 2

∴ The zeroes of x² - 6x + 8 are 4 & 2.

x² + 4ix – 4 = 0 

Given x² + 4ix – 4 = 0 

⇒ x² + 4ix + 4(–1) = 0 

We have i² = –1 

By substituting –1 = i² in the above equation, we get 

⇒ x² + 4ix + 4i² = 0 

⇒ x² + 2ix + 2ix + 4i² = 0 

⇒ x(x + 2i) + 2i(x + 2i) = 0 

⇒ (x + 2i)(x + 2i) = 0 

⇒ (x + 2i)² = 0 

⇒ x + 2i = 0 

∴ x = –2i (double root) 

Thus, the roots of the given equation are –2i and –2i.

Given as x² + 4ix – 4 = 0

x² + 4ix + 4(–1) = 0 [as we know, i² = –1]

Therefore by substituting –1 = i² in the above equation, we get

x² + 4ix + 4i² = 0

x² + 2ix + 2ix + 4i² = 0

x(x + 2i) + 2i(x + 2i) = 0

(x + 2i) (x + 2i) = 0

(x + 2i)² = 0

x + 2i = 0

x = –2i, -2i

Therefore, the roots of the given equation are –2i, –2i

ix² – x + 12i = 0 

Given ix² – x + 12i = 0 

⇒ ix² + x(–1) + 12i = 0 

We have i² = –1 

By substituting –1 = i² in the above equation, we get 

 ix²+ xi² + 12i = 0 

⇒ i(x² + ix + 12) = 0 

⇒ x² + ix + 12 = 0 

⇒ x² + ix – 12(–1) = 0 

⇒ x² + ix – 12i² = 0 [∵ i² = –1] 

⇒ x² – 3ix + 4ix – 12i² = 0 

⇒ x(x – 3i) + 4i(x – 3i) = 0 

⇒ (x – 3i)(x + 4i) = 0 

⇒ x – 3i = 0 or x + 4i = 0 

∴ x = 3i or –4i 

Thus, the roots of the given equation are 3i and –4i.

Given as ix² – 4x – 4i = 0

ix² + 4x(–1) – 4i = 0 [as we know, i² = –1]

Therefore by substituting –1 = i² in the above equation, we get

ix² + 4xi² – 4i = 0

i(x² + 4ix – 4) = 0

x² + 4ix – 4 = 0

x² + 4ix + 4(–1) = 0

x² + 4ix + 4i² = 0 [Since, i² = –1]

x² + 2ix + 2ix + 4i² = 0

x(x + 2i) + 2i(x + 2i) = 0

(x + 2i) (x + 2i) = 0

(x + 2i)² = 0

x + 2i = 0

x = –2i, -2i

Therefore, the roots of the given equation are –2i, –2i