CRE 1 - Solving Quadratic Equations | Algebra - Quadratic Equations

Answer: Option D

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Explanation :

For an equation to be called quadratic, the highest power of the variable should be 2.

Equations in options (a) and (c) are not quadratic equations because in 
(a) highest power of x is fractional and in 
(c) highest power of x is 4.

For option (b), (x – 2)(x + 5) = x² + 4

⇒ x² + 5x – 2x -10 = x² + 4

⇒ 3x – 14 = 0

which is not a quadratic equation but a linear equaiton.

For option (d), (x + 3)(2x – 1) = x² + 7

⇒ 2x² + 6x - x – 3 = x² + 7

⇒ x² + 5x – 10 = 0

which is clearly a quadratic equation.

Hence, option (d).

Answer: Option A

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Explanation :

Given, 2x² – 3x – 2 = 0

Here, we need to find two numbers such that their sum = coefficient of x i.e., -3 and their product = (coefficient of x²) × (contant term) = 2 × - 2 = - 4.

The two such numbers are -4 and 1. 

Now, splitting the middle term, we get

⇒ 2x² – 4x + x – 2 = 0

⇒ 2x(x – 2) + (x – 2) = 0

⇒ (x - 2)(2x + 1) = 0

Either (x - 2) = 0 or (2x + 1) = 0

∴ x = 2 or -1/2.

Hence, option (a).

Answer: Option B

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Explanation :

3p² – 7pq + 2q² = 0

Dividing the whole equation with q².

$3{\left(\frac{p}{q}\right)}^2-7\left(\frac{p}{q}\right)+2=0$

Let p/q = t

⇒ 3t² – 7t + 2 = 0

Splitting the middle term.

⇒ 3t² – 6t - t + 2 = 0

⇒ 3t(t – 2) - (t - 2) = 0

⇒ (3t - 1)(t - 2)= 0

⇒ t = p/q = 1/3 or 2.

Hence, option (b).

Answer: Option B

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Explanation :

Let the son’s present age be x years.

So, father’s present age = (4x + 3) years

Also, product of their ages i.e., x(4x + 3) = 162

⇒ 4x² + 3x – 162 = 0 

⇒ 4x² + 27x - 24x – 162 = 0

⇒ x(4x + 27) - 6(4x – 27) = 0

⇒ (4x + 27)(x - 6) = 0

∴ Either x = 6 or -27/4 

Since, x can not be negative, x = 6 is the only solution.

∴ Son’s age = 6 years and Father’s age = 4x + 3 = 27 years.

Hence, option (b).

Answer: Option B

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Explanation :

Let the rational number be x.

Then, $x+\frac{1}{x}=\frac{25}{12}$

⇒ $\frac{x^2+1}{x}=\frac{25}{12}$ 

⇒ 12x² – 25x + 12 = 0

⇒ 12x² – 16x – 9x + 12 = 0 

⇒ (4x – 3)(3x – 4) = 0

⇒ x = 3/4 or x = 4/3.

Hence, the required number is 3/4 or 4/3.

Hence, option (b).

Answer: Option B

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Explanation :

Suppose, $\sqrt{12+\sqrt{12+\sqrt{12+\dots }}}$  = x

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⇒ $\sqrt{12+x}$ = x

⇒ 12 + x = x²

⇒ x² - x – 12 = 0

⇒ x² – 4x + 3x – 12 = 0

⇒ (x – 4)(x + 3) = 0

⇒ x = 4 or -3.

Since $\sqrt{12+\sqrt{12+\sqrt{12+\dots }}}$  is a positive number, x must also be positive. Hence, only x = 4 will be accepted.

Hence, option (b).

Answer: Option A

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Explanation :

Let x = $\sqrt{3\sqrt{3\sqrt{3}\dots .\infty }}$ 

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⇒ x² = 3x 

⇒ x = 0, 3

Since $\sqrt{3\sqrt{3\sqrt{3}\dots .\infty }}$  is a positive number, x can not be 0. Hence, only x = 3 will be accepted.

Hence, option (a).

Answer: Option D

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Explanation :

Given, a + b = 9 …(1)

and $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$ …(2)

⇒ $\frac{a+b}{ab}=\frac{1}{2}$

⇒ $\frac{9}{ab}=\frac{1}{2}$

⇒ ab = 18 …(3)

From (1) and (3), a(9 – a) = 18

⇒ a² – 9a + 18 = 0 

⇒ (a – 6)(a – 3) = 0

⇒ a = 6 or 3 

∴ When a = 6, b = 3 or when a = 3, b = 6.

Hence the numbers are 6, 3.

Hence, option (d).