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IBPS PO Mains 5 Nov 2023 Paper

Question Numbers: 147-149

(x - 2)^{2} = (-3x^{2}) + 2^{2} + 25x - P

(10y^{2} - 3^{2}y + 2/3)(3) + 10y = 0

One factor of first equation is 5.

Question : 149 of 155

Marks: +1, -0

$15(3/4)$
$13(3/4)$
$15(3/7)$
$15(3/5)$
$15(3/8)$

Solution:

Concept:

Solve the given quadratic equation, identify its smaller root, then multiply it by the largest single-digit prime number.

Explanation:

The equation is:

(x - 2)^2 = -3x^2 + 4 + 25x - 45

Expand the left side:

x^2 - 4x + 4 = -3x^2 + 25x - 41

Bring all terms to one side:

4x^2 - 29x + 45 = 0

Factorize:

(4x - 9)(x - 5) = 0

So,

x = \frac{9}{4}

x = 5

The smaller root is

\frac{9}{4}

The largest single-digit prime number is

7

Multiply the smaller root by

7

\frac{9}{4} \times 7 = \frac{63}{4} = 15\frac{3}{4}

Answer:

15\frac{3}{4}

i.e., Option A.

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