Identify the standard form of the given equation $(4 x - 5) (4 x + 5) = 0$ .

16 x^{2} - 25 = 0

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16 x^{3} - 25 = 0

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16 x^{2} - 2 = 0

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6 x^{2} - 25 = 0

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Solution

The correct option is A $16 x^{2} - 25 = 0$
$(4 x + 5) (4 x - 5) = 0$
$4 x (4 x - 5) + 5 (4 x - 5) = 0$
$16 x^{2} - 20 x + 20 x - 25 = 0$
$16 x^{2} - 25 = 0$ which is in standard form of quadratic equation

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Similar questions

Consider the quadratic polynomial $16 x^{2} - 50 x + 25$ . Which of the following quadratic equations will have roots equal to the zeros of the given quadratic polynomial?

(i) $16 x^{2} - 50 x = 0$

(ii) $4 (2 x - 3)^{2} - (2 x - 3) - 14 = 0$

(iii) $16 x^{2} - 50 x + 81 = 25 + 2 x$

Q. Solve the quadratic equation:

16 x^{2} - 24 x = 0

Q. Solve

6 x^{2} - 5 x - 25 = 0

Solve the following equation.

16 x^{2} - 24 x + 9 = 0

Q. Find the roots of the equation

16 x^{2} - 27 x - 10 = 0

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Identify the standard form of the given equation(4x−5)(4x+5)=0.

The correct option is A 16x2−25=0(4x+5)(4x−5)=04x(4x−5)+5(4x−5)=016x2−20x+20x−25=016x2−25=0 which is in standard form of quadratic equation

Identify the standard form of the given equation $(4 x - 5) (4 x + 5) = 0$ .

The correct option is A $16 x^{2} - 25 = 0$
$(4 x + 5) (4 x - 5) = 0$
$4 x (4 x - 5) + 5 (4 x - 5) = 0$
$16 x^{2} - 20 x + 20 x - 25 = 0$
$16 x^{2} - 25 = 0$ which is in standard form of quadratic equation