Two of the steps in the derivation of the quadratic formula are shown below. Step 6: StartFraction b squared minus 4 a c Over 4 a squared EndFraction = (x + StartFraction b Over 2 a EndFraction) squared Step 7: StartFraction plus or minus StartRoot b squared minus 4 a c EndRoot Over 1 a EndFraction = x + StartFraction b Over 2 a EndFraction Which operation is performed in the derivation of the quadratic formula moving from Step 6 to Step 7? subtracting StartFraction b Over 2 a EndFraction from both sides of the equation squaring both sides of the equation taking the square root of both sides of the equation taking the square root of the discriminant

Step-by-step explanation:

From the step given step 6

(b² — 4ac) / 4a² = (x + b/2a)²

Mistake in question it is (x + b/2a)²

Step 7 given

±√(b² —4ac) /2a = x + b/2a

Mistake in question it is over 2a and not 1a.

1. The step taken from step 6 to step 7 is taking square roots of both sides

(b² — 4ac) / 4a² = (x + b/2a)²

Taking square of both sodes

√(b²—4ac) / √4a² = √(x + b/2a)²

√(b²—4ac) / 2a = x + b/2a

This is the required step 7.

Now, subtracting b/2a from both sides

√(b²—4ac) / 2a - b/2a= x + b/2a -b/2a

√(b²—4ac) / 2a — b/2a = x

(√(b²—4ac)  — b)/2a = x

x = [—b ± √(b²—4ac)] / 2a

So, this is the required formula method

The discriminant is D = b²—4ac.

17 1/3

step-by-step explanation:

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.