Exercise 43 Page 196

As we solve the equation, let's check the steps Tama and Jonathan took and compare our results with theirs. First, both Tama and Jonathan are correct to use the discriminant to help determine the number of solutions of the given quadratic equation. 2 &Quadratic equation: && ax^2+ bx+ c=0 &Discriminant:&&Δ= b^2-4 a c However, Tama made a mistake in identifying the coefficients. The equation needs to be in standard form.

3x^2-5x=7

SubEqn

LHS-7=RHS-7

3x^2-5x-7=0

Jonathan uses this form to correctly identify the coefficients. 3x^2-5x-7= 3x^2+( - 5)x+( -7) We see that a= 3, b= - 5, and c= - 7. Jonathan uses these coefficients to correctly find the discriminant.

Δ=b^2-4ac

SubstituteValues

Substitute values

Δ=( - 5)^2-4( 3)( -7)

▼

Simplify right-hand side

CalcPow

Calculate power

Δ=25-4( 3)( -7)

Multiply

Multiply

Δ=25-12( -7)

MultPosNeg

a(- b)=- a * b

Δ=25-(-84)

SubNeg

a-(- b)=a+b

Δ=25+84

AddTerms

Add terms

Δ=109

In the final step, Jonathan uses the correct discriminant to correctly draw a conclusion about the number of solutions. Therefore Jonathan is correct, but Tama is not.