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Can you explain the formula method for solving quadratic equations?

We can use the formula method (for solving quadratic equations) to find 'roots' or values of x that satisfy or 'work out' for a given quadratic equation of an unknown variable (say x.) The formula is:

x=-b(+or- sqrt[b2-4ac])/2a

Note that the 'plus or minus' can give us 2 possible values or 'roots' for the unknown 'x'. These may be 2 positive roots, 2 negative roots, or a negative and a positive root. These roots are the coordinates where a curve/line intersects with the x axis (we know that y=0 on the x-axis already.)

We may compare our quadratic equation to the general format (ax2+bx+c) to obtain the values for a, b, and c, which are coefficients of x (c is the coefficient of x0 which equals 1.)

Our 2 values may then be substituted back into our original equation to show that the 2 sides 'match' and thus the equation is valid. We let the quadratic equation equal zero to display that the 2 sides are balanced or 'homogeneous'.

Example:

Solve the quadratic equation 3x2+9x+3 via the formula method.

Firstly, we must compare the above quadratic equation with the general format (ax2+bx+c) to obtain values for the coefficients of x. We can see that a=3, b=9, and c=3. Our general formula:

x=-b(+or- sqrt[b2-4ac])/2a

is thus

x=-9(+or- sqrt[(9)2-4(3)(3)])/2(3)

So that by solving for x, x=-0.381 (3 d.p.) and x=-2.618 (3 d.p.). We obtained these answers by adding and subtracting the square root terms (respectively) and performing the arithmetic.

We can check that these are correct by equating the quadratic to zero and substituting in our x values:

3(-0.381)2+9(-0.381)+3=0.0064833

3(-2.618)2+9(-2.618)+3=-0.000228

Thus our roots are correct! The equations do not equal zero exactly as we have rounded our roots to 3 decimal places.

Daniel  M. GCSE Maths tutor, GCSE Physics tutor, GCSE Biology tutor, ...

6 months ago

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