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nae
07-05-2006, 10:51 AM
This equation can be solved by using either the factoring method or the quadratic formula p^2+6p+4=0 ? true or false i think it's true you can use the factoring method but not the quadratic is this correct?

royhaas
07-05-2006, 11:04 AM

galactus
07-05-2006, 11:16 AM
One little trick you might find helpful.

Check the discriminant. b^{2}-4ac

If it's not a perfect square, it's not factorable.

a=1, b=6, c=4

jonboy
07-06-2006, 11:25 AM
This equation can be solved by using either the factoring method or the quadratic formula p^2+6p+4=0 ? true or false i think it's true you can use the factoring method but not the quadratic is this correct?

If you think that \bold p^2+6p+4=0 can be solved by just using the factoring method, then the answer would be false because both factoring and quadratic methods have to be solvable w/p^2+6p+4=0 since or means both are efficient.

Just work it out and see:

Is there any number that multiply to give you 4 and add up to six?

2*2=4, 2+2=4 which is unequivalent to 6
4*1=4 4+1=5 which is unequivalent to 6

So NO you cannot solve this equation by using the factoring method.

x=\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}

x=\frac{{ - 6 \pm \sqrt {6^2 - 4(1)(4)} }}{{2(1)}}

x=\frac{{ - 6 \pm \sqrt {20} }}{2}

x\approx.8 &\approx - 8.2

Since you can put \bold p^2+6p+4=0 in the form ax^2 + bx + c = 0 in which a is not=0, yes it can be factored by using the Quadratic Formula.

So would the answer be False or True?