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?

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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

You are contradicting yourself.

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

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.

But the Quadratic 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?

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.

But the Quadratic 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?

mad_mathematician

07-06-2006, 02:28 PM

you can use the formula but you cant factorise it. use the discriminant (b^2 - 4ac) to ind out whether it has solutions (if the discriminant is negative then it doesnt, otherwise it does). Provided that a quadratic equation has real solutions, you will ALWAYS be able to solve it using the formula, and sometimes you'll be able to factorise if you prefer that method (if the discriminant is a square number, then it factorises)

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