# a levels maths-quadratic

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I can't figure out how to do these 3 questions

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

3)An object is thrown vertically upwards so that its height h metres above the ground at time t seconds is given by h=20t−5t2+1.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

3)An object is thrown vertically upwards so that its height h metres above the ground at time t seconds is given by h=20t−5t2+1.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

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#2

1) If the equation has real roots, then the discriminant (b

a = 1

b = k

c= 2k-3 (as 2k is equal to an integer)

Therefore, the discriminant is the root of k

= k

Factorising this equation gives k=2 or k=6

2)x

The equation would look like (x

which simplifies to y

Solving this equation gives y as -3 + √5 and -3 -√5

y = x

^{2}-4ac) should equal to 0.a = 1

b = k

c= 2k-3 (as 2k is equal to an integer)

Therefore, the discriminant is the root of k

^{2 }- (4*1*2k-3)= k

^{2}- 8k + 12Factorising this equation gives k=2 or k=6

2)x

^{4 }+ 6x^{2 }+ 4 uses a hidden quadratic. Imagine y was equal to x^{2}:The equation would look like (x

^{2})^{2}+ 6(x^{2}) + 4which simplifies to y

^{2}+ 6y + 4.Solving this equation gives y as -3 + √5 and -3 -√5

y = x

^{2}so rooting both would give x. You cannot root them both as -3 +/- √5 are both negative, so therefore the equation has no real roots.
Last edited by JAMEH; 9 months ago

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#3

(Original post by

I can't figure out how to do these 3 questions

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

**student144**)I can't figure out how to do these 3 questions

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

For the first one, for equal roots, it means we require this to have a single root. What's the condition on the discriminant? Apply it here and solve the new equation you derive for k.

3)An object is thrown vertically upwards so that its height h metres above the ground at time t seconds is given by h=20t−5t2+1.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

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#4

Do you have a textbook? I'm on EDEXCEL and our textbook has notes on the discriminant and examples which should show you how.

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#5

1) equal roots means the discriminant {b^2-4ac is equal to zero}

therefore substitute the values in the equation

to give k^2-4(1)(2k-3)=0

this gives you a quadratic equation then find the diff. values of k

3) equate the quadratic equation to 0 then solve.

therefore substitute the values in the equation

to give k^2-4(1)(2k-3)=0

this gives you a quadratic equation then find the diff. values of k

3) equate the quadratic equation to 0 then solve.

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(Original post by

1) If the equation has real roots, then the discriminant (b

a = 1

b = k

c= 2k-3 (as 2k is equal to an integer)

Therefore, the discriminant is the root of k

= k

Factorising this equation gives k=2 or k=6

2)x

The equation would look like (x

which simplifies to y

Solving this equation gives y as -3 + √5 and -3 -√5

y = x

**JAMEH**)1) If the equation has real roots, then the discriminant (b

^{2}-4ac) should equal to 0.a = 1

b = k

c= 2k-3 (as 2k is equal to an integer)

Therefore, the discriminant is the root of k

^{2 }- (4*1*2k-3)= k

^{2}- 8k + 12Factorising this equation gives k=2 or k=6

2)x

^{4 }+ 6x^{2 }+ 4 uses a hidden quadratic. Imagine y was equal to x^{2}:The equation would look like (x

^{2})^{2}+ 6(x^{2}) + 4which simplifies to y

^{2}+ 6y + 4.Solving this equation gives y as -3 + √5 and -3 -√5

y = x

^{2}so rooting both would give x. You cannot root them both as -3 +/- √5 are both negative, so therefore the equation has no real roots.
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#9

(Original post by

I can't figure out how to do these 3 questions

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

3)An object is thrown vertically upwards so that its height h metres above the ground at time t seconds is given by h=20t−5t2+1.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

**student144**)I can't figure out how to do these 3 questions

1)The quadratic equation x2+kx+2k−3=0 has equal roots.

The possible value(s) of k are?

2)ow many real roots does the equation x4+6x2+4=0 have?

3)An object is thrown vertically upwards so that its height h metres above the ground at time t seconds is given by h=20t−5t2+1.

After how many seconds does it hit the ground? Give your answer correct to 2 d.p.

You should get around 4.05 seconds

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#10

The first line is a quadratic, which is equal to zero for the question.

What methods can we use to find roots of such quadratic equations?

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