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#1
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.
0
9 months ago
#2
1) If the equation has real roots, then the discriminant (b2 -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 k2 - (4*1*2k-3)
= k2- 8k + 12
Factorising this equation gives k=2 or k=6

2)x4 + 6x2 + 4 uses a hidden quadratic. Imagine y was equal to x2:

The equation would look like (x2)2 + 6(x2) + 4
which simplifies to y2 + 6y + 4.
Solving this equation gives y as -3 + √5 and -3 -√5
y = x2 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
0
9 months ago
#3
(Original post by 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?
These are to do with the discriminant. Look at your notes on how the discriminant dictates the number of real roots, and use those conditions here.

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.
On the ground level, what is the value of h?
0
9 months ago
#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.
0
9 months ago
#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.
0
#6
(Original post by JAMEH)
1) If the equation has real roots, then the discriminant (b2 -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 k2 - (4*1*2k-3)
= k2- 8k + 12
Factorising this equation gives k=2 or k=6

2)x4 + 6x2 + 4 uses a hidden quadratic. Imagine y was equal to x2:

The equation would look like (x2)2 + 6(x2) + 4
which simplifies to y2 + 6y + 4.
Solving this equation gives y as -3 + √5 and -3 -√5
y = x2 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.
thank you very much!!!
0
#7
got 81 seconds
third line is 5(t+4)^2 -16 +1
0
9 months ago
#8
(Original post by student144)
thank you very much!!!
no problem!
0
9 months ago
#9
(Original post by 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.
3) Just solve it to equal 0 using qaudratic formula.

You should get around 4.05 seconds
0
9 months ago
#10
(Original post by student144)
got 81 seconds
third line is 5(t+4)^2 -16 +1 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?
0
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