Maths Question Help Please

Original post by Blackrose06

I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

what does the discriminant tell you about a quadratic equation (specifically about its roots)?

Reply 2

TypicalNerd

18

Original post by Blackrose06

I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

Start by calculating the discriminant using the given formula. b = 4, a = 1 and c = 9.

You should know of the following rules:

If b^2 - 4ac > 0, then the quadratic has 2 real roots

If b^2 - 4ac = 0, then the quadratic has 1 real root

If b^2 - 4ac < 0, then the quadratic has 0 real roots

Reply 3

Ignorabimus

9

If Δ is negative, what is happening?

Spoiler

Reply 4

OP

So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis

Reply 5

TypicalNerd

18

Original post by Blackrose06

So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis

Correct. Though to actually get the mark for writing that statement, I’m fairly certain you would have to show you have actually calculated the discriminant first.

Reply 6

OP

(4)^2 - 4(1)(9) = -20

(edited 8 months ago)

Reply 7

Original post by Blackrose06

So because the Discriminant is negative, the graph would have no roots, therefore it doesn't touch the X-Axis

To be on the safe side, I would say "because the discriminant is negative, the quadratic has no real roots, therefore..."

Reply 8

OP

Ok, thanks guys this has really helped me out (GCSE-> A-level transition work is confusing)

Reply 9

Original post by Blackrose06

Ok, thanks guys this has really helped me out (GCSE-> A-level transition work is confusing)

Is this holiday work before you return for A levels? This sort of stuff is good to reinforce your GCSE algebra and check that you have a good grasp on the concepts before things get more difficult.

Reply 10

OP

The discriminant - Using the discriminant to determine the number of roots - National 5 Maths Revision - BBC Bitesize

Original post by davros

Is this holiday work before you return for A levels? This sort of stuff is good to reinforce your GCSE algebra and check that you have a good grasp on the concepts before things get more difficult.

Yeah. This didn't come up in my GCSE lessons though.

Reply 11

Ignorabimus

9

(edited 8 months ago)

Web capture_3-8-2023_205918_www.bbc.co.uk.jpeg53.7KB

Reply 12

otinme

Original post by Blackrose06

I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

To determine whether the graph of the equation y = x² + 4x + 9 intersects the x-axis, we need to find the x-values (roots) at which the function y equals zero. In other words, we need to solve for x when y = 0. The graph intersects the x-axis at the points where y is equal to zero.

To find the x-values when y = 0, we set the equation equal to zero:

0 = x² + 4x + 9

Now, let's try to find the roots of this quadratic equation. We can use the quadratic formula, which is given by:

x = (-b ± √(b² - 4ac)) / 2a

Where the quadratic equation is in the form of ax² + bx + c = 0. In this case, a = 1, b = 4, and c = 9:

x = (-(4) ± √((4)² - 4(1)(9))) / 2(1)
x = (-4 ± √(16 - 36)) / 2
x = (-4 ± √(-20)) / 2

The expression inside the square root, √(-20), is negative. This means the quadratic equation does not have any real roots, and therefore, it does not intersect the x-axis.

The graph of y = x² + 4x + 9 is a parabola that opens upwards because the coefficient of x² (1) is positive. As a result, it sits entirely above the x-axis and never crosses it, confirming that it does not have any real x-intercepts or roots. (If you need more guidance or have similar problems you can pm me).

Reply 13

π/2=Σk!/(2k+1)!!

5

Original post by Blackrose06

I have never come across a question like this before so I have no idea where to start.
Explain why the graph of the equation y = x²+ 4x + 9 does not intersect the x-axis.

Use the discriminant Δ = b2– 4ac

The discriminant is less than 0, b=4 ,a =1 and c =9. The discriminant is (4^2) - 4(1)(9) = 16-36 = -20 since the square root of -20 is 2isqrt5, there are no real solutions and thus, the function doesn't intersect the x-axis.

Reply 14

Ignorabimus

9

Original post by π/2=Σk!/(2k+1)!!

The discriminant is less than 0, b=4 ,a =1 and c =9. The discriminant is (4^2) - 4(1)(9) = 16-36 = -20 since the square root of -20 is 2isqrt5, there are no real solutions and thus, the function doesn't intersect the x-axis.

This was answered a week ago, no need to bump it😐.

Reply 15