Question

1.) Find the value of the discriminant and the the number of real solutions of

Answer 1

Hola!

Standard form of an Eqn. is
ax² + bx + c = 0

Discriminant = b² - 4ac

When,

→ b² - 4ac > 0 -- [ real and distinct roots. ]
→ b² - 4ac < 0 -- [ no real roots, Imaginary numbers. ]
→ b² - 4ac = 0 -- [ real and equal roots. ]

According to Question,

1.] x² - 8x + 7 = 0

here,
→ a = 1
→ b = -8
→ c = 7

Discriminant = b² - 4ac

= (-8)² - 4 × (1) × (7)

= 64 - 28

= 36
[ Since, D > 0 -- Eqn has two real roots ]

Thus,
We have 2 real solutions.

2.] 2x² + 4x + 2 = 0

here,
→ a = 2
→ b = 4
→ c = 2

Discriminant = b² - 4ac

= (4)² - 4 × (2) × (2)

= 16 - 16

= 0
[ Since, D = 0 -- Eqn has equal roots ]

Thus,
We have 1 real solution.

hope it helps!

Answer 2

Hi there!

When we have an equation standard form...
$a {x}^{2} + bx + c = 0$
...the formula of the discriminant is
D = b^2 - 4ac

When
D > 0 we have two real solutions
D = 0 we have one real solutions
D < 0 we don't have real solutions

1.) Find the value of the discriminant and the the number of real solutions of
x^2-8x+7=0

Plug in the values from the equation into the formula of the discriminant

$( - 8) {}^{2} - 4 \times 1 \times 7 = 64 - 28 = 36$

D > 0 and therefore we have two real solutions.

2.) Find the value of the discriminant and the number of real solutions of
2x^2+4x+2=0

Again, plug in the values from the equation into the formula of the discriminant.

${4}^{2} - 4 \times 2 \times 2 = 16 - 16 = 0$

D = 0 and therefore we have one real solution.

~ Hope this helps you.