How do I solve this problem?

Question

sq rt (t+3) + sq rt (2t+7) = 1

Answer 1

√(t + 3) + √(2t + 7) = 1
So √(2t + 7) = 1 - √(t + 3)
Square both sides
2t + 7 = 1 - 2√(t + 3) + t + 3
So t + 3 = -2√(t + 3)

Square both sides again
t² + 6t + 9 = 4(t + 3)
t² + 2t - 3 = 0
(t +3)(t - 1) = 0
t = -3 or 1

Test

When t = -3
√(t + 3) + √(2t + 7)
= √(-3 + 3) + √(-6 + 7)
= 0 + 1
= 1 YESSSSSSSSSSSSSSSS!!

When t = 1
√(t + 3) + √(2t + 7)
= √(1 + 3) + √(2 + 7)
= 2 + 3
= 5
≠1 NOOOOOOOOOOOOOO!!!!!

So the only solution is t = -3

(The extra solutions come from the fact that other numbers can come into the solution through squaring and so the solutions must be throroughly tested)

joruaishi... was lucky because
(√(t + 3) + √(2t + 7))² = t + 3 + 2√[(t + 3)(2t + 7)] + 2t + 7
≠ (t + 3) + (2t + 7)

That is like saying (√4 + √9)² = 4 + 9 = 13
when in fact (√4 + √9)² = (2 + 3)² = 25

The differewnce is 12 = 2 √4 * √9
His worked because t = -3 rendered one of the surds zero

Answer 2

t= -3

Answer 3

You square both sides of the equation to get (t+3) + 2t+7=1..so that's 3t+10..subtract 10 from both side..3t=-9..so the answer is t=-3.
if you plug it back in.. -3+3 + -6+7..square root of zero is zero..and the square root of 1 is 1..so it's correct.

Answer 4

sq rt (t+3) + sq rt (2t+7) = 1 square each side
t+3+2t+7+sqrt(t+3)(2t+7)=1
3t+10+sqrt(t+3)(2t+7)=1
sqrt(2t^2+13t+21)=-3t-9 square both sides
2t^2+13t+21=9t^2+54t+81 subtract LHS
0=7t^2+41t+60
t=(-41+/-sqrt(41^2+4*7*60))/14
t=(-41+/-sqrt(3361))/14
t=-41/14+sqrt(3361)/14
t=-41-sqrt(3361)/14

Answer 5

square both sides of the equation, you'll have only one root left, of (t+3)*(2t+7) )

Move all non-root terms to one side and square it again - you will get quadratic equation with 2 roots.

make sure both roots work in the initial equation, i.e. you do not negative numbers under root.

Answer 6

I got t = 1, -3 using intel's method

Answer 7

Do ur own homework